Answer:
a) Let X the random variable that represent the blood pressure for people of a population, and for this case we know the distribution for X is given by:
[tex]X \sim N(128,23)[/tex]
Where [tex]\mu=128[/tex] and [tex]\sigma=23[/tex]
b) [tex]P(X\geq 135)=P(\frac{X-\mu}{\sigma}\geq \frac{135-\mu}{\sigma})=P(Z\geq \frac{135-128}{23})=P(Z\geq 0.304)[/tex]
And we can find this probability using the complement rule:
[tex]P(Z\geq 0.304)=1-P(Z<0.304)[/tex]
And in order to find this probabilities we can use tables for the normal standard distribution, excel or a calculator.
[tex]P(Z\geq 0.304)=1-P(Z<0.304)= 1-0.619=0.381 [/tex]
Step-by-step explanation:
Previous concepts
Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".
The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".
Part a
Let X the random variable that represent the blood pressure for people of a population, and for this case we know the distribution for X is given by:
[tex]X \sim N(128,23)[/tex]
Where [tex]\mu=128[/tex] and [tex]\sigma=23[/tex]
Part b
We are interested on this probability
[tex]P(X\geq 135)[/tex]
And the best way to solve this problem is using the normal standard distribution and the z score given by:
[tex]z=\frac{x-\mu}{\sigma}[/tex]
If we apply this formula to our probability we got this:
[tex]P(X\geq 135)=P(\frac{X-\mu}{\sigma}\geq \frac{135-\mu}{\sigma})=P(Z\geq \frac{135-128}{23})=P(Z\geq 0.304)[/tex]
And we can find this probability using the complement rule:
[tex]P(Z\geq 0.304)=1-P(Z<0.304)[/tex]
And in order to find this probabilities we can use tables for the normal standard distribution, excel or a calculator.
[tex]P(Z\geq 0.304)=1-P(Z<0.304)= 1-0.619=0.381 [/tex]
In this context, the random variable is the blood pressure of people in China. The probability that a person in China has a blood pressure of 135 mmHg or more is about 38.21%, calculated using the Z-score and standard Z-table.
Explanation:The subject of this question is the study of normal distribution and probability in statistics, a branch of mathematics.
a.) The random variable in this context is the blood pressure of people in China.
b.) To find the probability that a person in China has a blood pressure of 135 mmHg or more, we need to convert this to a Z-score. The Z-score is calculated by subtracting the mean from the individual score and then dividing by the standard deviation. Therefore, Z = (135 - 128) / 23 = 0.30.
Using a standard Z-table, the probability corresponding to Z=0.30 is about 0.6179. However, because we want the probability of a person having a blood pressure that's 135 mmHg or more (greater than the mean), we must subtract this value from 1. Thus, the probability is about 1 - 0.6179 = 0.3821, or 38.21%.
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Suppose the standard deviation of a normal population is known to be 3, and H0 asserts that the mean is equal to 12. A random sample of size 36 drawn from the population yields a sample mean 12.95. For H1: mean > 12 and alpha 0.05, will you reject the claim?
Answer:
[tex]z=\frac{12.95-12}{\frac{3}{\sqrt{36}}}=1.9[/tex]
[tex]p_v =P(z>1.9)=0.0287[/tex]
If we compare the p value and the significance level given [tex]\alpha=0.05[/tex] we see that [tex]p_v<\alpha[/tex] so we can conclude that we have enough evidence to reject the null hypothesis, so we can conclude that the true mean is higher than 12 at 5% of signficance.
Step-by-step explanation:
Data given and notation
[tex]\bar X=12.95[/tex] represent the sample mean
[tex]\sigma=3[/tex] represent the population standard deviation for the sample
[tex]n=36[/tex] sample size
[tex]\mu_o =12[/tex] represent the value that we want to test
[tex]\alpha=0.05[/tex] represent the significance level for the hypothesis test.
z would represent the statistic (variable of interest)
[tex]p_v[/tex] represent the p value for the test (variable of interest)
State the null and alternative hypotheses.
We need to conduct a hypothesis in order to check if the mean is equal to 12, the system of hypothesis would be:
Null hypothesis:[tex]\mu \leq 12[/tex]
Alternative hypothesis:[tex]\mu > 12[/tex]
Since we know the population deviation, is better apply a z test to compare the actual mean to the reference value, and the statistic is given by:
[tex]z=\frac{\bar X-\mu_o}{\frac{\sigma}{\sqrt{n}}}[/tex] (1)
z-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".
Calculate the statistic
We can replace in formula (1) the info given like this:
[tex]z=\frac{12.95-12}{\frac{3}{\sqrt{36}}}=1.9[/tex]
P-value
Since is a right tailed test the p value would be:
[tex]p_v =P(z>1.9)=0.0287[/tex]
Conclusion
If we compare the p value and the significance level given [tex]\alpha=0.05[/tex] we see that [tex]p_v<\alpha[/tex] so we can conclude that we have enough evidence to reject the null hypothesis, so we can conclude that the true mean is higher than 12 at 5% of signficance.
Answer:
Yes, we will reject the claim population mean = 12.
Step-by-step explanation:
It is provided that Standard deviation, [tex]\sigma[/tex] = 3 and sample mean,Xbar = 12.95 .
Let, Null Hypothesis,[tex]H_0[/tex] : mean, [tex]\mu[/tex] = 12
Alternate Hypothesis,[tex]H_1[/tex] : mean, [tex]\mu[/tex] > 12
Since Population is Normal so our Test Statistics will be:
[tex]\frac{Xbar-\mu}{\frac{\sigma}{\sqrt{n} } }[/tex] follows standard normal,N(0,1)
Here, sample size, n = 36.
Test Statistics = [tex]\frac{12.95-12}{\frac{3}{\sqrt{36} } }[/tex] = 1.9
So, at 5% level of significance z % table gives the critical value of 1.6449 and our test statistics is higher than this as 1.6449 < 1.9. So,we have sufficient evidence to reject null hypothesis or accept [tex]H_1[/tex] as our test statistics falls in the rejection region because it is more than 1.6449.
Hence we conclude after testing that we will reject claim of Population mean, μ = 12.
Abby is buying a widescreen TV that she will hang on the wall between two windows. The windows are 36 inches apart, and wide screen TVs are approximately twice as wide as they are tall. Of the following, which is the longest that the diagonal of a widescreen TV can measure and still fit between the windows
Answer:
D < 40.2 inches
Step-by-step explanation:
The maximum width of the TV must be 36 inches. Since TVs are approximately twice as wide as they are tall, the maximum height is 18 inches.
The diagonal of a TV can be determined as a function of its width (w) and height (h) as follows:
[tex]d^2=h^2+w^2\\d=\sqrt{18^2+36^2}\\d= 40.2\ in[/tex]
Therefore, the diagonal must be at most 40.2 inches.
Since the answer choices were not provided with the question, you should choose the biggest value that is under 40.2 inches.
The maximum diagonal size of the widescreen TV that can fit between two windows 36 inches apart is slightly more than 40 inches, given that the TV has an aspect ratio where the width is about twice the height.
Let's denote the TV's width as w and the height as h. Given that widescreen TVs are about twice as wide as they are tall, we can express the width as w = 2h.
The diagonal d of the TV can be found using Pythagoras' theorem where d² = w² + h².
