Answer:
a) [tex] \bar X =117.8[/tex]
[tex] Median= \frac{117+118}{2}=117.5[/tex]
The mode on this case is the most repeated value 128 with a frequency of 3
b) [tex] Range = Max -Min = 150-88=62[/tex]
[tex] s^2 = 225.334[/tex]
[tex] s= \sqrt{225.334}= 15.011[/tex]
c) [tex] y \pm s[/tex]
[tex] Lower = 117.8 -15.011=102.809[/tex]
[tex] Upper = 117.8 +15.011=132.831[/tex]
[tex] y \pm 2s[/tex]
[tex] Lower = 117.8 -2*15.011=87.797[/tex]
[tex] Upper = 117.8 +2*15.011=147.842[/tex]
[tex] y \pm 3s[/tex]
[tex] Lower = 117.8 -3*15.011=72.787[/tex]
[tex] Upper = 117.8 +3*15.011=162.85[/tex]
d) For this case we can calculate the position where we have accumulated 70% of the data below.
50*0.7 = 35
So on the position 35th from the dataset ordered we see that the value is 128 and this value would represent the 70th percentile on this case.
Step-by-step explanation:
For this case we consider the following data:
128,119,95,97,124,128,142,98,108,120,113,109,124,132,97,138,133,136,120,112,146,128,103,135,114,109,100,111,131,113,124,131,133,131,88,118,116,98,112,138,100,112,111,150,117,122,97,116,92,122
Part a
For this case we can calculate the mean with the following formula:
[tex] \bar X = \frac{\sum_{i=1}^{50} X_i}{50}[/tex]
And after replace we got [tex] \bar X =117.8[/tex]
In order to calculate the median first we order the dataset and we got:
88 92 95 97 97 97 98 98 100 100 103 108 109 109 111 111 112 112 112 113 113 114 116 116 117 118 119 120 120 122 122 124 124 124 128 128 128 131 131 131 132 133 133 135 136 138 138 142 146 150
The median would be the average between the position 25 and 26 from the data ordered and we got:
[tex] Median= \frac{117+118}{2}=117.5[/tex]
The mode on this case is the most repeated value 128 with a frequency of 3
Part b
the range is defined as the difference between the maximun and minimum so we got:
[tex] Range = Max -Min = 150-88=62[/tex]
The sample variance can be calculated with this formula:
[tex] s^2 = \frac{\sum_{i=1}^n (X_i -\bar X)^2}{n-1}[/tex]
And after calculate we got: [tex] s^2 = 225.334[/tex]
And the deviation is just the square root of the variance and we got:
[tex] s= \sqrt{225.334}= 15.011[/tex]
Part c
For this case we can construct the interval with 1 , 2 and 3 deviation from the mean like this:
[tex] y \pm s[/tex]
[tex] Lower = 117.8 -15.011=102.809[/tex]
[tex] Upper = 117.8 +15.011=132.831[/tex]
[tex] y \pm 2s[/tex]
[tex] Lower = 117.8 -2*15.011=87.797[/tex]
[tex] Upper = 117.8 +2*15.011=147.842[/tex]
[tex] y \pm 3s[/tex]
[tex] Lower = 117.8 -3*15.011=72.787[/tex]
[tex] Upper = 117.8 +3*15.011=162.85[/tex]
Part d
For this case we can calculate the position where we have accumulated 70% of the data below.
50*0.7 = 35
So on the position 35th from the dataset ordered we see that the value is 128 and this value would represent the 70th percentile on this case.
Two planes left the airport traveling in the same direction. The distance Plane A traveled is modeled by the function d(t)=290t where d represents distance in miles and t represents time in hours. Plane B traveled a total of 540 miles in 2 hours. How does the distance Plane A traveled in 1 hour compare to the distance Plane B traveled in 1 hour? The distance Plane A traveled in 1 h is greater than the distance Plane B traveled in 1 h. The distance Plane A traveled in 1 h is less than the distance Plane B traveled in 1 h. The distance Plane A traveled in 1 h is equal to the distance Plane B traveled in 1 h.
