a. The probability is 0.30. The advertisement seems to have a positive effect on customer purchases.
b. The market share is 0.20. The advertisement could potentially increase the company's market share
c. The conditional purchase probability of 0.333. The second advertisement seems to have had a slightly bigger effect on customer purchases.
a. We are asked to find the probability of an individual purchasing the product given that the individual recalls seeing the advertisement, i.e., P(B|S).
Using the formula for conditional probability:
P(B|S) = P(B∩S) / P(S)
Given:
P(B∩S) = 0.12
P(S) = 0.40
So, P(B|S)
= 0.12 / 0.40
= 0.30
Seeing the advertisement increases the probability that an individual will purchase the product from 0.20 (P(B)) to 0.30 (P(B|S)). Therefore, the advertisement seems to have a positive effect on customer purchases.
As a decision maker, if the cost of the advertisement is reasonable, it would be recommended to continue the advertisement since it increases the likelihood of product purchases.
b. The company's market share can be estimated by considering the probability of individuals purchasing the company's soap product and the probability of individuals purchasing from competitors.
The probability of an individual purchasing from competitors
= 1 - P(B)
= 1 - 0.20
= 0.80
Therefore, the company's market share is:
Market Share = P(B)
= 0.20
Continuing the advertisement could potentially increase the company's market share since the advertisement has a positive effect on customer purchases, as shown in part (a).
c. For the other advertisement:
Given:
P(S) = 0.30
P(B∩S) = 0.10
Using the formula for conditional probability:
P(B|S) = P(B∩S) / P(S)
So, P(B|S)
= 0.10 / 0.30
= 0.333
Comparing the two advertisements, the first advertisement had a conditional purchase probability of 0.30, while the second advertisement had a conditional purchase probability of 0.333. Therefore, the second advertisement seems to have had a slightly bigger effect on customer purchases.
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The advertisements increase the chance for an individual to purchase the product. The second advertisement yields a higher likelihood of purchase, implying it's more effective.
Explanation:The problem is related to conditional probability. To solve the question:
The probability of an individual purchasing the product given that the individual recalls seeing the advertisement is calculated by P(B|S) = P(B∩S) / P(S) = .12 / .40 = .30 or 30%. This is greater than the overall probability of purchasing the product, P(B) = .20 or 20%. Meaning the advertisement does increase the probability for an individual to purchase the product.The estimated market share of the company is simply P(B) = .20 or 20%. As advertisement increases the probability of purchase, it would likely increase the market share as long as the cost of advertising does not outweigh the added revenue.For the other advertisement, P(B|S) = P(B∩S) / P(S) = .10 / .30 = .33 or 33%, which is higher than the first advertisement. Therefore, the other advertisement has a larger effect on customer purchases.Learn more about Conditional Probability here:https://brainly.com/question/32171649
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f left parenthesis x right parenthesis equals StartFraction 16 x squared Over x Superscript 4 Baseline plus 64 EndFractionf(x)=16x2 x4+64(a) Is the pointleft parenthesis negative 2 StartRoot 2 EndRoot comma 1 right parenthesis−22,1on the graph of f?(b) Ifx equals 2 commax=2,what is f(x)? What point is on the graph of f?(c) If f left parenthesis x right parenthesis equals 1 commaf(x)=1, what is x? What point(s) is (are) on the graph of f?(d) What is the domain of f?(e) List the x-intercepts, if any, of the graph of f.(f) List the y-intercept, if there is one, of the graph of f.
Answer:
a) Yes
b) (2,0.8)
c)[tex](2\sqrt2,1), (-2\sqrt2,1)[/tex]
d) [tex]x \in (-\infty,\infty)[/tex]
e) (0,0)
f) (0,0)
Step-by-step explanation:
We are given the following function in the question:
[tex]f(x) = \displaystyle\frac{16x^2}{x^4 + 64}[/tex]
a) We have to check whether given point lies on the function or not.
[tex](-2\sqrt2,1)\\\\f(-2\sqrt2) = \displaystyle\frac{16(-2\sqrt2)^2}{(-2\sqrt2)^4 + 64} = \frac{128}{128} = 1[/tex]
b) Find value of f(x) at x = 2
[tex]f(2) = \displaystyle\frac{16(2)^2}{(2)^4 + 64} =\frac{64}{80}= 0.8[/tex]
Thus, (2,0.8) lies on the graph of given function.
c) We have to find the value of x, when f(x) = 1
[tex]1 = \displaystyle\frac{16x^2}{x^4 + 64}\\\\x^4 -16x^2 + 64 = 0\\(x^2-8)^2 = 0\\x^2 - 8 = 0\\x = \pm 2\sqrt2[/tex]
[tex](2\sqrt2,1), (-2\sqrt2,1)[/tex] lies on he graph of function.
d) Domain is the collection of all values of x for which the function is defined.
[tex]x \in (-\infty,\infty)[/tex]
e) x-intercepts
This is the value of x such that the function is zero.
[tex]0 = \displaystyle\frac{16x^2}{x^4 + 64}\\\\16x^2 = 0\\x = 0[/tex]
f) y-intercept
It is the value of function when x is zero.
[tex]f(0) = \displaystyle\frac{16(0)^2}{(0)^4 + 64} = 0[/tex]
The function passes trough origin.
The solution involves determining if a specific coordinates exist on the graph, calculating the function value for specific x-values, determining x-values for a specific function value, finding the domain of the function, and identifying the x and y intercepts of the function.
Explanation:The function is f(x) = 16x²/(x⁴ + 64).
To check if the point (-2√2, 1) is on the graph, substitute x = -2√2 into f(x). If f(-2√2) equals 1, the point is on the graph.For x = 2, substitute x = 2 into f(x) to get f(2). The point on the graph for this x-value will be (2, f(2)).To find x when f(x) = 1, set f(x) equal to 1 and solve for x. The points on the graph will be (x, 1), where x are the solutions to the equation.The domain of the function f is all real numbers except for x-values that make the denominator zero. Solve x⁴ + 64 = 0 to find x-values to exclude from the domain.To find the x-intercepts of the graph, set f(x) equal to zero and solve for x.The y-intercept of the graph is the value of f(x) at x = 0, which is f(0).Learn more about Function Analysis here:https://brainly.com/question/34156091
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The route used by a certain motorist in commuting to work contains two intersections with traffic signals. The probability that he must stop at the first signal is 0.35, the analogous probability for the second signal is 0.55, and the probability that he must stop at at least one of the two signals is 0.75.
What is the probability that he must stop:
A. at both signals?
B. at the first signal but not at the second one?
C. at exactly one signal?
Answer:
a) 0.15
b) 0.2
c) 0.6
Step-by-step explanation:
We are given the following in the question:
A: Stopping at first signal
B: Stopping at second signal
P(A) = 0.35
P(B) = 0.55
Probability that he must stop at at least one of the two signals is 0.75
[tex]P(A\cup B) = 0.75[/tex]
a) P(at both signals)
[tex]P(A\cup B) = P(A) + P(B) - P(A\cap B)\\0.75 = 0.35 + 0.55 - P(A\cap B)\\P(A\cap B) = 0.35 + 0.55 - 0.75 = 0.15[/tex]
0.15 is the probability that motorist stops at both signals.
b) P(at the first signal but not at the second one)
[tex]P(A\cap B') = P(A) - P(A\cap B)\\P(A\cap B') = 0.35 - 0.15 = 0.2[/tex]
0.2 is the probability that motorist stops at the first signal but not at the second one.
c) P(at exactly one signal)
[tex]P(A\cap B') + P(A\cap 'B) = P(A\cup B) - P(A\cap B) \\P(A\cap B') + P(A\cap 'B) = 0.75 - 0.15 = 0.6[/tex]
0.6 is the probability that the motorist stops at exactly one signal.