Substituting 2h for w, we get d² = (2h)² + h² which simplifies to d² = 4h² + h² and further to d² = 5h².
Thus, d = h√5.
If the space between windows is 36 inches, this would be the maximum width of the TV. Therefore, 36 = 2h which means that h = 18 inches. Using this height in the diagonal equation, we get d = 18√5 which is approximately 40.2 inches. This means the longest diagonal of the widescreen TV that can fit between the windows is slightly more than 40 inches.
you have 200 grams of a radioactive kind of iodine how much will ne left after 120 days if its half life is 60 days?
By definition, the half life is a quantity is the time it takes to lose half of that quantity.
So, if the half life is 60 days, it means that after 60 days you have lost half the initial amount, i.e. you're left with 100 grams.
After another 60 days, you've lost, again, half of that amount, so you're left with 50 grams.
In other words, every time a half life period passes, you're left with half the quantity you had at the beginning of that period.
So, after two half-life periods (i.e. 120 days), you'll have half of the half of the initial quantity, i.e. one quarter.
It is well-known that lack of sleep impairs concentration and alertness, and this might be due partly to late night food consumption. A 2015 study took 44 people aged 21 to 50 and gave them unlimited access to food and drink during the day, but allowed them only 4 hours of sleep per night for 3 consecutive nights. On the fourth night, all participants again had to stay up until 4 am, but this time participants were randomized into two groups; one group (A) was only given access to water from 10 pm until their bedtime at 4 am while the other group (B) still had unlimited access to food and drink for all hours. The group (A) performed significantly better on tests of reaction time and had fewer attention lapses than group (B).a. What are the explanatory and response variables? b. Is this nu observational study or a randomized experiment? c. Can we conclude that eating late at night worsens some of the typical effeets of alcep deprivation (reaction time and attention lapses)? d. Are there likely to be confounding variables? Why or why not?
Answer: a) Food and performance b) Randomized Experiment c) Yes d) Age, type of food
Step-by-step explanation:
a) The explanatory variable in this case is the food and sleep hours given to the groups and the response variable is the performance or the reaction time.
b) It is a randomized experiment as the people are given food and water and is under the influence of the study head.
c) From the detailed summary given, we can conclude that eating late worsens the effects of alcep depreviation as the group A was not given food and performed better than the group B who which had access to food and water.
d.Confounding variables are the age 21-50, the gender of people who took part in study and the type of food. Because changing these will result in different results.
The study examines the effects of eating late at night on sleep-deprived individuals' reaction time and attention lapses. It was a randomized experiment that concluded that eating late at night worsens these effects. Confounding variables may exist.
Explanation:a. The explanatory variable in this study is the access to food and drink at night, and the response variables are the participants' reaction time and attention lapses.
b. This is a randomized experiment because the participants were randomly assigned to either group A (water only) or group B (unlimited access to food and drink).
c. Based on the results of the study, we can conclude that eating late at night worsens some of the typical effects of sleep deprivation, specifically reaction time and attention lapses.
d. There are likely to be confounding variables in this study, such as individual differences in metabolism or other lifestyle factors that could impact the participants' performance.
1) Let f(x)=ax2+bx+c for some value of a, b and c. f intersects the x-axis when x=−2 or x=3, and f( 1 3 )=−25. Find the values of a, b and c and sketch the graph of f(x).
2) A right prism has a base that is an equilateral triangle. The height of the prism is equal to the height of the base. If the volume of the prism is 81, what are the lengths of the sides of the base?
thank u sm
Answer:
1) a = -⅙, b = ⅙, c = 1
2) 6 units
Step-by-step explanation:
1) f(x) = ax² + bx + c
Given the roots, we can write this as:
f(x) = a (x + 2) (x − 3)
We know that f(13) = -25, so we can plug this in to find a:
-25 = a (13 + 2) (13 − 3)
-25 = 150a
a = -⅙
Therefore, the factored form is:
f(x) = -⅙ (x + 2) (x − 3)
Distributing:
f(x) = -⅙ (x² − x − 6)
f(x) = -⅙ x² + ⅙ x + 1
Graph: desmos.com/calculator/6m6tjoodvb
2) Volume of a right prism is area of the base times the height.
V = Ah
The base is an equilateral triangle. Area of a triangle is one half the base times height.
V = ½ ab h
The height of the triangle is the same as the height of the prism.
V = ½ bh²
In an equilateral triangle, the height is equal to half the base times the square root of 3.
V = ½ b (½√3 b)²
V = ⅜ b³
Given that V = 81, solve for b.
81 = ⅜ b³
216 = b³
b = 6
In the United States in 1986, 48.7% of persons age 25-pluswere males. Of these males 23.8% were college graduates. In addition, 20.5% of all persons (malesand females) were college graduates.A) What proportion of persons 25-plus were female college graduates?B) What proportion of females 25-plus were college graduates?
Answer:
A) 8.9%
B.) 17.35%
Step-by-step explanation:
Let X be the total number of persons age 25-plus
This means 48.7% of X are males.
Number of females become:
(100 - 48.7)x = 51.3%X
IF 23.8% of these males were graduates, then it means
23.8% * 48.7%X of these males are graduates.
0.238 * 0.487 X = 0.116X males are graduates.
That is 11.6% of persons are male College graduates.
If the question states that 20.5% of male and female were college graduates, then to answer question A, to get the number of female college graduates, it becomes:
20.5% - 11.6% = 8.9%
That means 8.9% of persons were female college graduates. .
If 53.1% of persons are females, let y be the percentage of these females that are College graduates.
Then it implies
Y * 51.3 = 8.9
Y = 8.9/51.3
Y = 0.1735 = 17.35%
This means the proportion of females that are College graduates = 17.35%
In this exercise we have to use the knowledge of statistics to describe the results in percentage, in this way we can describe as:
A) [tex]8.9\%[/tex]
B) [tex]17.35\%[/tex]
Knowing that the information given at the beginning of the utterance is:
Let X be the total number of persons age 25-plusThis means 48.7% of X are males.Calculating the number of women:
[tex](100 - 48.7)X = 51.3\%X[/tex]
[tex]20.5\% - 11.6\% = 8.9\% [/tex]
If 53.1% of human being happen woman, let y exist the allotment of these woman that exist College graduates. Then it indicate that:
[tex]Y * 51.3 = 8.9\\ Y = 8.9/51.3\\ Y = 0.1735 = 17.35\%[/tex]
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An urn contains 5 red, 6 blue, and 8 green balls. If a set of 3 balls is randomly selected, what is the probability that each of the balls will be (a) of the same color(b) of all different colors
Repeat the experiment under the assumption that whenever a ball is selected, its color is noted and it is then replaced in the urn before the next selection. This is known as sampling with replacement. What is the probability that each of the balls will be:
(c) of the same color
(d) of all different colors
The probability that each of the balls will be of the same color can be calculated using combinations, while the probability that each of the balls will be of all different colors can be calculated using permutations. When sampling with replacement, the probabilities of selecting each color remain the same in each selection.
Explanation:(a) Probability that each of the balls will be of the same color:
This can be calculated using the concept of combinations. There are a total of 19 balls in the urn. Let's calculate the number of ways to choose all 3 balls of the same color:
Picking all 3 red balls: There are 5 red balls, so the number of ways is C(5, 3).Picking all 3 blue balls: There are 6 blue balls, so the number of ways is C(6, 3).Picking all 3 green balls: There are 8 green balls, so the number of ways is C(8, 3).Summing up the number of ways for each color, we get the total number of ways to pick 3 balls of the same color. The probability will be this number divided by the total number of ways to pick any 3 balls from the urn, which is C(19, 3).