Answer:
The distance Plane A traveled in 1 h is greater than the distance Plane B traveled in 1 h.
Step-by-step explanation:
The equation of the distance traveled by Plane A is
[tex]d(t) = 290t[/tex]
The plane B traveled 540 miles in 2 hours.
So in 1 hour, plane B traveled 540/2 = 270 miles:
How does the distance Plane A traveled in 1 hour compare to the distance Plane B traveled in 1 hour?
Plane A:
d(1) = 290*1 = 290
Plane A traveled 290 miles in 1 hour.
Plane B travaled 270 miles in 1 hour.
So the correct answer is:
The distance Plane A traveled in 1 h is greater than the distance Plane B traveled in 1 h.
Answer:
The distance Plane A traveled in 1 h is greater than the distance Plane B traveled in 1 h.
Step-by-step explanation:
The Bay of Fundy in Canada has the largest tides in the world. The difference between low and high water levels is 20 meters. At a particular point the depth of the water, y meters, is given as a function of time, t, in hours since midnight by y = D + A cos(B(t ? C)).
a) What is the value of B? Assume the time between successive high tides is 12.7 hours. Give an exact answer.
b) What is the physical meaning of C?
The value of B is determined by the equation 2π / 12.7, which corresponds to the tide's period. The variable C represents the time delay from midnight to the first high tide, which is a phase shift in the function.
Explanation:The Bay of Fundy tidal pattern can be modeled using a cosine function. Since tides go through a complete cycle (360 degrees or 2π radians) every 12.7 hours, the value of B, the frequency, can be determined by dividing 2π by the period of the tide in hours.
Therefore, B = 2π / 12.7.
The variable C in the equation represents a phase shift. In this context, a phase shift refers to a horizontal shift of the cosine function, which corresponds to a time delay or advance of the tides. The meaning of C is the time delay between midnight and the first high tide of the day.
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You are about to take the road test for your driver's license. You hear that only 38% of candidates pass the test the first time, but the percentage rises to 76% on subsequent requests. Estimate the average number of tests drivers take in order to get a license. Your simulation should use at least 20 runs.
Answer:
Answers may vary but will most likely be close to 2.
Step-by-step explanation
Given:first test:38%
second test:76%
SIMULATION FIRST TEST
Randomly select a 2-digit number.
If the digit is between 00 and 35 then you passed the test,else you did not pass the test.
SIMULATION SECOND TEST
Randomly select a 2-digit number.
if the digit is between 00 and 75 then you passed the test,else you did not pass the test.
SIMULATION TRIAL
Perform the simulation of the first test.if you did not pass the first test then perform the simulation of the second test.
Record the number of trials needed to pass the first or second test.
Repeat 20 times and take the average of the 20 recorded number of trials
(what is the sum of recorded values divided by 20).
Note:you will most likely obtain a result of about two trials needed.
The estimated average number of tests drivers take to get a license is 1.9
To estimate the average number of tests drivers take to get a license, we can simulate the process using a simple random experiment. We will assume that each test attempt is independent of the others and that the probabilities of passing are 38% on the first attempt and 76% on subsequent attempts. Here's how we can perform the simulation:
1. For each driver, simulate the number of tests they take by generating a random number between 0 and 1. If the number is less than or equal to 0.38 (38%), the driver passes on the first attempt. Otherwise, they proceed to a second attempt.
2. For the second attempt and any subsequent attempts, generate another random number. If the number is less than or equal to 0.76 (76%), the driver passes.
3. Repeat this process until the driver passes the test. Record the number of attempts it took.
4. Perform this simulation for 20 drivers to get a sample size that will help estimate the average number of tests taken.
5. Calculate the average number of tests by summing the number of tests taken by all drivers and dividing by the number of drivers (20).
Let's perform the simulation:
1. For the first attempt, generate a random number for each of the 20 drivers and check if it's less than or equal to 0.38.
2. For any subsequent attempts, generate a random number for each driver who did not pass on the first attempt and check if it's less than or equal to 0.76.