Final answer:
To calculate various probabilities related to stopping at traffic signals, we use given values to determine a 15% chance of stopping at both signals, a 20% chance of stopping at the first but not the second, and a 60% chance of stopping at exactly one signal.
Explanation:
The question involves calculating probabilities of stopping at traffic signals. Given are the probabilities of stopping at the first (0.35) and second (0.55) signals, and the probability of stopping at at least one signal (0.75). Using these, we can find the probabilities for various scenarios.
A. Probability of stopping at both signals:
To find this, we use the formula: P(A and B) = P(A) + P(B) - P(A or B). Here, P(A or B) is the probability of stopping at least at one signal, which is given as 0.75. Thus, the calculation would be 0.35 + 0.55 - 0.75 = 0.15. So, there is a 15% chance of stopping at both signals.
B. Probability of stopping at the first signal but not the second one:
This can be calculated by subtracting the probability of stopping at both signals from the probability of stopping at the first signal: 0.35 - 0.15 = 0.20. Therefore, there is a 20% chance of stopping at the first signal but not the second.
C. Probability of stopping at exactly one signal:
This involves adding the probabilities of stopping only at the first signal or only at the second signal. We've already calculated the first part as 0.20. For the second part, subtract the probability of stopping at both signals from stopping at the second signal: 0.55 - 0.15 = 0.40. Adding these together, 0.20 + 0.40 = 0.60, there is a 60% chance of stopping at exactly one signal.
Almost all medical schools in the United States require students to take the Medical College Admission Test (MCAT). A new version of the exam was introduced in spring 2015 and is intended to shift the focus from what applicants know to how well they can use what they know. One result of the change is that the scale on which the exam is graded has been modified, with the total score of the four sections on the test ranging from 472 to 528 . In spring 2015 the mean score was 500.0 with a standard deviation of 10.6 . Use Table A to find the answers to the two questions.
(a) What proportion of students taking the MCAT had a score over 519 ?
(b) Compute the proportion and then enter the answer as a percentage rounded to two decimal places.
Answer:
(a) The proportion of students taking the MCAT had a score over 519 is 0.0367.
(b) The percentage of students who scored more than 519 in the MCAT is 3.67%.
Step-by-step explanation:
Assuming that the sample size or the number of students taking the MCAT in 2015 is large, the sampling distribution of scores follow a normal distribution.
Let X = score of a student
Given:
Mean = [tex]\mu=500[/tex]
Standard deviation = [tex]\sigma=10.6[/tex]
(a)
Compute the probability of students who scored more than 519 as follows:
[tex]P(X>519)=P(\frac{X-\mu}{\sigma}> \frac{519-500}{10.6})\\=P(Z>1.7925)\\=1-P(Z<1.7925)[/tex]
Use the z-table to determine the probability.
[tex]P(X>519)=P(\frac{X-\mu}{\sigma}> \frac{519-500}{10.6})\\=P(Z>1.7925)\\=1-P(Z<1.7925)\\=1-0.9633\\=0.0367[/tex]
Thus, the probability of students who scored more than 519 is 0.0367.
(b)
Convert the probability of students who scored more than 519 to percentage
[tex]=0.0367\times100\\=3.67\%[/tex]
Thus, the percentage of students who scored more than 519 is 3.67%.
Final answer:
The z-score for 519 is approximately 1.79, which correlates to about 3.67% of students scoring higher than 519.
Explanation:
To find the proportion of students scoring over 519 on the MCAT, first, we must calculate the z-score. The z-score tells us how many standard deviations an element is from the mean. Since the mean score is 500.0 and the standard deviation is 10.6, the z-score formula is z = (X - μ) / σ, where X is the score, μ is the mean, and σ is the standard deviation. For a score of 519, the z-score would be z = (519 - 500.0) / 10.6 ≈ 1.79. To find the proportion of students scoring above this z-score, we would look up the z-score in a standard normal distribution table (Table A), or use a statistical software to find that the area to the right of z=1.79. This area corresponds to the proportion of students scoring higher than 519. Since standard normal distribution tables are commonly used and we don't have one provided here, let's assume that the area to the right of z=1.79 is approximately 3.67%.
Two particles, A and B, are in uniform circular motion about a common center. The acceleration of particle A is 4.9 times that of particle B. The period of particle B is 2.4 times the period of particle A. The ratio of the radius of the motion of particle A to that of particle B is closest to
The ratio of the radius of particle A to that of particle B in uniform circular motion is approximately 2.437.
Explanation:To find the ratio of the radius of particle A to that of particle B in uniform circular motion, let's first consider the equations of motion for uniform circular motion. The centripetal acceleration, a_c, is given by the equation a_c = v^2/r, where v is the velocity and r is the radius of the circle. The period of motion, T, is the time it takes for one complete revolution around the circle. Given that the acceleration of particle A is 4.9 times that of particle B, we can write the equation a_A = 4.9 * a_B. Also, the period of particle B is 2.4 times the period of particle A, so we can write the equation T_B = 2.4 * T_A.
Next, we can use the equations of motion to express the velocity and period in terms of the acceleration and radius. From the equation a_c = v^2/r, we can rearrange it to solve for v: v = sqrt(a_c * r). By substituting this expression for v into the equation T = 2 * pi * r / v, we can solve for the period T in terms of a_c and r. Plugging these expressions for the velocities and periods of particles A and B into the equations a_A = 4.9 * a_B and T_B = 2.4 * T_A, we can form an equation that relates the radii of the two particles: sqrt(a_A * r_A) = 4.9 * sqrt(a_B * r_B) and 2 * pi * r_B / sqrt(a_B * r_B) = 2.4 * (2 * pi * r_A / sqrt(a_A * r_A)). Simplifying these equations, we can solve for the ratio of the radii r_A/r_B and find that it is approximately 2.437.
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To find the ratio of the radius of motion between particles A and B, we can use the equations of uniform circular motion and the given information. The ratio is approximately 2.21.
Explanation:To find the ratio of the radius of motion between particles A and B, we need to analyze the given information. Let's denote the acceleration of particle B as aB and the acceleration of particle A as aA. We're told that aA is 4.9 times aB, and the period of particle B is 2.4 times the period of particle A.
From the equations of uniform circular motion, we know that the acceleration is given by a = (4π2)/T2, where T is the period. Since aA = 4.9aB and TB = 2.4TA, we can set up the following equation:
(4π2)/TA2 = 4.9(4π2)/TB2
After canceling out common terms, we'll find that (TA2) / (TB2) = 4.9. Taking the square root of both sides, we get TA / TB = √4.9 = 2.21. Since the period is inversely proportional to the angular velocity, we can conclude that the ratio of the radii of motion is approximately 2.21.