(b) Probability that each of the balls will be of all different colors:
This can be calculated using the concept of permutations. There are a total of 19 balls in the urn. Let's calculate the number of ways to choose 1 ball of each color:
Picking 1 red ball: There are 5 red balls, so the number of ways is 5.Picking 1 blue ball: There are 6 blue balls, so the number of ways is 6.Picking 1 green ball: There are 8 green balls, so the number of ways is 8.Since these events are independent, we multiply the number of ways for each color. The probability will be this number divided by the total number of ways to pick any 3 balls from the urn, which is C(19, 3).
(c) Probability that each of the balls will be of the same color (sampling with replacement):
When sampling with replacement, each ball has the same probability of being chosen in each selection. So the probability of selecting 1 red ball, 1 blue ball, and 1 green ball is the product of the probabilities of selecting each color. The probability for each color is the number of balls of that color divided by the total number of balls in the urn. Therefore, the probability will be (5/19) * (6/19) * (8/19).
(d) Probability that each of the balls will be of all different colors (sampling with replacement):
When sampling with replacement, the probability of each color being chosen in each selection remains the same. Therefore, the probability of selecting 1 red ball, 1 blue ball, and 1 green ball is the product of the probabilities of selecting each color. The probability for each color is the number of balls of that color divided by the total number of balls in the urn. Therefore, the probability will be (5/19) * (6/19) * (8/19).
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The intelligence quotients (IQs) of 16 students from one area of a city showed a mean of 107 and a standard deviation of 10, while the IQs of 14 students from another area of the city showed a mean of 112 and a standard deviation of 8. Is there a significant difference between the IQs of the two groups at significance level of 0.01. What is the alternative hypothesis?
Answer:
No, there is no significant difference between the IQs of the two groups.Alternative Hypothesis is that the two groups have different IQs os students.Step-by-step explanation:
We are provided that IQs of 16 students from one area of a city had a mean of 107 and a standard deviation of 10 while the IQs of 14 students from another area of the city had a mean of 112 and a standard deviation of 8.
And we have to check that is there a significant difference between the IQs of the two groups.
Firstly let, Null Hypothesis, [tex]H_0[/tex] : The two groups have same IQs { [tex]\mu_1 = \mu_2[/tex] }
Alternate Hypothesis, [tex]H_1[/tex] : The two groups have different IQs{ [tex]\mu_1 \neq \mu_2[/tex]}
Since we don't know about population standard deviations;
The test statistics we will use here will be ;
[tex]\frac{(X_1bar - X_2bar)- (\mu_1 - \mu_2) }{s_p\sqrt{\frac{1}{n_1}+\frac{1}{n_2} } }[/tex] follows t distribution with [tex](n_1 + n_2 -2)[/tex] degree
of freedom { [tex]t_n__1 +n_2 - 2[/tex] }
Here, [tex]X_1bar[/tex] = 107 [tex]X_2bar[/tex] = 112 [tex]s_1[/tex] = 10 [tex]s_2[/tex] = 8
[tex]n_1[/tex] = 16 [tex]n_2[/tex] = 14
[tex]s_p[/tex] = [tex]\sqrt{\frac{(n_1 - 1)*s_1^{2} + (n_2 -1)*s_2^{2} }{(n_1 + n_2 -2)} }[/tex] = 9.1261
Test statistics = [tex]\frac{(107-112) - 0}{9.1261*\sqrt{\frac{1}{16} +\frac{1}{14} } }[/tex] follows [tex]t_2_8[/tex]
= -1.50
Now at 1% level of significance t table is giving the critical value of -2.467 and our test statistics is higher than this means it does not fall in the rejection region so we will accept our null hypothesis and conclude that there is no significant difference between the IQs of the two groups.
This question can be addressed by conducting a two-sample t-test to determine if there is a significant difference between the mean intelligence quotients of students from two areas of a city. The steps include calculating pooled standard deviation, followed by standard error, calculating the t-score, and comparing it to a critical value. You can either reject or fail to reject the null hypothesis based on these results.
Explanation:The question is asking if there is a significant difference in the mean intelligence quotients (IQs) of students between two areas of a city. Conducting a two-sample t-test can address this. First, let's define the null and alternative hypotheses.
Null Hypothesis (H₀): There is no significant difference between the two sets of IQ scores (mean1 = mean2).Alternative Hypothesis (Hᵃ): There is a significant difference between the two sets of IQ scores (mean1 ≠ mean2).To perform this test, follow these steps:
Calculate the pooled standard deviation for the two samples.Use it to determine the standard error of the difference between the two means.Use these to calculate the t-score.Compare the t-score with the critical value for the t-distribution with a significance level of 0.01. If the t-score exceeds the critical value, you reject the null hypothesis, i.e., there's a significant difference in IQs between the areas of the city. While if it is less, the null hypothesis is not rejected, thus, no significant difference.Remember, depending on the specific results, we may or may not find enough evidence to support the alternative hypothesis.
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Write down the general zeroth order linear ordinary differential equation. Write down the general solution.
The zeroth derivative of a function [tex]y(x)[/tex] is simply the function itself, so the zeroth order linear ODE takes the general form
[tex]y(x)=f(x)[/tex]
whose solution is [tex]f(x)[/tex].
What is the probability that someone who brings a laptop on vacation also uses a cell phone to stay connected?
Answer:
1/2
Step-by-step explanation:
well imagine there are only 2 people who has a laptop or phone. so it would be 1/2 because there are 2 people with a laptop and phone, and they are asking who uses a cell phone to stay connected which is 1.
Use the net as an aid to compute the surface area of the triangular prism. Show your work.
Check the picture below.
as we can see, the triangular prism is really just 3 rectangles and 2 triangles stacked up to each other at the edges, so if we simply get the area of each of those five figures and sum them up, that's the surface area of the triangular prism.
[tex]\bf \stackrel{\textit{three rectangles' area}}{(17\cdot 11)+(16\cdot 11)+(17\cdot 11)}~~+~~\stackrel{\textit{two triangles' area}}{\cfrac{1}{2}(16)(15) + \cfrac{1}{2}(16)(15)} \\\\\\ 187+176+187+120+120\implies 790[/tex]
Calculate the sample standard deviation and sample variance for the following frequency distribution of heart rates for a sample of American adults. If necessary. round to one more decimal place than the largest number of decimal places given in the data. Heart Rates in Beats per Minute Class Frequency 61-6613 67-72 10 73-78 3 79-8411 85-90 3
Answer:
[tex] \bar X = \frac{\sum_{i=1}^5 x_i f_i}{n} = \frac{2906}{40}= 72.65[/tex]
[tex] s^2 = \frac{213856 -\frac{2906}{40}}{40-1}=70.131[/tex]
[tex] s = \sqrt{70.131}= 8.374[/tex]
Step-by-step explanation:
For this case we can calculate the expected value with the following table"
Class Midpoint(xi) Freq. (fi) xi fi xi^2 * fi
61-66 63.5 13 825.5 52419.5
67-72 69.5 10 695 48302.5
73-78 75.5 3 226.5 17100.75
79-84 81.5 11 896.5 73064.75
85-90 87.5 3 262.5 22968.75
________________________________________________
Total 40 2906 213856
For this case the midpoint is calculated as the average between the minimum and maximum point for each class.