3. Record the number of attempts for each driver.
4. Sum the total number of attempts and divide by 20 to find the average.
Now, let's calculate the average number of tests based on the simulation: Assuming we have performed the simulation and recorded the number of attempts for each driver, we would have a list of numbers. For example, it might look something like this:
Driver 1: 1 attempt
Driver 2: 2 attempts
Driver 3: 1 attempt
Driver 20: 3 attempts
Let's say after performing the simulation, we have the following counts for each number of attempts:
1 attempt: 8 drivers
2 attempts: 7 drivers
3 attempts: 4 drivers
4 attempts: 1 driver
Now, we calculate the average number of tests taken by these 20 drivers:
Average = (8 * 1 + 7 * 2 + 4 * 3 + 1 * 4) / 20
Average = (8 + 14 + 12 + 4) / 20
Average = 38 / 20
Average = 1.9
The original price of a toy boat was $50. The boat is marked up 15% before it’s sold. What is the selling price of the boat?
Answer: the selling price of the boat is $57.5
Step-by-step explanation:
The original price of a toy boat was $50. The boat is marked up 15% before it’s sold. This means that the amount by which the original price of the toy boat was increased would be
15/100 × 50 = 0.15 × 50 = $7.5
The selling price of the toy boat would be the sum of its original price and the amount by which it was marked up. It becomes.
50 + 7.5 = $57.5
An apple juice producer buys all his apples from a conglomerate of apple growers in one northwestern state. The amount of juice obtained from each of these apples is approximately normally distributed with a mean of 2.25 ounces and a standard deviation of 0.15 ounce. Between what two values (in ounces) symmetrically distributed around the population mean will 80 percent of the apples fall?
A. [2.13, 2.37]
B. [2.10, 2.40]
C. [2.06, 2.44]
D. [1.95, 2.55]
Answer:
[tex]z=-1.28<\frac{a-2.25}{0.15}[/tex]
And if we solve for a we got
[tex]a=2.25 -1.28*0.15=2.058[/tex]
So the value of height that separates the bottom 10% of data from the top 90% is 2.06.
For the upper limit since the distribution is symmetrical we can do this:
[tex]z=1.28<\frac{a-2.25}{0.15}[/tex]
And if we solve for a we got
[tex]a=2.25 +1.28*0.15=2.442[/tex]
So the value of height that separates the bottom 90% of data from the top 10% is 2.44.
And the best answer for this case would be:
C. [2.06, 2.44]
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 amount of juice of a population, and for this case we know the distribution for X is given by:
[tex]X \sim N(2.25,0.15)[/tex]
Where [tex]\mu=2.25[/tex] and [tex]\sigma=0.15[/tex]
For this case we want the limits for the middle 80% values of the distribution. so then we need 100-80= 20% of the area in the tails and 10% on each tail since the distribution is symmetrical.
We can use this condition for the lower limits
[tex]P(X>a)=0.9[/tex] (a)
[tex]P(X<a)=0.1[/tex] (b)
Both conditions are equivalent on this case. We can use the z score again in order to find the value a.
As we can see on the figure attached the z value that satisfy the condition with 0.1 of the area on the left and 0.9 of the area on the right it's z=-1.28. On this case P(Z<-1.28)=0.1 and P(z>-1.28)=0.9
If we use condition (b) from previous we have this:
[tex]P(X<a)=P(\frac{X-\mu}{\sigma}<\frac{a-\mu}{\sigma})=0.1[/tex]
[tex]P(z<\frac{a-\mu}{\sigma})=0.1[/tex]
But we know which value of z satisfy the previous equation so then we can do this:
[tex]z=-1.28<\frac{a-2.25}{0.15}[/tex]
And if we solve for a we got
[tex]a=2.25 -1.28*0.15=2.058[/tex]
So the value of height that separates the bottom 10% of data from the top 90% is 2.06.
For the upper limit since the distribution is symmetrical we can do this:
[tex]z=1.28<\frac{a-2.25}{0.15}[/tex]
And if we solve for a we got
[tex]a=2.25 +1.28*0.15=2.442[/tex]
So the value of height that separates the bottom 90% of data from the top 10% is 2.44.