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The following six independent length measurements were made (in feet) for a line: 736.352, 736.363, 736.375, 736.324, 736.358, and 736.383. Determine the standard deviation of the measurements.
Answer:
Assuming population data
[tex] \sigma = \sqrt{0.000354}=0.0188[/tex]
Assuming sample data
[tex] s = \sqrt{0.000425}=0.0206[/tex]
Step-by-step explanation:
For this case we have the following data given:
736.352, 736.363, 736.375, 736.324, 736.358, and 736.383.
The first step in order to calculate the standard deviation is calculate the mean.
Assuming population data
[tex]\mu = \frac{\sum_{i=1}^6 X_i}{6}[/tex]
The value for the mean would be:
[tex]\mu = \frac{736.352+736.363+736.375+736.324+736.358+736.383}{6}=736.359[/tex]
And the population variance would be given by:
[tex] \sigma^2 = \frac{\sum_{i=1}^6 (x_i-\bar x)}{6}[/tex]
And we got [tex] \sigma^2 =0.000354[/tex]
And the deviation would be just the square root of the variance:
[tex] \sigma = \sqrt{0.000354}=0.0188[/tex]
Assuming sample data
[tex]\bar X = \frac{\sum_{i=1}^6 X_i}{6}[/tex]
The value for the mean would be:
[tex]\bar X = \frac{736.352+736.363+736.375+736.324+736.358+736.383}{6}=736.359[/tex]
And the population variance would be given by:
[tex] s^2 = \frac{\sum_{i=1}^6 (x_i-\bar x)}{6-1}[/tex]
And we got [tex] s^2 =0.000425[/tex]
And the deviation would be just the square root of the variance:
[tex] s = \sqrt{0.000425}=0.0206[/tex]
To determine the standard deviation of given length measurements, calculate the mean, find squared differences, average them, and take the square root. The standard deviation of the measurements is 0.01879 feet.
Determining the Standard Deviation of Measurements
To find the standard deviation of the given length measurements, follow these steps:
Calculate the mean of the measurements.
Find the squared differences between each measurement and the mean.
Compute the average of the squared differences.
Take the square root of that average.
Step-by-Step Calculation
→ List of measurements: 736.352, 736.363, 736.375, 736.324, 736.358, 736.383.
→ Calculate the mean:
= (736.352 + 736.363 + 736.375 + 736.324 + 736.358 + 736.383) / 6
= 736.3591667
Find the squared differences:
(736.352 - 736.3591667)^2 = 0.00005256(736.363 - 736.3591667)^2 = 0.00001464(736.375 - 736.3591667)^2 = 0.00024811(736.324 - 736.3591667)^2 = 0.00123264(736.358 - 736.3591667)^2 = 0.00000136(736.383 - 736.3591667)^2 = 0.00057044→ Calculate the average of the squared differences:
= (0.00005256 + 0.00001464 + 0.00024811 + 0.00123264 + 0.00000136 + 0.00057044) / 6
= 0.000353292
Take the square root to find the standard deviation:
= √(0.000353292)
= 0.01879 feet
what are the common factors for 54,24,18
Answer:
Step-by-step explanation:
We find what number we multiply by another number to get 54, 24, 18
54: 1, 2, 3, 6, 9, 18, 27, 54
24: 1, 2, 3, 4, 6, 8, 12, 24
18: 1, 2, 3, 6, 9, 18
Now the numbers that repeat in these sets are the common factors
We got 1, 2, 3, and 6, which make these our common factors
Answer:1,2,3,6
Step-by-step explanation:
The factors of 54,24 and 18 are 1,2,3,6
Four candidates are to be interviewed for a job. Two of them, numbered 1 and 2, are qualified, and the other two, numbered 3 and 4, are not. The candidates are interviewed at random, and the first qualified candidate interviewed will be hired. The outcomes are the sequences of candidates that are interviewed. So one outcome is 2, and another is 431.
a. List all the possible outcomes.
b. Let A be the event that only one candidate is interviewed. List the outcomes in A.
c. Let B be the event that three candidates are interviewed. List the outcomes in B.
d. Let C be the event that candidate 3 is interviewed. List the outcomes in C.
e. Let D be the event that candidate 2 is not interviewed. List the outcomes in D
f. Let E be the event that candidate 4 is interviewed. Are A and E mutually exclusive? How about B and E, C and E, D and E?
Answer:
a) [tex]\Omega=\{1, 2, 31, 32, 41, 42, 341, 342, 431, 432\}[/tex]
b) A={1, 2}
c) B={341, 342, 431, 432}
d) C={31, 32, 341, 342, 431, 432}
e) D={1, 31, 41, 341, 431}
f) E={41, 42, 341, 342, 431, 432}
Step-by-step explanation:
We know that four candidates are to be interviewed for a job. Two of them, numbered 1 and 2, are qualified, and the other two, numbered 3 and 4, are not.
a) We get a set of all possible outcomes:
[tex]\Omega=\{1, 2, 31, 32, 41, 42, 341, 342, 431, 432\}[/tex]
b) Let A be the event that only one candidate is interviewed.
We get a set of all possible outcomes:
A={1, 2}
c) Let B be the event that three candidates are interviewed.
We get a set of all possible outcomes:
B={341, 342, 431, 432}
d) Let C be the event that candidate 3 is interviewed.
We get a set of all possible outcomes:
C={31, 32, 341, 342, 431, 432}
e) Let D be the event that candidate 2 is not interviewed.
We get a set of all possible outcomes:
D={1, 31, 41, 341, 431}
f) Let E be the event that candidate 4 is interviewed.
We get a set of all possible outcomes:
E={41, 42, 341, 342, 431, 432}
We conclude that the events A and E are mutually exclusive.
We conclude that the events B and E are not mutually exclusive.
We conclude that the events C and E are not mutually exclusive.
We conclude that the events D and E are not mutually exclusive.
A survey reveals that each customer spends an average of 35 minutes with a standard deviation of 10 minutes in a department store. Assuming the distribution is normal, what is the probability a customer spends less than 30 minutes in the department store?
Answer:
[tex]P(X<30)=P(\frac{X-\mu}{\sigma}<\frac{30-\mu}{\sigma})=P(Z<\frac{30-35}{10})=P(Z<-0.5)[/tex]
And we can find this probability using the normal standard table or excel and we got:
[tex]P(Z<-0.5)=0.309[/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".
Solution to the problem
Let X the random variable that represent the time spent for each customer of a population, and for this case we know the distribution for X is given by:
[tex]X \sim N(35,10)[/tex]
Where [tex]\mu=35[/tex] and [tex]\sigma=10[/tex]
We are interested on this probability
[tex]P(X<30)[/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<30)=P(\frac{X-\mu}{\sigma}<\frac{30-\mu}{\sigma})=P(Z<\frac{30-35}{10})=P(Z<-0.5)[/tex]
And we can find this probability using the normal standard table or excel and we got:
[tex]P(Z<-0.5)=0.309[/tex]
And the excel code for this case would be : "=NORM.DIST(-0.5,0,1,TRUE)"
To calculate the probability that a customer spends less than 30 minutes in the store, we apply the Z score formula (Z = (X - μ)/σ), resulting in a Z score of -0.5. Using a Z-table, we find the corresponding probability to be approximately 30.85%.