The expected value can be calculated with the following formula:
[tex] \bar X = \frac{\sum_{i=1}^5 x_i f_i}{n} = \frac{2906}{40}= 72.65[/tex]
For this case n =40 represent the total number of obervations given,
And for the sample variance we can use the following formula:
[tex] s^2 = \frac{\sum x^2_i f_i -\frac{\sum x_i f_i}{n}}{n-1}[/tex]
And if we replace we got:
[tex] s^2 = \frac{213856 -\frac{2906}{40}}{40-1}=70.131[/tex]
And for the deviation we take the square root:
[tex] s = \sqrt{70.131}= 8.374[/tex]
To calculate the sample standard deviation and sample variance, first calculate the sample mean, then calculate the sample variance, and finally find the square root of the sample variance to get the sample standard deviation.
Explanation:To calculate the sample standard deviation and sample variance for the given frequency distribution of heart rates, we need to follow these steps:
Create a chart to organize the data, frequencies, relative frequencies, and cumulative relative frequencies. Calculate the sample mean (average) by multiplying each heart rate value by its frequency, summing those products, and dividing by the total number of observations. Calculate the sample variance by finding the squared difference between each heart rate value and the mean, multiplying each squared difference by its frequency, summing those products, and dividing by the total number of observations minus 1. Calculate the sample standard deviation by taking the square root of the sample variance.
Using the provided data, the sample standard deviation and sample variance can be calculated as follows:
Sample mean = (65 * 13 + 69.5 * 10 + 75.5 * 3 + 81.5 * 11 + 87.5 * 3) / (13 + 10 + 3 + 11 + 3) ≈ 72.74
Sample variance = [(65 - 72.74)² * 13 + (69.5 - 72.74)² * 10 + (75.5 - 72.74)² * 3 + (81.5 - 72.74)² * 11 + (87.5 - 72.74)² * 3] / (13 + 10 + 3 + 11 + 3 - 1) ≈ 54.21
Sample standard deviation = √(54.21) ≈ 7.36
Suppose that diameters of a new species of apple have a bell-shaped distribution with a mean of 7.25 cm and a standard deviation of 0.42 cm. Using the empirical rule, what percentage of the apples have diameters that are between 6.41cm and 8.09 cm?
Answer:
95%
Step-by-step explanation:
Upper limit = 8.09 cm
Lower limit = 6.41 cm
Distribution mean = 7.25 cm
Standard deviation = 0.42 cm
The number of standard deviations from the mean of the upper and lower limits are, respectively:
[tex]N_U=\frac{U-M}{SD} =\frac{8.09-7.25}{0.42}=2 \\N_L=\frac{M-L}{SD} =\frac{7.25-6.41}{0.42}=2[/tex]
Both limits are two standard deviations away from the mean.
According to the empirical rule, in normal distributions, 95% of the data falls within two standard deviations of the mean. Therefore, 95% of the apples have diameters that are between 6.41cm and 8.09 cm.
At a Superbowl party, you and your friend are both eying the last slice of pizza. To settle the matter, you agree on the following dice game: each of you is going to roll one die; if the highest number rolled by either one of you is a 1, 2, 3 or 4, then Player 1 wins. If the highest number is a 5 or a 6, then Player 2 wins. Assuming that you really want that last slice of pizza would you rather be Player 1 or Player 2 to maximize your chance of winning? Explain your choice.
Material 1 has Young’s modulus Y1 and density rho1, material 2 has Young’s modulus Y2 and density rho2, and material 3 has Young’s modulus Y3 and density rho3. If Y1 > Y2 > Y3 and if rho1 < rho2 < rho3, which material has the highest speed of sound? Group of answer choices
Answer:
v_s,1 > v_s,2 > v_s,3
Step-by-step explanation:
Given:
- Material 1:
modulus of elasticity = E_1
density of material = p_1
- Material 2:
modulus of elasticity = E_2
density of material = p_2
- Material 3:
modulus of elasticity = E_3
density of material = p_3
- E_1 > E_2 > E_3
- p_1 < p_2 < p_3
Find:
- Which material has highest speed of sound from highest to lowest:
Solution:
- The relationship between velocity of sound in a material with its elastic modulus and density is:
v_s = sqrt ( E / p )
- Since , v_s is proportional to E^0.5 and inversely proportional to p^0.5, then we have:
E_1 > E_2 > E_3
E_1^0.5 > E_2^0.5 > E_3^0.5
and p_1 < p_2 < p_3
p_1^0.5 < p_2^0.5 < p_3^0.5
Divide the two: (E_1 / p_1)^0.5 > (E_1 / p_1)^0.5 > (E_1 / p_1)^0.5
Hence, v_s,1 > v_s,2 > v_s,3
A pharmaceutical company receives large shipments of aspirin tablets. The acceptance sampling plan is to randomly select and test 60 tablets, then accept the whole batch if there is only one or none that doesn't meet the required specifications. If one shipment of 7000 aspirin tablets actually has a 4% rate of defects, what is the probability that this whole shipment will be accepted? Will almost all such shipments be accepted, or will many be rejected?
Answer:
the probability to be accepted is 0.302 (30.2%) (many will be rejected)
Step-by-step explanation:
assuming that the rate of 4% applies to the 60 tablets then since each tablet behaves independently, the random variable X= number of tablets with defects out of 60 tablets has a binomial distribution , where:
p(X)=n!/((n-x)!*x!)*p^x*(1-p)^(n-x)
where
n= total number of tablets tested = 60
x = number of defective tablets
p= probability to be defective = 0.04
then in order to be accepted x≤0 , then the probability that the batch is accepted Pa is
Pa=P(x≤1) = P(0) + P(1) = (1-p)^n + n*p*(1-p)^(n-1)
replacing values
Pa= (1-p)^n + n*p*(1-p)^(n-1) = 0.96^60 + 60*0.04*0.96^59 = 0.302 (30.2%)
then the probability to be accepted is 0.302 (30.2%) (many will be rejected)
Recursive definitions for subsets of binary strings.Give a recursive definition for the specified subset of the binary strings. A string r should be in the recursively defined set if and only if r has the property described. The set S is the set of all binary strings that are palindromes. A string is a palindrome if it is equal to its reverse. For example, 0110 and 11011 are both palindromes.
Answer:
Step-by-step explanation:
A binary string with 2n+1 number of zeros, then you can get a binary string with 2n(+1)+1 = 2n+3 number of zeros either by adding 2 zeros or 2 1's at any of the available 2n+2 positions. Way of making each of these two choices are (2n+2)22. So, basically if b2n+12n+1 is the number of binary string with 2n+1 zeros then your
b2n+32n+3 = 2 (2n+2)22 b2n+12n+1
your second case is basically the fact that if you have string of length n ending with zero than you can the string of length n+1 ending with zero by:
1. Either placing a 1 in available n places (because you can't place it at the end)
2. or by placing a zero in available n+1 places.