And the best answer for this case would be:
C. [2.06, 2.44]
student records suggest that the population of students spends an average of 6.30 hours per week playing organized sports. The population's standard deviation is 2.10 hours per week. Based on a sample of 49 students, Healthy Lifestyles Incorporated (HLI) would like to apply the central limit theorem to make various estimates.
A) What is the chance HLI will find a sample mean between 5.5 and 7.1 hours?
B) Calculate the probability that the sample mean will be between 5.9 and 6.7 hours.
Answer:
a) 99.24% chance HLI will find a sample mean between 5.5 and 7.1 hours.
b) 81.64% probability that the sample mean will be between 5.9 and 6.7 hours.
Step-by-step explanation:
To solve this question, it is important to know the Normal probability distribution and the Central Limit Theorem
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 = 6.3, \sigma = 2.1, n = 49, s = \frac{2.1}{\sqrt{49}} = 0.3[/tex]
A) What is the chance HLI will find a sample mean between 5.5 and 7.1 hours?
This is the pvalue of Z when X = 7.1 subtracted by the pvalue of Z when X = 5.5.
By the Central Limit Theorem, the formula for Z is:
[tex]Z = \frac{X - \mu}{s}[/tex]
X = 7.1
[tex]Z = \frac{7.1 - 6.3}{0.3}[/tex]
[tex]Z = 2.67[/tex]
[tex]Z = 2.67[/tex] has a pvalue of 0.9962
X = 5.5
[tex]Z = \frac{5.5 - 6.3}{0.3}[/tex]
[tex]Z = -2.67[/tex]
[tex]Z = -2.67[/tex] has a pvalue of 0.0038
So there is a 0.9962 - 0.0038 = 0.9924 = 99.24% chance HLI will find a sample mean between 5.5 and 7.1 hours.
B) Calculate the probability that the sample mean will be between 5.9 and 6.7 hours.
This is the pvalue of Z when X = 6.7 subtracted by the pvalue of Z when X = 5.9
X = 6.7
[tex]Z = \frac{6.7 - 6.3}{0.3}[/tex]
[tex]Z = 1.33[/tex]
[tex]Z = 1.33[/tex] has a pvalue of 0.9082
X = 5.9
[tex]Z = \frac{5.9 - 6.3}{0.3}[/tex]
[tex]Z = -1.33[/tex]
[tex]Z = -1.33[/tex] has a pvalue of 0.0918.
So there is a 0.9082 - 0.0918 = 0.8164 = 81.64% probability that the sample mean will be between 5.9 and 6.7 hours.
One percent of a certain model of television have defective speakers. Suppose 500 televisions of this model are ready to ship. Find an approximate probability that 5 to 8 televisions (inclusive) in the shipment have defective speakers (round off to second decimal place).
Answer:
49.34% probability that 5 to 8 televisions (inclusive) in the shipment have defective speakers
Step-by-step explanation:
For each television, there are only two possible outcomes. Either they have defective speakers, or they do not. The probabilities of each television having defective speakers are independent from each other. 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.
In this problem we have that:
[tex]p = 0.01, n = 500[/tex]
Find an approximate probability that 5 to 8 televisions (inclusive) in the shipment have defective speakers
This is
[tex]P(5 \leq X \leq 8) = P(X = 5) + P(X = 6) + P(X = 7) + P(X = 8)[/tex]
So
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
[tex]P(X = 5) = C_{500,5}.(0.01)^{5}.(0.99)^{495} = 0.1764[/tex]
[tex]P(X = 6) = C_{500,6}.(0.01)^{6}.(0.99)^{494} = 0.1470[/tex]
[tex]P(X = 7) = C_{500,7}.(0.01)^{7}.(0.99)^{493} = 0.1048[/tex]
[tex]P(X = 8) = C_{500,8}.(0.01)^{8}.(0.99)^{492} = 0.0652[/tex]
[tex]P(5 \leq X \leq 8) = P(X = 5) + P(X = 6) + P(X = 7) + P(X = 8) = 0.1764 + 0.1470 + 0.1048 + 0.0652 = 0.4934[/tex]
49.34% probability that 5 to 8 televisions (inclusive) in the shipment have defective speakers
The approximate probability that 5 to 8 out of 500 televisions have defective speakers is 0.49. This was calculated using the binomial distribution with n = 500 and p = 0.01. We summed P(X = 5), P(X = 6), P(X = 7), and P(X = 8) to find the result.