Explanation:This problem involves understanding the concept of a normal distribution, including the mean and standard deviation. The mean here is the average time customers spend in the store, which is 35 minutes, and the standard deviation is 10 minutes. The question is asking us to find the probability that a customer spends less than 30 minutes in the store.
We can calculate this using the concept of a Z score, which is given by the formula Z = (X - μ)/σ where X is the value we're interested in, μ is the mean, and σ is the standard deviation. Plugging in the values, we get Z = (30 - 35)/10 = -0.5. Looking this value up in a Z-table, we find that the probability a customer spends less than 30 minutes in the department store is approximately 0.3085, or 30.85%.
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A survey of 400 non-fatal accidents showed that 109 involved the use of a cell phone. Construct a 99% confidence interval for the proportion of fatal accidents that involved the use of a cell phone
Answer:
(0.215,0.33)
Step-by-step explanation:
The 99% confidence interval can be calculated as
[tex]p- z_{\frac{\alpha }{2} } \sqrt{\frac{pq}{n} } <P<p+z_{\frac{\alpha }{2} } \sqrt{\frac{pq}{n} }[/tex]
Where p is the estimated sample proportion that can be calculated as
p=x/n
where x=109 and n=400
p=109/400=0.2725
q=1-p=1-0.273=0.7275
[tex]z_{\frac{\alpha }{2} } =z_{\frac{\0.01 }{2} }=z_{0.005}=2.5758[/tex]
The 99% confidence interval is
[tex]0.2725-2.5758 \sqrt{\frac{0.2725(0.7275)}{400} } <P<0.2725+2.5758 \sqrt{\frac{0.2725(0.7275)}{400} }[/tex]
0.2725-2.5758(0.022262 )< P < 0.2725+2.5758(0.022262)
02725-0.057343 < P < 0.2725+0.057343
0.215157 < P < 0.329843
Rounding the obtained answer to three decimal places
0.215 < P < 0.33
Thus, the 99% confidence interval for the proportion of fatal accidents that involved the use of a cell phone is (0.215,0.33).
We are 99% confident that population the proportion of fatal accidents that involved the use of a cell phone will lie in this interval (0.215,0.33).
The time to fly between New York City and Chicago is uniformly distributed with a minimum of 120 minutes and a maximum of 150 minutes.
What is the probability that a flight is between 125 and 140 minutes?
A. 1.00.
B. 0.50.
C. 0.33.
D. 0.12.
E. 0.15
Answer:
B. 0.50.
Step-by-step explanation:
An uniform probability is a case of probability in which each outcome is equally as likely.
For this situation, we have a lower limit of the distribution that we call a and an upper limit that we call b.
The probability of a measure X being between two values c and d, in which d is larger than c, is given by the following formula:
[tex]P(c \leq X \leq d) = \frac{d - c}{b - a}[/tex]
Uniformly distributed with a minimum of 120 minutes and a maximum of 150 minutes.
This means that [tex]a = 120, b = 150[/tex]
What is the probability that a flight is between 125 and 140 minutes?
This is
[tex]P(125 \leq X \leq 140) = \frac{140 - 125}{150 - 120} = 0.5[/tex]
So the correct answer is:
B. 0.50.
What odds should a person give in favor of the following events? (a) A card chosen at random from a 52-card deck is an ace. (b) Two heads will turn up when a coin is tossed twice. (c) Boxcars (two sixes) will turn up when two dice are rolled
Answer:
(a) 7.69%
(b) 25%
(c) 2.78%
Step-by-step explanation:
(a)
In a deck of 52 cards there are 4 aces.
The odds in favor or the probability of selecting an ace is:
[tex]P(Ace) = \frac{Number\ of\ aces}{Number\ of\ cards\ in\ total}\\ =\frac{4}{52}\\ =0.076923\\\approx7.69\%[/tex]
Thus, the probability of selecting an ace from a random deck of 562 cards is 7.69%.
(b)
The outcomes of each toss of a coin is independent of the other, since the result of the previous toss does not affect the result of the current toss.
The probability that both the tosses will end up in heads is:
[tex]P(2\ Heads)=P(1^{st}\ Head)\times P(2^{nd}\ Head)\\=\frac{1}{2}\times \frac{1}{2}\\ =\frac{1}{4}\\ =0.25\ or\ 25\%\\[/tex]
Thus, the probability that both the tosses will end up in heads is 25%.
(c)
The sample space of two dice consists of 36 outcomes in total.
Out of these 36 outcomes there is only 1 Boxcar, i.e. two sixes.
The probability of a boxcar when two dice are rolled is:
[tex]P(Boxcar)=\frac{Favorable\ outcomes}{Total\ no.\ of\ outcomes}\\= \frac{1}{36}\\ =0.027777\\\approx2.78\%[/tex]
Thus, the probability of a boxcar when two dice are rolled is 2.78%.
Final answer:
To find the odds in favor of specific events in probability theory, one must compare the number of successful outcomes to the number of unsuccessful ones. For selecting an ace from a deck of cards, the odds are 1:12; for getting two heads from two coin tosses, the odds are 1:3; and for rolling two sixes with two dice, the odds are 1:35.
Explanation:
The question asks for the odds in favor of several different probabilistic events, which relate to the field of probability theory within mathematics. Here's how to calculate the odds for each of the requested scenarios:
(a) Odds in favor of a card being an ace: There are 4 aces in a standard 52-card deck. The odds in favor are the number of ways the event can occur (4 aces) to the number of ways the event can fail to occur (52 - 4 = 48 non-aces), which simplifies to 1:12.
(b) Odds in favor of two heads when a coin is tossed twice: The probability of getting a head on one coin toss is 1/2, and since the two tosses are independent, the probability of getting two heads is (1/2) * (1/2) = 1/4. The odds in favor are calculated by taking the probability of the event occurring (1 chance) against the probability of it not occurring (3 chances), which gives us odds of 1:3.
(c) Odds in favor of rolling boxcars (two sixes) with two dice: Each die has a 1/6 chance of rolling a six, so the probability of rolling two sixes is (1/6) * (1/6) = 1/36. The odds in favor are the number of successful outcomes (1) against the number of all other outcomes (35), resulting in odds of 1:35.
Simplify 25 ^1/2 using the radical form
To simplify 25 to the power of 1/2 using the radical form, we equate this to the square root of 25. Since 5 squared is 25, the square root of 25 is 5, hence 25^1/2 equals 5.
Explanation:To simplify 25 to the power of 1/2 using the radical form, we look at the properties of exponents and radicals. Exponentiation and radicals are inverse operations, so the expression 251/2 is equivalent to the square root of 25 because squaring a number and then taking the square root of it returns the original number.
So, 251/2 = √25. Since the square of 5 is 25, the square root of 25 is 5. Therefore, 251/2 = 5.
A line has a slope of -3/5. Which ordered pairs could be points on a parallel line? Select two options.