0 ϵ P
x ϵ P → 1x ϵ P , x1 ϵ P
x' ϵ P,x'' ϵ P → xx'x''ϵ P
A recursive definition for the set of binary string palindromes starts with the base cases '0' and '1'. Other palindromes can be obtained by nesting a palindrome between '0' and '0' or '1' and '1'.
A recursive definition for the set S, consisting of all binary strings that are palindromes, would be defined by two rules:
For the base cases, both '0' and '1' are in S. This covers the palindromes of length 1.
The inductive step would be: If 'P' is a string in S, then both '0P0' and '1P1' are in S. This allows us to generate palindromes of increasing lengths all the way to infinity.
By this definition, a string is a palindrome if it is the same when read from left to right and right to left. It starts with the simplest cases (single digit palindromes) and then defines how to build larger examples based on smaller ones.
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Which of the following measurements are in their most appropriate form:
(A) 7.425 ± 3.2 m
(B) 9,876,543,210 ± 21,648 years
(C) 6.541 ×103 ± 43 seconds
(D) 2.222 ×10−3 ± 2.2 ×10-5 radians
(E) (0.00 ± 0.04) kg
Answer:
(E) (0.00 + or - 0.04) kg
Step-by-step explanation:
(0.00 + or - 0.04) kg gives the most appropriate form of mass measurement.
Lower bound of the mass = 0.00 - 0.04 = -0.04 kg
Upper bound of the mass = 0.00 + 0.04 = 0.04 kg
The mass lies between -0.04 kg and 0.04 kg
Waiting in line. A quality - control manager at an amusement park feels that the amount of time that people spend waiting in line for the American Eagle roller coaster is too long. To determinate if a new loading/unloading procedure if efective in reducing wait time in line, he measured the amount of time (in minutes) people are waiting in line for 7 days. After implementing the new procedure, he again measures the amount of time in minutes and people are waiting in line 7 days and obtains the following data.
Wait time before new procedure
Day
Mon Tues Wed Thurs Fri Sat Sat Sun Sun
11.6 25.9 20.0 38.2 57.3 32.1 81.8 57.1 62.8
Wait time after new procedure
10.7 28.3 19.2 35.9 59.2 31.8 75.3 54.9 62.0
test the claim that the new loading/unloading procedure is effective in reducing wait time (H0: µd=0 and H1: µd<0)at α=.05 level of significance. Note: A normal probability plot and boxplot of the data indicate that the differences are approximately normally distributed with no outliers (use the classical approach and the p-value approach).
Answer:
No
explanation:
given:
n=9
[tex]\alpha[/tex]=0.05
see the attachment
Determine the sample mean of the differences. The mean is the sum of all values divided by the number of values.
d=0.9-2.4+0.8+...+6.5+2.2+0.8/9
=1.0556
The variance is the sum of squared deviations from the mean divided by n-1. The standard deviation is the square root of the variance. Determine the sample standard deviation of the differences:
s_d=√(0.9-1.0556)^2+...+(0.8-1.0556)^2/9-1
=2.6
CLASSICAL APPROACH :
Given claim: new procedure reduces [tex]u_{d}[/tex] > 0
The claim is either the null hypothesis or the alternative hypothesis The null hypothesis and the alternative hypothesis state the opposite of each other The null hypothesis needs to contain an equality
[tex]H_{0}:u_{d}=0\\ H_{1}:u_{d}>0[/tex]
Determine the value of the test statistic
t=d-[tex]u_{d}[/tex]/s_d/√n
=1.220
Determine the critical value from the Student T distribution table in the appendix in the row with d_f = n- 1 = 9-1 = 8 and in the column with [tex]\alpha[/tex] = 0.05 t =1.860
The rejection region then contains all values larger than 1.860
If the value of the test statistic is within the failed region, then the null hypothesis is failed
1.220 < 1.860 failed H_0
There is not sufficient evidence to support the claim that the new loading/unloading procedure is effective in reducing the wait time.
P VALUE APPROACH:
Given claim: new procedure reduces [tex]u_{d}[/tex] > 0
The claim is either the null hypothesis or the alternative hypothesis. The null hypothesis and the alternative hypothesis state the opposite of each other. The null hypothesis needs to contain an equality.
[tex]H_{0}:u_{d}=0\\ H_{1}:u_{d}>0[/tex]
Determine the value of the test statistic:
t=d-[tex]u_{d}[/tex]/s_d/√n
The P-value is the probability of obtaining the value of the test statistic, or a value more extreme, assuming that the null hypothesis is true. The P-value is the number (or interval) in the column title of the Students T distribution in the appendix containing the t-value in the row d_f = n-1 = 9-1 = 8
0.10 < P < 0.15
If the P-value is less than the significance level, reject the null hypothesis.
P > 0.05 failed H_0
There is not sufficient evidence to support the claim that the new loading unloading procedure is effective in reducing the wait time.
The J.O. Supplies Company buys calculators from a non-US supplier. The probability of a defective calculator is 10 percent. If 3 calculators are selected at random, what is the probability that one of the calculators will be defective
Answer:
There is a 24.3% probability that one of the calculators will be defective.
Step-by-step explanation:
For each calculator, there are only two possible outcomes. Either it is defective, or it is not. So we use the binomial probability distribution to solve this problem.
Binomial probability distribution
The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
In which [tex]C_{n,x}[/tex] is the number of different combinations of x objects from a set of n elements, given by the following formula.
[tex]C_{n,x} = \frac{n!}{x!(n-x)!}[/tex]
And p is the probability of X happening.
The probability of a defective calculator is 10 percent.
This means that [tex]p = 0.1[/tex]
If 3 calculators are selected at random, what is the probability that one of the calculators will be defective
This is P(X = 1) when n = 3. So
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
[tex]P(X = 1) = C_{3,1}.(0.1)^{3}.(0.9)^{2} = 0.243[/tex]
There is a 24.3% probability that one of the calculators will be defective.
I need help please!!!!
Answer:
x = 14.48
Step-by-step explanation:
first we have to see that we have the measurements from all sides
and we know that the angle between side 21 and 20 is 90 degrees
well to start we have to know the relationships between angles, legs and the hypotenuse.
a: adjacent
o: opposite
h: hypotenuse
sin α = o/h
cos α= a/h
tan α = o/a
let's take the left angle as α
sin α = 21/29
α = sin^-1 (21/29)
α = sin^-1 (0.7241)
α = 46.397
Now we do the same with the smaller triangle
tan α = o/a
sin 46.397 = x/20
0.724 = x/20
0.724 * 20 = x
14.48 = x
x = 14.48
if we want to check it we can do the same procedure with the other angle
Define the folowing terms. (a) Experimental unit (b) Treatment (c) Response variable (d) Factor (e) Placebo ( Confounding (a) Define experimental unit. Choose the correct answer below. O A. A person, object, or some other well-defined item upon which a treatment is applied ? B. The quantitative or qualitative variable or which the experimenter wishes to determine how, its value is affected by the explanatory vanable O C. Any combination of the values of the factors (explanatory variables) O D. An innocuous medication, such as a sugar tablet, that looks, tastes, and smells like the experimental medication (b) Define treatment. Choose the correct answer below. O A. The number of individuals in the experiment O B. The quantitative or qualitative variable for which the experimenter wishes to determine how its value is affected by the explanatory variable O C. Any combination of the values of the factors (explanatory variables) ? D. A controlled study to determine the effect varying one or more explanatory variables or factors has on a response variable (c) Define response variable. Choose the correct answer below. O A. The quantitative or qualitative variable for which the experimenter wishes to determine how its value is affected by the explanatory variable O B. The effect of two factors (explanatory variables on the response variable) cannot be distinguished O C. An innocuous medication, such as a sugar tablet, that looks, tastes, and smells like the experimental medication O D. The variable whose effect on the response variable is to be assessed by the experimenter
Answer:A) A
B) B
C) A
Step-by-step explanation:Experimental unit is a unit of statistical analysis.