To find the approximate probability that 5 to 8 televisions out of 500 have defective speakers, we can use the binomial distribution.
Let X be the random variable representing the number of defective televisions in the shipment. Since 1% of the televisions are defective, the probability of a television being defective is p = 0.01, and the sample size is n = 500.The probability mass function for a binomial distribution is given by:
[tex]P(X = k) = C(n, k) * p^k * (1 - p)^(n - k)[/tex]
where C(n, k) is the binomial coefficient.
We need to calculate P(5 ≤ X ≤ 8), which is P(X = 5) + P(X = 6) + P(X = 7) + P(X = 8).
Using a binomial calculator or statistical software, we get:
P(X = 5) ≈ 0.1755P(X = 6) ≈ 0.1447P(X = 7) ≈ 0.1040P(X = 8) ≈ 0.0681Summing these probabilities gives:
P(5 ≤ X ≤ 8) ≈ 0.1755 + 0.1447 + 0.1040 + 0.0681
≈ 0.4923
Therefore, the approximate probability that 5 to 8 televisions have defective speakers is 0.49 (rounded to two decimal places)
Solve the following systems of equations using the matrix method: a. 3x1 + 2x2 + 4x3 = 5 2x1 + 5x2 + 3x3 = 17 7x1 + 2x2 + 2x3 = 11 b. x − y − z = 0 30x + 40y = 12 30x + 50z = 12 c. 4x1 + 2x2 + x3 + 5x4 = 0 3x1 + x2 + 4x3 + 7x4 = 1 2x1 + 3x2 + x3 + 6x4 = 1 3x1 + x2 + x3 + 3x4 = 4
Answer:
(a) x1 = 11/13, x2 = 50/13, x3 = -17/13
(b) x = 54/235, y = 6/47, z = 24/235
(c) x1 = 22/9, x2 =164/9, x3 = 139/9, x4 = -37/3
Step-by-step explanation:
Gaussian Elimination Method was the matrix method used in solving the system of equations.
It is done by writing the equations given in an augmented form, this is shown in the attachment. The coefficients of each variable is taken to form a matrix.
Row operations are then performed on the augmented matrix. This operation can be addition, subtraction, multiplication, or division.
For convenience, Row is written as R1, Row 2 as R2, and so on
R2 - R3 means Subtract Row 3 from Row 2, and so on.
The step by step operations for each question are shown in the attachment.
The matrix method involves representing the systems of equations as matrices, and then using matrix operations or the inverse matrix method to solve for the variables. This method can only be used when the system has a unique solution.
Explanation:Using the matrix method to solve systems of equations involves first representing the system as a matrix. For example, the first system of equations can be represented as a matrix: [[3,2,4], [2,5,3], [7,2,2]][[x1],[x2],[x3]] = [[5], [17], [11]]. Similarly, the second and third systems can be written in matrix form. Then you can use various matrix operations or the inverse matrix method to solve for the variables. Note that this method is used only when the system has a unique solution - that is, when the coefficient matrix is invertible.
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Employment data at a large company reveal that 59 % of the workers are married, that 43 % are college graduates, and that 1/3 of the college graduates are married. What is the probability that a randomly chosen worker is: a) neither married nor a college graduate? Answer = % b) married but not a college graduate? Answer = % c) married or a college graduate?
Answer:
a. 74.63%
b. 33.63%
c. 87.67%
Step-by-step explanation:
If 59% (0.59) of the workers are married, then It means (100-59 = 41%) of the workers are not married.
If 43% (0.43) of the workers are College graduates, then it means (100-43= 57%) of the workers are not college graduates.
If 1/3 of college graduates are married, it means portion of graduate that are married = 1/3 * 43% = 1/3 * 0.43 = 0.1433.