The question is missing the options. The options are:
(A) (-8, 8) and (2, 2)
(B) (-5, -1) and (0, 2)
(C) (-3, 6) and (6.-9)
(D) (-2, 1) and (3,-2)
(E) (0, 2) and (5,5)
Answer:
Options (A) and (D)
Step-by-step explanation:
Given:
A line with slope (m) = [tex]-\frac{3}{5}[/tex]
Now, a parallel line to the given line will have the same slope.
So, let us check each of the given options.
Option (A)
(-8, 8) and (2, 2)
The slope of line passing through two points [tex](x_1,y_1)\ and\ (x_2,y_2)[/tex] is given as:
[tex]m=\dfrac{y_2-y_1}{x_2-x_1}[/tex]
Now, the slope of a line passing through (-8, 8) and (2, 2) is given as:
[tex]m_1=\frac{2-8}{2-(-8)}=\frac{-6}{10}=-\frac{3}{5}[/tex]
So, [tex]m=m_1[/tex]
Therefore, option (A) is correct.
Option (B): (-5, -1) and (0, 2)
The slope of a line passing through (-5, -1) and (0, 2) is given as:
[tex]m_2=\frac{2-(-1)}{0-(-5)}=\frac{3}{5}[/tex]
So, [tex]m\ne m_2[/tex]
Therefore, option (B) is not correct.
Option (C): (-3, 6) and (6, -9)
The slope of a line passing through (-3, 6) and (6, -9) is given as:
[tex]m_3=\frac{-9-6}{6-(-3)}=\frac{-15}{9}=-\frac{5}{3}\ne m[/tex]
Therefore, option (C) is not correct.
Option (D): (-2, 1) and (3, -2)
The slope of a line passing through (-2, 1) and (3, -2) is given as:
[tex]m_4=\frac{-2-1}{3-(-2)}=-\frac{3}{5}=m[/tex]
Therefore, option (D) is correct.
Option (E): (0, 2) and (5, 5)
The slope of a line passing through (0, 2) and (5, 5) is given as:
[tex]m_5=\frac{5-2}{5-0}=\frac{3}{5}\ne m[/tex]
Therefore, option (E) is not correct.
Hence, only options (A) and (D) are correct.
You have 200 dice in a bag. One of the dice has a six on all sides so it will land on a six every time you roll it. The other 199 are normal dice with six sides, each with a different number. You randomly pick one of the dice from the bag and roll it three times. It lands on six all three times. What is the probability it is the die that always lands on six and what is the probability it is a normal die?
Answer:
Step-by-step explanation:
There are 200 dice out of which 199 are fair
Prob for 6 in one special die = 1 and
Prob for 6 in other die = 1/6
A1- drawing special die and A2 = drawing any other die
A1 and A2 are mutually exclusive and exhaustive
P(A1) = 1/200 and P(A2) = 199/200
B = getting 6
i) Required probability
= P(A1/B) = [tex]\frac{P(A1B)}{P(A1B)+P(A2B)} \\[/tex]
P(A1B) = [tex]\frac{1}{200} *1 = \frac{1}{200}[/tex]
P(A2B) = [tex]\frac{199}{200}*\frac{1}{6}=\frac{199}{1200}[/tex]
P(B) = [tex]\frac{205}{1200} =\frac{41}{240}[/tex]
P(A1/B) = [tex]\frac{1/200}{41/240} =\frac{7}{205}[/tex]
P(A2/B) = [tex]\frac{199/1200}{41/240} =\frac{199}{205}[/tex]
Same-sex unions increasingly become heated political issue. the 2006 GSS asked respondents opinions on homosexual relations. five response categories ranged from always wrong to not wrong at all. see the following frequency distributions. at what level is this variable measured? Homosexual Relations Frequency Percentage Cumulative Percentage Always wrong 467 50.2 50.2Almost always wrong 41 4.4 54.6 Sometimes wrong 76 8.2 62.8Not wrong at all 346 37.2 100.0 Total 930 100.0
Answer:
Ordinal level
Step-by-step explanation:
The variable of interest is opinion on homosexual relations and the frequency distribution for opinion on homosexual relations is given.
The opinion of people is categorized from wrong to not wrong at all. There exists order in the categorizes and measurement of variable indicates the ordinal level of measurement.
Thus, variable is measured at ordinal level.
Explain what a P-value is. What is the criterion for rejecting the null hypothesis using the P-value approach?
Answer:
P-value or Probability value is the exact percentage where test statistics lie.The criterion for rejecting the null hypothesis using the P-value approach is that if P-value < Level of significance , then we reject our null hypothesis
Step-by-step explanation:
P-value or Probability value is the exact percentage where test statistics lie.
It also tells the probability of obtaining extreme results corresponding to our level of significance keeping in state that our null hypothesis is true or correct.
The criterion for rejecting the null hypothesis using the P-value approach is that if P-value < Level of significance , then we reject our null hypothesis i.e.
Suppose P-value is 2.33% and Level of significance is 5%, then we will reject our null hypothesis as 2.33% < 5%.
On the other hand, if P-value > Level of significance , then we cannot reject or accept our null hypothesis.
A p-value is a measure of the strength of evidence against the null hypothesis in a statistical hypothesis test. The criterion for rejecting the null hypothesis using the p-value approach is to compare the calculated p-value to a predetermined significance level.
Explanation:A p-value is a measure of the strength of evidence against the null hypothesis in a statistical hypothesis test. It represents the probability of observing a test statistic as extreme or more extreme than the one observed, assuming that the null hypothesis is true. In other words, it quantifies how likely it is for the observed data to have occurred by chance if the null hypothesis is true.
The criterion for rejecting the null hypothesis using the p-value approach is to compare the calculated p-value to a predetermined significance level (usually denoted as alpha). If the p-value is smaller than alpha, we reject the null hypothesis, indicating that there is enough evidence to support the alternative hypothesis. Alternatively, if the p-value is greater than or equal to alpha, we fail to reject the null hypothesis, suggesting that there is not enough evidence to support the alternative hypothesis.
In 1990 the Department of Natural Resources released 1000 splake (a crossbreed of fish) into a lake. In 1997 the population of splake in the lake was estimated to be 3000. Using the Malthusian law for population growth, estimate the population of splake in the lake in the year 2020.
Using the Malthusian law for population growth and the given data, the estimated population of splake in a lake in 2020 is approximately 21,485.
Explanation:The question involves estimating the population of splake in a lake in 2020 using the Malthusian law for population growth. The Malthusian law indicates that populations grow exponentially under ideal conditions. Given that the population increased from 1000 to 3000 splake between 1990 and 1997, we can calculate the rate of growth and then apply this rate to predict the population in 2020.
To begin, we identify the years of growth as 1997 - 1990 = 7 years. The formula for exponential growth is P = P0ert, where P is the final population, P0 is the initial population, r is the rate of growth, and t is the time in years. With P = 3000, P0 = 1000, and t = 7, we can solve for r.
3000 = 1000e7r, which simplifies to 3 = e7r. Taking the natural logarithm of both sides gives us ln(3) = 7r, and solving for r gives r ~ 0.1487. Now, to find the population in 2020, which is 30 years from 1990, we use the formula with P0 = 1000, r = 0.1487, and t = 30: P = 1000e0.1487*30.