It is also a member in a set entities.
b) Treatment is a combination of factor levels. It is an independent variable and can be manipulated by the one doing the experiment.
c) Response variables is an independent variable in which changes can be made to effect a change in the result.
Factor is a circumstance that engenders to a result.
e) Placebo is an inert substance or treatment which delivers a therapeutic value.
The experimental terms are defined as follows:
(a) The definition of an Experimental Unit is A. A person, object, or some other well-defined item upon which a treatment is applied.
(b) The definition of treatment is C. Any combination of the values of the factors (explanatory variables).
(c) The definition of a response variable is A. The quantitative or qualitative variable for which the experimenter wishes to determine how its value is affected by the explanatory variable.
(d) The definition of a factor is A. A variable whose effect on the response variable is to be assessed by the experimenter.
(e) The definition of a Placebo is C. An innocuous medication, such as a sugar tablet, that looks, tastes, and smells like the experimental medication.
(f) The definition of Confounding is when A. The effect of two factors (explanatory variables on the response variable) cannot be distinguished.
A confounding factor affects the dependent and independent variables with spurious effects.
Thus, the terms have been correctly defined with the right options.
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Samples of laboratory glass are in small, lightpackaging or heavy, large packaging. Suppose that 2% and 1% of thesample shipped in small and large packages, respectively, breakduring transit. (a) If 60% of the samples are shipped in largepackages and 40% are shipped in small packages, what proportion ofsamples break during shipment? (b) Also, if a sample breaks duringshipment, what is the probability that it was shipped in a smallpackage?
Answer:
a) 1.4% of the samples break during shipment
b) the probability is 4/7 ( 57.14%)
Step-by-step explanation:
a) defining the event B= the sample of laboratory glass breaks , then the probability is:
P(B)= probability that sample is shipped in small packaging * probability that the sample breaks given that was shipped in small packaging + probability that sample is shipped in large packaging * probability that the sample breaks given that was shipped in large packaging = 0.40* 0.02 + 0.60*0.01 = 0.014
b) we can use the theorem of Bayes for conditional probability. Then defining the event S= the sample is shipped in small packaging . Thus we have
P(S/B)= P(S∩B)/P(B) = 0.40* 0.02 / 0.014= 4/7 ( 57.14%)
where
P(S∩B)= probability that sample is shipped in small packaging and it breaks
P(S/B)= probability that sample was shipped in small packaging given that is broken
A data set that consists of 33 numbers has a minimum value of 19 and a maximum value of 71. Determine the class boundaries using the 2 Superscript k Baseline greater than or equals n rule if the data are:
a) discrete
b) continuous
Answer:
b)continous
Step-by-step explanation:
A continuous data is a finite number within a chosen range example temperature range.
Matt is a software engineer writing a script involving 6 tasks. Each must be done one after the other. Let ti be the time for the ith task. These times have a certain structure:
•Any 3 adjacent tasks will take half as long as the next two tasks.
•The second task takes 1 second.
•The fourth task takes 10 seconds.
a) Write an augmented matrix for the system of equations describing the length of each task.
b) Reduce this augmented matrix to reduced echelon form.
c) Suppose he knows additionally that the sixth task takes 20 seconds and the first three tasks will run in 50 seconds. Write the extra rows that you would add to your answer in b) to take account of this new information.
d) Solve the system of equations in c).
Answer:
Let [tex]t_i[/tex] be the time for the [tex]i[/tex]th task.
We know these times have a certain structure:
Any 3 adjacent tasks will take half as long as the next two tasks.In the form of an equations we have
[tex]t_1+t_2+t_3=\frac{1}{2}t_4+\frac{1}{2}t_5 \\\\t_2+t_3+t_4=\frac{1}{2}t_5+\frac{1}{2}t_6[/tex]
The second task takes 1 second [tex]t_2=1[/tex]The fourth task takes 10 seconds [tex]t_4=10[/tex]So, we have the following system of equations:
[tex]t_1+t_2+t_3-\frac{1}{2}t_4-\frac{1}{2}t_5=0 \\\\t_2+t_3+t_4-\frac{1}{2}t_5-\frac{1}{2}t_6=0\\\\t_2=1\\\\t_4=10[/tex]
a) An augmented matrix for a system of equations is a matrix of numbers in which each row represents the constants from one equation (both the coefficients and the constant on the other side of the equal sign) and each column represents all the coefficients for a single variable.
Here is the augmented matrix for this system.
[tex]\left[ \begin{array}{cccccc|c} 1 & 1 & 1 & - \frac{1}{2} & - \frac{1}{2} & 0 & 0 \\\\ 0 & 1 & 1 & 1 & - \frac{1}{2} & - \frac{1}{2} & 0 \\\\ 0 & 1 & 0 & 0 & 0 & 0 & 1 \\\\ 0 & 0 & 0 & 1 & 0 & 0 & 10 \end{array} \right][/tex]
b) To reduce this augmented matrix to reduced echelon form, you must use these row operations.
Subtract row 2 from row 1 [tex]\left(R_1=R_1-R_2\right)[/tex].Subtract row 2 from row 3 [tex]\left(R_3=R_3-R_2\right)[/tex].Add row 3 to row 2 [tex]\left(R_2=R_2+R_3\right)[/tex].Multiply row 3 by −1 [tex]\left({R}_{{3}}=-{1}\cdot{R}_{{3}}\right)[/tex].Add row 4 multiplied by [tex]\frac{3}{2}[/tex] to row 1 [tex]\left(R_1=R_1+\left(\frac{3}{2}\right)R_4\right)[/tex].Subtract row 4 from row 3 [tex]\left(R_3=R_3-R_4\right)[/tex].Here is the reduced echelon form for the augmented matrix.
[tex]\left[ \begin{array}{ccccccc} 1 & 0 & 0 & 0 & 0 & \frac{1}{2} & 15 \\\\ 0 & 1 & 0 & 0 & 0 & 0 & 1 \\\\ 0 & 0 & 1 & 0 & - \frac{1}{2} & - \frac{1}{2} & -11 \\\\ 0 & 0 & 0 & 1 & 0 & 0 & 10 \end{array} \right][/tex]
c) The additional rows are
[tex]\begin{array}{ccccccc} 0 & 0 & 0 & 0 & 0 & 1 & 20 \\\\ 1 & 1 & 1 & 0 & 0 & 0 & 50 \end{array} \right[/tex]
and the augmented matrix is
[tex]\left[ \begin{array}{ccccccc} 1 & 0 & 0 & 0 & 0 & \frac{1}{2} & 15 \\\\ 0 & 1 & 0 & 0 & 0 & 0 & 1 \\\\ 0 & 0 & 1 & 0 & - \frac{1}{2} & - \frac{1}{2} & -11 \\\\ 0 & 0 & 0 & 1 & 0 & 0 & 10 \\\\ 0 & 0 & 0 & 0 & 0 & 1 & 20 \\\\ 1 & 1 & 1 & 0 & 0 & 0 & 50 \end{array} \right][/tex]
d) To solve the system you must use these row operations.