For question a, Probability that the worker is neither married nor a college graduate becomes:
= (probability of not married) + (probability of not a graduate) - (probability of not married * not a graduate)
= 0.41 + 0.57 - (0.41*0.57) = 0.98 - 0.2337
= 0.7463 = 74.63%
For question b, probability that the worker is married but not a college graduate becomes:
=(probability of married) * (probability of not a graduate.)
= 0.59 * 0.57
= 0.3363 = 33.63%
For question c, probability that the worker is either married or a college graduate becomes:
=probability of marriage + probability of graduate - (probability of married and graduate)
= 0.59 + 0.43 - (0.1433)
= 0.8767. = 87.67%
The probability that a randomly chosen worker is neither married nor a college graduate is 0%. The probability that a worker is married but not a college graduate is approximately 34%. The probability that a worker is married or a college graduate is approximately 75%.
Explanation:To calculate the probabilities, we need to use the formulas for conditional probability. Let's solve each part:
a) To find the probability that a randomly chosen worker is neither married nor a college graduate, we can subtract the probability that the worker is married and the probability that the worker is a college graduate from 1. The probability of being married is 59%, and the probability of being a college graduate is 43%. Using the formula, 1 - 0.59 - 0.43 = 0.
b) To find the probability that a worker is married but not a college graduate, we can multiply the probability of being married and the probability of not being a college graduate. The probability of being married is 59%, and the probability of not being a college graduate is 57% (100% - 43%). Using the formula, 0.59 * 0.57 = 0.3363 (approximately 34%).
c) To find the probability that a worker is married or a college graduate, we can add the probabilities of being married and being a college graduate, and then subtract the probability of being both married and a college graduate to avoid double counting. The probability of being married is 59%, the probability of being a college graduate is 43%, and the probability of being both married and a college graduate is 1/3 of the college graduates (1/3 * 43%). Using the formula, 0.59 + 0.43 - (1/3 * 0.43) = 0.745 (approximately 75%).
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The median weight of a boy whose age is between 0 and 36 months can be approximated by the function w (t )equals 8.38 plus 1.51 t minus 0.0069 t squared plus 0.000254 t cubed, where t is measured in months and w is measured in pounds. Use this approximation to find the following for a boy with median weight in parts a) through c) below. a) The rate of change of weight with respect to time.
Answer:
a) W'(t) = 1.51 -0.0138t + 0.000762t² pounds/day
Step-by-step explanation:
The median weight of a boy with age between 0 and 36 months is given by:
[tex]W(t) = 8.38+1.51t-0.0069t^2 + 0.000254t^3[/tex]
To find the rate of change of weight with respect to time, that is, the change in weight measured in pounds caused by a unit change in time, measured in days, simply derivate the weight function over time:
[tex]\frac{d(W(t)}{dt}= W'(t) = 1.51-0.0138t + 0.000762t^2[/tex]
The rate of change is 1.51 -0.0138t + 0.000762t² pounds/day.
Slow response times by paramedics, firefighters, and policemen can have serious consequences for accident victims. In the case of life-threatening injuries, victims generally need medical attention within 8 minutes of the accident. Several cities have begun to monitor emergency response times. In one such city, the mean response time to all accidents involving life-threatening injuries last year was m = 6.7 minutes. Emergency personnel arrived within 8 minutes on 78% of all calls involving life-threatening injuries last year. The city manager shares this information and encourages these first responders to do better. At the end of the year, the city manager selects an SRS of 400 calls involving life-threatening injuries and examines the response times. Awful accidents (a) State hypotheses for a significance test to determine whether first responders are arriving within 8 minutes of the call more often. Be sure to define the parameter of interest. (b) Describe a Type I error and a Type II error in this setting and explain the consequences of each. (c) Which is more serious in this setting: a Type I error or a Type II error? Justify your answer.
Answer:
a)
[tex]H_0: \pi\geq0.78\\\\H_a: \pi<0.78[/tex]
b) The Type I error occurs when we reject a null hypothesis that is actually true. In this case, it means we conclude that the arrival time have improved, when it didn't.
The Type II error occurs when we accept a null hypothesis that is actually false. In this case, although the arrival times have really improved, the evidence from the sample was not enough to show that improvement.
c) In this case, the Type I error is more serious, because it gives the wrong impression of improvement and no further actions will be taken to reduce the times.