Upon calculation, the predicted population of splake in the year 2020 is approximately 21,485.
g If there are 52 cards in a deck with four suits (hearts, clubs, diamonds, and spades), how many ways can you select 5 diamonds and 3 clubs?
Answer:
The number of ways to select 5 diamonds and 3 clubs is 368,082.
Step-by-step explanation:
In a standard deck of 52 cards there are 4 suits each consisting of 13 cards.
Compute the probability of selecting 5 diamonds and 3 clubs as follows:
The number of ways of selecting 0 cards from 13 hearts is:
[tex]{13\choose 0}=\frac{13!}{0!\times(13-0)!} =\frac{13!}{13!}=1[/tex]
The number of ways of selecting 3 cards from 13 clubs is:
[tex]{13\choose 3}=\frac{13!}{3!\times(13-3)!} =\frac{13!}{13!\times10!}=286[/tex]
The number of ways of selecting 5 cards from 13 diamonds is:
[tex]{13\choose 5}=\frac{13!}{5!\times(13-5)!} =\frac{13!}{13!\times8!}=1287[/tex]
The number of ways of selecting 0 cards from 13 spades is:
[tex]{13\choose 0}=\frac{13!}{0!\times(13-0)!} =\frac{13!}{13!}=1[/tex]
Compute the number of ways to select 5 diamonds and 3 clubs as:
[tex]{13\choose0}\times{13\choose3}\times{13\choose5}\times{13\choose0} = 1\times286\times1287\times1=368082[/tex]
Thus, the number of ways to select 5 diamonds and 3 clubs is 368,082.
A stone is dropped into a lake, creating a circular ripple that travels outward at a speed of 35 cm/s.
Express the radius r of this circle as a function of the time t (in seconds).
Answer:
The radius r of this circle as a function of the time t :
[tex]r(t)=35\times t[/tex]
Step-by-step explanation:
Speed of the circular ripple = S = 35 cm/s
Radius of the ripple at time t = r
[tex]Speed=\frac{Distance}{Time} [/tex]
[tex]35cm/s=\frac{r}{t}[/tex]
[tex]r=35 cm/s \times t[/tex]
The radius r of this circle as a function of the time t :
[tex]r(t)=35\times t[/tex]
The radius r of this circle is a function of the time t (in seconds) is 35t.
Given that
A stone is dropped into a lake, creating a circular ripple that travels outward at a speed of 35 cm/s.
We have to determineThe radius r of this circle is a function of the time t (in seconds).
According to the questionSpeed of the circular ripple = S = 35 cm/s
Radius of the ripple at time t = r
Then
The radius r of this circle is a function of the time t (in seconds) is determined by the following formula;
[tex]\rm Speed = \dfrac{Distance}{Time}[/tex]
Substitute all the values in the formula;
[tex]\rm Speed = \dfrac{Distance}{Time}\\ \\ 35 = \dfrac{r}{t}\\ \\ r = 35 \times t\\ \\ r = 35t[/tex]
Hence, The radius r of this circle is a function of the time t (in seconds) is 35t.
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How many names and binary predicates would a language like the first need in order to say everything you can say in the second?
Answer:The same number of names and 4 predicates
Step-by-step explanation:
Answer:
The same number of names and 4 predicates
Step-by-step explanation:
Please help!!!!!!!!
Answer:
a) 62
b) 24
Step-by-step explanation:
For A, add the students who watched only one movie: 18+24+20=62
For B, look at how many students only watched Star Wars: 24
A box contains 11 two-inch screws, of which 4 have a Phillips head and 7 have a regular head. Suppose that you select 3 screws randomly from the box with replacement. Find the probability there will be more than one Phillips head screw.
Answer:
The probability that there will be more than one Phillips head screw = 0.1803 .
Step-by-step explanation:
We are given that there are 11 two-inch screws in a box of which 4 have a Phillips head and 7 have a regular head.
We are selecting 3 screws randomly from the box with replacement, so the probability that there will be more than one Phillips head screw is given by :
Probability of selecting two Phillips head screw.Probability of selecting three Phillips head screw.Now P(selecting 2 Phillips head screw with replacement) is given by :
Selecting 2 Phillip head screw = [tex]\frac{4}{11}[/tex] * [tex]\frac{4}{11}[/tex] = [tex]\frac{16}{121}[/tex]
P(selecting three Phillips head screw) = [tex]\frac{4}{11}[/tex] * [tex]\frac{4}{11}[/tex] * [tex]\frac{4}{11}[/tex] = [tex]\frac{64}{1331}[/tex]
Therefore, Probability that there will be more than one Phillips head screw
= [tex]\frac{16}{121}[/tex] + [tex]\frac{64}{1331}[/tex] = [tex]\frac{240}{1331}[/tex] = 0.1803 .
The diameter of the dot produced by a printer is normally distributed with a mean diameter of 0.002 inch and a standard deviation of 0.0004 inch. A. What is the probability that the diameter of a dot exceeds 0.0026 inch? B. What is the probability that a diameter is between 0.0014 and 0.0026? C. What standard deviation of diameters is needed so that the probability in part (b) is 0.995?
Answer:
(a) 0.06681
(b) 0.86638
(c) [tex]\sigma[/tex] = 0.000214
Step-by-step explanation:
We are given that the diameter of the dot produced by a printer is normally distributed with a mean diameter of 0.002 inch and a standard deviation of 0.0004 inch i.e.; [tex]\mu[/tex] = 0.002 inch and [tex]\sigma[/tex] = 0.0004
Also, Z = [tex]\frac{X -\mu}{\sigma}[/tex] ~ N(0,1)
(a) Let X = diameter of a dot
P(X > 0.0026 inch) = P( [tex]\frac{X -\mu}{\sigma}[/tex] > [tex]\frac{0.0026 -0.002}{0.0004}[/tex] ) = P(Z > 1.5) = 1 - P(Z <= 1.5)
= 1 - 0.93319 = 0.06681
(b) P(0.0014 < X < 0.0026) = P(X < 0.0026) - P(X <= 0.0014)
P(X < 0.0026) = P( [tex]\frac{X -\mu}{\sigma}[/tex] < [tex]\frac{0.0026 -0.002}{0.0004}[/tex] ) = P(Z < 1.5) = 0.93319
P(X <= 0.0014) = P( [tex]\frac{X -\mu}{\sigma}[/tex] <= [tex]\frac{0.0014 -0.002}{0.0004}[/tex] ) = P(Z <= -1.5) = 1 - P(Z <= 1.5)
= 1 - 0.93319 = 0.06681
Therefore, P(0.0014 < X < 0.0026) = 0.93319 - 0.06681 = 0.86638 .