Subtract row 1 from row 6 [tex]\left(R_6=R_6-R_1\right)[/tex].Subtract row 2 from row 6 [tex]\left(R_6=R_6-R_2\right)[/tex].Subtract row 3 from row 6 [tex]\left(R_6=R_6-R_3\right)[/tex].Swap rows 5 and 6.Add row 5 to row 3 [tex]\left(R_3=R_3+R_5\right)[/tex].Multiply row 5 by 2 [tex]\left(R_5=\left(2\right)R_5\right)[/tex].Subtract row 6 multiplied by 1/2 from row 1 [tex]\left(R_1=R_1-\left(\frac{1}{2}\right)R_6\right)[/tex].Add row 6 multiplied by 1/2 to row 3 [tex]\left(R_3=R_3+\left(\frac{1}{2}\right)R_6\right)[/tex].[tex]\left[ \begin{array}{ccccccc} 1 & 0 & 0 & 0 & 0 & 0 & 5 \\\\ 0 & 1 & 0 & 0 & 0 & 0 & 1 \\\\ 0 & 0 & 1 & 0 & 0 & 0 & 44 \\\\ 0 & 0 & 0 & 1 & 0 & 0 & 10 \\\\ 0 & 0 & 0 & 0 & 1 & 0 & 90 \\\\ 0 & 0 & 0 & 0 & 0 & 1 & 20 \end{array} \right][/tex]
The solutions are: [tex](t_1,...,t_6)=(5,1,44,10,90,20)[/tex].
Ammonia at 70 F with a quality of 50% and a total mass of 4.5 lbm is in a rigid tank with an outlet valve at the bottom. How much liquid mass can be removed through the valve, assuming the temperature stays constant
Answer:
0.10865 killograms
Step-by-step explanation:
calculating the liquid mass of ammonia removed through the bottom value from a rigid tank at constant temperature.
Given:
temperature: [tex]T=70 F[/tex]
quality : 50% = 0.5
initial mass: [tex]m1= 4.5 lbm[/tex]
to find the removed liquid mass first we have to find total volume from which we can find remaining mass. as the tang is rigid the temperature and volume remains constant.
by taking the difference of mass we can determine the mass of liquid removed.
we have two phases at temperature [tex]T= 70 F[/tex] with specific volume for liquid [tex]vf=0.02631 ft^3/lbm[/tex] and specific volume for vapor is [tex]vg=2.3098 ft^3/lbm[/tex] .
The Volume in the initial state is given by, (Using definition of specific volume)
[tex]V= m1v1[/tex]
using [tex]v1=x(vf+vg)[/tex]
[tex]V= m1x(vf+vg)[/tex]
substituting [tex]m1= 4.5 lbm\\[/tex] , [tex]vf= 0.02631 ft^3/lbm[/tex] , [tex]vg=2.3098 ft^3/lbm[/tex]
we get
[tex]V= (4.5 lbm)(0.5)(0.02631 ft^3/lbm +2.3098 ft^3/lbm)[/tex]
finally [tex]V=5.2625 ft^3[/tex]
we know the formula to find liquid mass is
[tex]mass =density *volume[/tex]
density of ammonia is [tex]0.73 kg/m^3[/tex]
inserting the values into the formula we get the value for liquid mass removed through the valve.
[tex]m = (0.73 kg/m^3)*(5.25625 ft^3)[/tex]
the final answer is
[tex]m= 0.10865 kg[/tex]
The Maclaurin series expansion for the arctangent of x is defined for |x| ≤ 1 as arctan x = ∑ n=0 [infinity] (−1)n ______ 2n +1 x 2n+1 (a) Write out the first 4 terms (n = 0,...,3). (b) Starting with the simplest version, arctan x = x, add terms one at a time to estimate arctan(π/6). After each new term is added, comput
Answer:
a) [tex] n =0, \frac{(-1)^0}{2*0+1} x^{2*0+1}= x[/tex]
[tex] n =1, \frac{(-1)^1}{2*1+1} x^{2*1+1}= -\frac{x^3}{3}[/tex]
[tex] n =2, \frac{(-1)^2}{2*2+1} x^{2*2+1}= \frac{x^5}{5}[/tex]
[tex] n =3, \frac{(-1)^3}{2*3+1} x^{2*3+1}= -\frac{x^7}{7}[/tex]
b) n=0
[tex] arctan(\pi/6) \approx \pi/6 = 0.523599[/tex]
The real value for the expression is [tex] arctan (\pi/6) = 0.482348[/tex]
And if we replace into the formula of relative error we got:
[tex] \% error= \frac{|0.523599 -0.482348|}{0.482348} * 100= 8.55\%[/tex]
n =1
[tex] arctan(\pi/6) \approx \pi/6 -\frac{(pi/6)^3}{3} = 0.47576[/tex]
[tex] \% error= \frac{|0.47576 -0.482348|}{0.482348} * 100= 1.37\%[/tex]
n =2
[tex] arctan(\pi/6) \approx \pi/6 -\frac{(pi/6)^3}{3} +\frac{(pi/6)^5}{5} = 0.483631[/tex]
[tex] \% error= \frac{|0.483631 -0.482348|}{0.482348} * 100= 0.27\%[/tex]
n =3
[tex] arctan(\pi/6) \approx \pi/6 -\frac{(pi/6)^3}{3} +\frac{(pi/6)^5}{5}-\frac{(pi/6)^7}{7} = 0.48209[/tex]
[tex] \% error= \frac{|0.48209 -0.482348|}{0.482348} * 100= 0.05\%[/tex]
[tex] \arctan (\pi/6) = 0.48[/tex]
Step-by-step explanation:
Part a
the general term is given by:
[tex] a_n = \frac{(-1)^n}{2n+1} x^{2n+1}[/tex]
And if we replace n=0,1,2,3 we have the first four terms like this:
[tex] n =0, \frac{(-1)^0}{2*0+1} x^{2*0+1}= x[/tex]
[tex] n =1, \frac{(-1)^1}{2*1+1} x^{2*1+1}= -\frac{x^3}{3}[/tex]
[tex] n =2, \frac{(-1)^2}{2*2+1} x^{2*2+1}= \frac{x^5}{5}[/tex]
[tex] n =3, \frac{(-1)^3}{2*3+1} x^{2*3+1}= -\frac{x^7}{7}[/tex]
Part b
If we use the approximation [tex] arctan x \approx x[/tex] we got:
n=0
[tex] arctan(\pi/6) \approx \pi/6 = 0.523599[/tex]
The real value for the expression is [tex] arctan (\pi/6) = 0.482348[/tex]
And if we replace into the formula of relative error we got:
[tex] \% error= \frac{|0.523599 -0.482348|}{0.482348} * 100= 8.55\%[/tex]
If we add the terms for each value of n and we calculate the error we see this:
n =1
[tex] arctan(\pi/6) \approx \pi/6 -\frac{(pi/6)^3}{3} = 0.47576[/tex]
[tex] \% error= \frac{|0.47576 -0.482348|}{0.482348} * 100= 1.37\%[/tex]
n =2
[tex] arctan(\pi/6) \approx \pi/6 -\frac{(pi/6)^3}{3} +\frac{(pi/6)^5}{5} = 0.483631[/tex]
[tex] \% error= \frac{|0.483631 -0.482348|}{0.482348} * 100= 0.27\%[/tex]
n =3
[tex] arctan(\pi/6) \approx \pi/6 -\frac{(pi/6)^3}{3} +\frac{(pi/6)^5}{5}-\frac{(pi/6)^7}{7} = 0.48209[/tex]
[tex] \% error= \frac{|0.48209 -0.482348|}{0.482348} * 100= 0.05\%[/tex]
And thn we can conclude that the approximation is given by:
[tex] \arctan (\pi/6) = 0.48[/tex]
Rounded to 2 significant figures
Final answer:
The Maclaurin series for arctan x includes the first four terms: x, -x^3/3, x^5/5, and -x^7/7. To estimate arctan(π/6), we incrementally add these terms, leading to progressively better approximations.