Step-by-step explanation:
a) If you want to determine if the responders are arriving within 8 minutes of the call more often, you have to evaluate the proportion of accidents in which the arrival time is less than 8 minutes and compare it with the known proportion of π=0.78.
The sample parameter "p: proportion of accidents with arrival time of 8 minutes or less" will be used to test the hypothesis.
The null and alternative hypothesis will be:
[tex]H_0: \pi\geq0.78\\\\H_a: \pi<0.78[/tex]
Final answer:
The hypotheses for a significance test to determine whether first responders are arriving more often within 8 minutes are H0: p = 0.78 and HA: p > 0.78, with p representing the proportion of calls responded to within this timeframe. A Type I error involves mistakenly concluding an improvement, while a Type II error occurs by overlooking an actual improvement. The potential demotivating effect of a Type II error may render it more serious in this context.
Explanation:
The question seems to revolve around the concept of hypothesis testing in statistics and how it applies to the emergency response times in a city. The parameter of interest would be the proportion of emergency calls responded to within 8 minutes. So for part (a), the hypotheses could be stated as follows:
H0: p = 0.78 (The proportion of calls responded to within 8 minutes is 78% as it was last year.)
HA: p > 0.78 (The proportion of calls responded to within 8 minutes has increased from last year.)
In part (b), a Type I error would occur if the city concludes that the proportion of calls responded to within 8 minutes has increased when in reality, it has not. The consequence of a Type I error would be misallocating resources based on false success. A Type II error would occur if the city fails to recognize an actual improvement in response times. The consequence of this could lead to a lack of recognition and continued encouragement for first responders who have actually improved.
Part (c) asks which error is more serious. A Type II error may be considered more serious in this setting, as failing to acknowledge and react to an actual improvement could demotivate emergency personnel and affect future performances, possibly leading to life-threatening delays for accident victims.
Suppose each of 12 players rolls a pair of dice 3 times. Find the probability that at least 4 of the players will roll doubles at least once. (Answer correct to four decimal places.)
Answer:
Our answer is 0.8172
Step-by-step explanation:
P(doubles on a single roll of pair of dice) =(6/36) =1/6
therefore P(in 3 rolls of pair of dice at least one doubles)=1-P(none of roll shows a double)
=1-(1-1/6)3 =91/216
for 12 players this follows binomial distribution with parameter n=12 and p=91/216
probability that at least 4 of the players will get “doubles” at least once =P(X>=4)
=1-(P(X<=3)
=1-((₁₂ C0)×(91/216)⁰(125/216)¹²+(₁₂ C1)×(91/216)¹(125/216)¹¹+(₁₂ C2)×(91/216)²(125/216)¹⁰+(₁₂ C3)×(91/216)³(125/216)⁹)
=1-0.1828
=0.8172
What is the strength of an electric field that will balance the weight of a 9.0 gg plastic sphere that has been charged to -1.6 nCnC ? Express your answer to two significant figures and include the appropriate units.
Answer: The strength of an electric field is E = - 0,05.10⁹ N/C.
Step-by-step explanation: According to the question, the plastic sphere is in equilibrium in an electric field. This sugests that the forces acting on the sphere, which are Gravitational Force (Fg) and Electric Force (Fe) are also in equilibrium, denotating Fg=Fe.
As Fg = m . g, with m = 0,009kg and g= 9,8m/s², we have Fg = 0,0882N.
Knowing the value of Fe, the strength of the electric field can be calculated as
E = Fe/Q, in which Q is the electric charge.
E = (0,0882) / (-1,6·10⁻⁹)
E = - 0,05·10⁹N/C
John works at a restaurant and gets paid $80 per week plus $5 per tip. This week, John wants to earn at least $300. How many tips, x, must he make to reach his goal?
Answer: John must make at least 44 tips to reach his goal.
Step-by-step explanation:
Let x represent the number of tips that John must make this week to reach his goal.