(c) P(0.0014 < X < 0.0026) = 0.995
P( [tex]\frac{0.0014 -0.002}{\sigma}[/tex] < [tex]\frac{X -\mu}{\sigma}[/tex] < [tex]\frac{0.0026 -0.002}{\sigma}[/tex] ) = 0.995
P( [tex]\frac{ -0.0006}{\sigma}[/tex] < Z < [tex]\frac{0.0006}{\sigma}[/tex] ) = 0.995
P(Z < [tex]\frac{0.0006}{\sigma}[/tex] ) - P(Z <= [tex]\frac{-0.0006}{\sigma}[/tex] ) = 0.995
P(Z < [tex]\frac{0.0006}{\sigma}[/tex] ) - (1 - P(Z < [tex]\frac{0.0006}{\sigma}[/tex] ) ) = 0.995
2 * P(Z < [tex]\frac{0.0006}{\sigma}[/tex] ) - 1 = 0.995
P(Z < [tex]\frac{0.0006}{\sigma}[/tex] ) = 0.9975
On seeing the z table we observe that at critical value of x = 2.81 we get the probability area of 0.9975 i.e.;
[tex]\frac{0.0006}{\sigma}[/tex] = 2.81 ⇒ [tex]\sigma[/tex] = 0.000214
Therefore, 0.000214 standard deviation of diameters is needed so that the probability in part (b) is 0.995 .
The probabilities of obtaining a given diameter is found from the z-table,
given that the dot produced by the printer are normally distributed.
A. The probability that the diameter of a dot exceeds 0.0026 inch is 0.0668B. The probability that the diameter is between 0.0014 and 0.0026 inch is 0.8664C. The standard deviation needed for a probability of 0.995 is 2.135 × 10⁻⁴Reasons:
The mean diameter, μ = 0.002
The standard deviation, σ = 0.0004
A. The probability that the diameter exceeds 0.0026 inch
Solution;
[tex]\displaystyle z-score,\ Z= \mathbf{\dfrac{x-\mu }{\sigma }}[/tex]
At x = 0.0026 inch, we have;
[tex]\displaystyle Z=\dfrac{0.0026-0.002 }{0.0004 } = 1.5[/tex]
P(Z > 1.5) = 1 - P(Z < 1.5) = 1 - 0.9332 = 0.0668
The probability that the diameter of a dot exceeds 0.0026 inch = 0.0668
B. The probability that the diameter is less than 0.0026 inch = 0.9332
The z-score for a diameter of x = 0.0014 inch is given as follows;
[tex]\displaystyle Z=\mathbf{\dfrac{0.0014-0.002 }{0.0004 }} = -1.5[/tex]
P(Z < -1.5) = 0.0668
Therefore, the probability that the diameter is between 0.0014 and 0.0026 inch is given as follows;
P(0.0014 < x < 0.0026) = 0.9332 - 0.0668 = 0.8664
The probability that the diameter is between 0.0014 and 0.0026 = 0.8664
C. For the probability in part (b) to be 0.995, we have;
For a probability of 0.995, the z-score ≈ 2.575
[tex]\displaystyle P\left(Z < \dfrac{0.0026-0.002 }{\sigma } \right)- P\left(Z < \dfrac{0.0014-0.002 }{\sigma } \right)= 0.995[/tex]
Therefore;
[tex]\displaystyle \mathbf{ P\left(Z < \dfrac{0.0026-0.002 }{\sigma } \right)} = 0.995 + \frac{1 - 0.995}{2} = 0.9975[/tex]
From the z-table, we get;
[tex]\displaystyle P\left(Z < \dfrac{0.0026-0.002 }{\sigma } \right) = 0.9975[/tex]
The z-score with a probability of 0.9975 = 2.81
Which gives;
[tex]\displaystyle \left( \dfrac{0.0026-0.002 }{\sigma } \right) = 2.81[/tex]
[tex]\displaystyle \sigma = \left( \dfrac{0.0026-0.002 }{2.81} \right) = \mathbf{2.135 \times 10^{-4}}[/tex]
The standard deviation of the diameters needed so that the probability in part (b) is 0.995 is 2.135 × 10⁻⁴
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An adult male African elephant weighs about 9.07*10^3 kg. Compute how many times heavier an adult male blue whale is than an adult male African elephant(I.e., find the value of the ratio). Round your final answer to the nearest tenth.
Answer:
An adult male blue whale is 18.7 times heavier than an adult male African elephant.
Step-by-step explanation:
As the weight of an adult male African elephant weighs about
[tex]9.07\:\times\:10^3[/tex] kgAnd the weight of an adult blue whale is
[tex]1.7\:\times\:10^5[/tex] kgDetermining the ratio of adult male African elephant to the weight of an adult blue whale as:
[tex]\:\:\frac{1.7\:\times\:10^5}{9.07\:\times\:10^3}[/tex]
[tex]\mathrm{Apply\:exponent\:rule}:\quad \frac{x^a}{x^b}=x^{a-b}[/tex]
[tex]=\frac{10^{5-3}\times \:1.7}{9.07}[/tex]
[tex]=\frac{10^2\times \:1.7}{9.07}[/tex]
As [tex]10^2\times \:1.7=170[/tex], So
[tex]=\frac{170}{9.07}[/tex]
[tex]=18.74310\dots[/tex]
Round the answer to the nearest tenth
[tex]=18.7[/tex]
Therefore, an adult male blue whale is 18.7 times heavier than an adult male African elephant.
Keywords: ratio, nearest tenth
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Set up the integral that uses the method of disks/washers to find the volume V of the solid obtained by rotating the region bounded by the given curves about the specified lines.
y=3\sqrt(x), y = 3, x= 0
a.) about the line y = 3
b.) about the line x = 5
The integrals for the volume of the solids
a.) V = ∫[0 to 1] π(3 - 3√x)² dx
b.) V = ∫[0 to 27] π(5 - [tex]x^{2/3}[/tex]))² dx
We have,
To set up the integral using the method of disks/washers to find the volume V of the solid obtained by rotating the region bounded by the curves y = 3√x, y = 3, and x = 0 about the specified lines, follow these steps:
Given curves: y = 3√x, y = 3, x = 0
a.)
Rotating about the line y = 3:
- Draw the region bounded by the curves y = 3√x, y = 3, and x = 0.
- The solid will be formed by revolving this region around the line y = 3.
- For the method of disks/washers, consider a vertical slice (dx) of thickness dx at a distance x from the y-axis.
- The radius of the disk is the distance between the curve y = 3√x and the line y = 3, which is (3 - 3√x).
- The area of the disk is π(radius)^2 = π(3 - 3√x)².
- The volume of the infinitesimally thin disk is dV = π(3 - 3√x)² dx.
- Integrate the volume from x = 0 to x = (3/3)² = 1:
V = ∫[0 to 1] π(3 - 3√x)² dx
b)
Rotating about the line x = 5:
- Draw the region bounded by the curves y = 3√x, y = 3, and x = 0.
- The solid will be formed by revolving this region around the line x = 5.
- For the method of disks/washers, consider a vertical slice (dx) of thickness dx at a distance x from the y-axis.
- The radius of the disk is the distance between the line x = 5 and the curve y = 3√x, which is (5 - [tex]x^{2/3}[/tex]).
- The area of the disk is π(radius)^2 = π(5 - [tex]x^{2/3}[/tex])².
- The volume of the infinitesimally thin disk is dV = π(5 - [tex]x^{2/3}[/tex])² dx.
Integrate the volume from x = 0 to x = 3³ = 27:
V = ∫[0 to 27] π(5 - [tex]x^{2/3}[/tex])² dx
These integrals will give you the volumes of the solid obtained by rotating the region about the specified lines. You can evaluate these integrals to find the exact values of the volumes.