Explanation:
The Maclaurin series for the arctangent function is given by:
arctan x = ∑ n=0 [infinity] (−1)n/(2n +1) x2n+1
(a) Writing out the first 4 terms for n = 0 to 3, we get:
For n = 0: x
For n = 1: −x3/3
For n = 2: x5/5
For n = 3: −x7/7
The series starts with x and alternates between subtracting and adding subsequent odd-powered terms, each divided by the respective odd number.
(b) To estimate arctan(π/6), also known as arctan(1/(√3)), we add terms of the series one by one:
Simplest estimate: arctan(π/6) ≈ π/6
Adding the second term: arctan(π/6) ≈ π/6 - (π/6)3/3
Including the third term: arctan(π/6) ≈ π/6 - (π/6)3/3 + (π/6)5/5
Including the fourth term: arctan(π/6) ≈ π/6 - (π/6)3/3 + (π/6)5/5 - (π/6)7/7
By computing these sums, we get increasingly accurate estimates for the value of arctan(π/6).
The time until failure of an electronic device has an exponential distribution with mean 15 months. If a random sample of five such devices are tested, what is the probability that the first failure among the five devices occurs a after 9 months? b before 12 months?
Final answer:
To calculate the probability of the first failure of an electronic device occurring after 9 months and before 12 months, we use the exponential distribution formula. For part a, we raise the probability that one device surpasses 9 months to the power of five. For part b, we use the complementary probability that at least one device fails before 12 months.
Explanation:
The time-to-failure of an electronic device which follows an exponential distribution can be used to calculate probabilities for different time intervals. With a mean of 15 months, the rate [tex](\lambda\))[/tex] is the reciprocal of the mean, thus [tex]\(\lambda = 1/15\)[/tex]. Here's how we can calculate the requested probabilities:
Probability that the first failure occurs after 9 months: We use the exponential distribution property P(X > x) = e-\. [tex](\lambda x\))[/tex] Substituting the given values we get P(X > 9) = e-(1/15 × 9).Probability that the first failure occurs before 12 months: Again, using P(X < x) = 1 - e-[tex](\lambda x\))[/tex], we get P(X < 12) = 1 - e-(1/15 × 12).For a sample of five such devices, assuming independence, we require all five devices to surpass 9 months for the first failure to occur after that time, thus we raise the probability found in part a to the power of five. Similarly, for the first failure to occur before 12 months, at least one device must fail before that time, so we use the complementary probability of all devices lasting longer than 12 months.
Let's calculate these probabilities using the exponential distribution formula.
a. P(first failure after 9 months) = \((e-(1/15 × 9))5\) = 0.5488.b. P(first failure before 12 months) = 1 - \((e-(1/15 × 12))5\) = 0.5513.. During a year, the probability a structure will be damaged by an earthquake (A) is 0.02, that it will be damaged by a hurricane (B) is 0.03, and that it will be damaged by both is 0.007. What is the probability that it will not be damaged by a hurricane or an earthquake during that year?
Answer:
0.9506
Step-by-step explanation:
Pr(A) = 0.02
Pr(B) = 0.03
Pr(both) = 0.007
So,
Pr(Not A) = 1 - Pr(A)
= 1 - 0.02
= 0.98
Pr(Not B) = 1 - Pr(B)
= 1 - 0.03
= 0.97
Pr(Not by both) = 1 - Pr(both)
= 1 - 0.007
= 0.993
Thus,
Pr(Not B) or Pr(Not A) = 0.97 × 0.98
= 0.9056
∴ the probability that the house would not be damaged by a hurricane or an earthquake during the year is 0.9506.
Brian and Jennifer each take out a loan of X. Jennifer will repay her loan by making one payment of 800 at the end of year 10. Brian will repay his loan by making one payment of 1,120 at the end of year 10. The nominal semi-annual rate being charged to Jennifer is exactly one-half the nominal semi-annual rate being charged to Brian. Calculate X.
Answer:
$568.148
Step-by-step explanation:
We will relate the loan (X) with their nominal annual rate converted semiannually:
Jennifer's loan, [tex]X = 800(1+{\frac{j}{2}})^{-10*2}[/tex]
Brian's loan, [tex]X = 1,120(1+{\frac{2j}{2}})^{-10*2}[/tex]
Since Brain and Jennifer took the same amount loan, the two equations of semi annual rates can be combined thus:
[tex]X = 800(1+{\frac{j}{2}})^{-20}} = 1,120(1+{\frac{2j}{2}})^{-20}\\= {\frac{800}{1,120}}(1+{\frac{j}{2}})=(1+{\frac{2j}{2}})[/tex]
For simplicity, we will use "Y" to represent 0.714 (i.e: [tex]{\frac{800}{1,120}} = 0.714[/tex] )
Therefore, continuing with the equation above:
[tex]Y + {\frac{Yj}{2}}=1+j\\2Y+Yj=2+2j\\2Y+Yj-2j=2\\Yj-2j=2-2Y\\j(Y-2)=2-2Y\\j={\frac{2-2Y}{Y-2}}[/tex]
substituting the real value of Y (0.714) into the equation, we have:
[tex]j = {\frac{2-(2*0.714)}{0.714-2}}\\={\frac{2-1.428}{-1.286}}\\={\frac{0.572}{-1.286}}\\ =-0.445[/tex]
Solving for the value of X using j, we have:
[tex]X(1+{\frac{j}{2}})=800\\X=568.148[/tex]
This is a mathematical problem about compound interest and loan repayment. It creates two equations based on the given scenario. These equations can be solved to find the initial loan amount X.
Explanation:This question pertains to the concepts of compound interest and loan repayment in mathematics. Given the terms, we can use the formula for the future value of a loan: FV = P(1 + r/n) ^(nt). Here, FV is the future value of the loan, P is the principal loan amount, r is the annual interest rate, n is the number of times that interest is compounded per time t, and t is the time the money is invested for.
According to the problem, the nominal semi-annual rate charged to Jennifer is exactly half of the rate charged to Brian. Hence, if we denote the rate for Brian as 2r, the rate for Jennifer would be r. We know Jennifer's payment is 800 and Brian's is 1,120; these are the future values of their respective loans.
So, we get two equations from this problem:
1. X(1 + r/2)^(2*10) = 800
2. X(1 + 2r/2)^(2*10) = 1,120
You can solve these equations to find the value of X, which represents the amount they borrowed initially.
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