Thus,
The integrals for the volume of the solids
a.) V = ∫[0 to 1] π(3 - 3√x)² dx
b.) V = ∫[0 to 27] π(5 - [tex]x^{2/3}[/tex]))² dx
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The following histogram presents the amounts of silver (in parts per million) found in a sample of rocks. One rectangle from the histogram is missing. What is its height?
Answer:
The height of the missing rectangle is 0.15Explanation:
The image attached has the mentioned histogram.
Such histogram presents the relative frequencies for the clases [0,1], [1,2],[2,3], [4,5], and [5,6] Silver in ppm.
Only the rectangle for the class [3,4] is missing.
The height of each rectangle is the relative frequency of the corresponding class.
The relative frequencies must add 1, because each relative frequency is calculated dividing the absolute class frequency by the total number in the sample; hence, the sum of all the relative frequencies is equal to the total absolute class frequencies divided by the same number, yielding 1.
In consequence, you can sum all the known relative frequencies and subtract from 1 to get the missing relative frequency, which is the height of the missing rectangle.
1. Sum of the known relative frequencies:
0.2 + 0.3 + 0.15 + 0.1 + 0.1 = 0.852. Missing frequency:
1 - 0.85 = 0.153. Conclusion:
The height of the missing rectangle is 0.15b=2.35 + 0.25x
c=1.75+0.40x
In the equations above,b and c represent the price per pound,in dollars, of beef and chicken,respectively,x weeks after July 1 of last summer.What was the price per pound of beef when it was equal to the price per pound of chicken?
A.2.60
B.2.85
C.2.95
D.3.35
Answer:
D. 3.35
Step-by-step explanation:
First we need to form an equation and solve it to find the number of weeks when the prices were the same. Because the prices were the same we can say that b = c, and therefore form the equation:
2.35 + 0.25x = 1.75 + 0.4x - Now we nee to solve it and find x.
2.35 - 1.75 = 0.4x - 0.25x
0.6 = 0.15x
x = 0.6 ÷ 0.15
x = 4 weeks
So now we substitute x into the equation for beef and find the price.
b = 2.35 + (0.25 × 4)
b = 2.35 + 1
b = $3.35 per pound
The price per pound of beef when it was equal to the price of chicken is $3.35 per pound.
The prices of beef and chicken are represented by the following linear equations :
Price of beef, B :
B = 2.35 + 0.25x
Price of chicken, C :
C = 1.75 + 0.40x
Where x in both equations represents x weeks after July 1 of last summer :
Firstly :
We find the week in which the price of beef and chicken are the same :
Beef = Chicken
2.35 + 0.25x = 1.75 + 0.40x
We solve for x
2.35 - 1.75 = 0.40x - 0.25x
0.60 = 0.15x
x = 0.60 / 0.15
x = 4
Therefore, 4 weeks after July 1 of last summer, the price of beef and chicken were the same.
Therefore, the price per pound of beef in the 4th week is :
B = 2.35 + 0.25(4)
B = 2.35 + 1
B = 3.35
The price per pound of beef when it was equal to the price of chicken is $3.35 per pound.
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According to CNN business partner Careerbuilder, the average starting salary for accounting graduates in 2018 was at least $57,413. Suppose that the American Society for Certified Public Accountants planned to test this claim by randomly sampling 200 accountants who graduated in 2018. State the appropriate null and alternative hypotheses.
Answer:
Null hypothesis: The American Society for Certified Public Accountants says the average starting salary of accountants who graduated in 2018 is $57,413
Alternate hypothesis: The American Society for Certified Public Accountants says the average starting salary of accountants who graduated in 2018 is less than or equal to $57,413
Step-by-step explanation:
A null hypothesis is a statement from a population parameter that is subject to testing. It is expressed with equality.
An alternate hypothesis is also a statement from the population parameter that negates the null hypothesis. It is expressed with inequality
A campus deli serves 300 customers over its busy lunch period from 11:30 a.m. to 1:30 p.m. A quick count of the number of customers waiting in line and being served by the sandwich makers shows that an average of 10 customers are in process at any point in time. What is the average amount of time that a customer spends in process?
Answer:
4 minutes
Step-by-step explanation:
There are two hours from 11:30 a.m. to 1:30 p.m
The hourly rate of service is:
[tex]r=\frac{300}{2}=150\ customers/hour[/tex]
If the average number of customers in the system (n) is 10, the time that a customer spends in process is given by:
[tex]t=\frac{n}{r} =\frac{10}{150}=0.06667\ hours[/tex]
Converting it to minutes:
[tex]t= 0.066667\ hours*\frac{60\ minutes}{1\ hour}\\t=4\ minutes[/tex]
A customer spends, on average, 4 minutes in process.
The amount of corn chips dispensed into a 10-ounce bag by the dispensing machine has been identified at possessing a normal distribution with a mean of 10.5 ounces and a standard deviation of 0.2 ounces (these are the population parameters). Suppose a sample of 100 bags of chips were randomly selected from this dispensing machine. Find the probability that the sample mean weight of these 100 bags is less than 10.45 ounces. (Hint: think of this in terms of a sampling distribution with sample size
Answer:
0.62% probability that the sample mean weight of these 100 bags is less than 10.45 ounces.
Step-by-step explanation:
To solve this question, the concepts of the normal probability distribution and the central limit theorem are important.
Normal probability distribution
Problems of normally distributed samples can be solved using the z-score formula.
In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore of a measure X is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.
Central Limit Theorem
The Central Limit Theorem estabilishes that, for a random variable X, with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], a large sample size can be approximated to a normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]\frac{\sigma}{\sqrt{n}}[/tex]
In this problem, we have that:
[tex]\mu = 10.5, \sigma = 0.2, n = 100, s = \frac{0.2}{\sqrt{100}} = 0.02[/tex]
Find the probability that the sample mean weight of these 100 bags is less than 10.45 ounces
This is the pvalue of Z when X = 10.45. So
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{10.45 - 10.5}{0.02}[/tex]
[tex]Z = -2.5[/tex]
[tex]Z = -2.5[/tex] has a pvalue of 0.0062.
So there is a 0.62% probability that the sample mean weight of these 100 bags is less than 10.45 ounces.
The probability of the sample mean weight being less than 10.45 ounces can be found by calculating the Z-score and referencing a standard normal distribution table. The calculated Z-score (-2.5) corresponds to a probability of approximately 0.62%.
Explanation:The problem is about determining the probability that the sample mean weight of corn chip bags is less than 10.45 ounces. This is a problem of finding a probability in a sampling distribution when the population parameters are known. Given the data, we can use the Central Limit Theorem, which states that if the sample size is large enough (usually >30), the sampling distribution approximates a normal distribution.
To solve this, you can use the formula Z = (X - μ) / (σ/√n), where X is the sample mean, μ is the population mean, σ is the population standard deviation, and n is the sample size.
Plugging in the given values: Z = (10.45 - 10.5) / (0.2 / √100) = -2.5. The Z-score tells us how many standard deviations away our data point is from the mean. To find the probability that the Z is less than -2.5, you can refer to a standard normal distribution table or use statistical software. According to the Z table, the probability is approximately 0.0062 or 0.62% that the sample mean weight of these 100 bags is less than 10.45 ounces.
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