John works at a restaurant and gets paid $80 per week plus $5 per tip. This means that the total amount that John would earn in a week if he makes x tips would be
80 + 5x
This week, John wants to earn at least $300. This is expressed as
80 + 5x ≥ 300
5x ≥ 300 - 80
5x ≥ 220
x ≥ 220/5
x ≥ 44
In F, < CFD = < DFE, m< BFA = 4x, m< AFE = 3x + 12, and BE and AC are diameters.
Answer:
m arc DC=48°
Step-by-step explanation:
Angles in a Circle
We know [tex]\overline{BE}[/tex] and [tex]\overline{AC}[/tex] are diameters, so
[tex]m\angle BFA+m\angle AFE=180^o[/tex].
Since [tex]m\angle AFE=3x+12[/tex] and [tex]m\angle BFA=4x[/tex]
We set the equation
[tex]3x+12+4x=180^o[/tex]
Solving
[tex]7x=180-12=168[/tex]
[tex]x=24^o[/tex]
Thus
[tex]m\angle BFA=4(24)=96^o[/tex]
We also know
[tex]m\angle CFD\cong m\angle DFE[/tex]
Being [tex]\angle CFE[/tex] opposite to [tex]\angle BFA[/tex]
Then [tex]m\angle CFE=96^o[/tex]
It's divided in two equal angles [tex]\angle CFD\ and\ \angle DFE[/tex], thus m arc DC is half of 96°:
m arc DC=48°
First option
Consider the following types of data that were obtained from a random sample of 49 credit card accounts. Identify all the averages (mean, median, or mode) that can be used to summarize the data. (Select all that apply.)
(a) Outstanding balance on each account.
A. mode
B. median
C. mean
Answer:
All options( option A, option B and option C)
Step-by-step explanation:
Mean, median and mode are used to summarize the data. Mode can be calculated for both quantitative and qualitative data but mean and median cannot be calculated for qualitative data.
Here outstanding balance on each account represents quantitative data and mode, median and mean all can summarize the quantitative data. So, mean, median and mode each can be used to summarize the data of outstanding balance on each account.
What are the real and complex solutions of the polynomial equation? x^3-64=0
The real and complex solutions of the cubic equation [tex]x^3-64=0[/tex] are x=4 (real solution) and x= -2+2i√3, x= -2-2i√3 (complex solutions). This was found using the difference of cubes formula.
Explanation:The polynomial equation asked in the question is [tex]x^3-64=0,[/tex] which is a cubic equation rather than a quadratic equation. Hence we need to use a different method to solve it rather than the quadratic formula. Here we can use the difference of cubes formula, which indicates [tex]a^3-b^3[/tex] can be factored as [tex](a-b)(a^2+ab+b^2).[/tex] For this equation, the 'a' term is x (because [tex]x^3 = a^3[/tex]) and the 'b' term is 4 (because 4^3 = 64 which is b^3).
Following this formula, we factor the equation as [tex](x-4)(x^2+4x+16)=0.[/tex] Since this equation is set to equal zero, either the first factor equals zero (which gives us a solution x=4) or the second factor equals zero. After using the quadratic equation for the second factor, it has no real roots since its discriminant [tex](b^2-4ac = 4^2 - 4*1*16 = 16 - 64 = -48)[/tex]is negative. However, it has complex roots, which are -2+2i√3 and -2-2i√3.
So, the real and complex solutions of the polynomial equation [tex]x^3-64=0[/tex]are x=4, x= -2+2i√3, x= -2-2i√3.
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The real solution for the equation x^3-64=0 is 4. The complex solutions are -2 + 2i√3 and -2 - 2i√3. Therefore, the complete solutions are {4, -2 + 2i√3, -2 - 2i√3}.
Explanation:The given equation is x3-64=0. First, we can rewrite this equation as x3=64. This can be solved by taking the cube root of both sides, which gives us x = 4. Thus, 4 is the real solution.
To find the complex solutions, we need to use the fact that every non-zero number has three cube roots. The other two solutions can be found using the formula:
x = -2 + 2i√3
x= -2 - 2i√3
Therefore, the complete solution set of the equation x3-64=0 is {4, -2 + 2i√3, -2 - 2i√3}
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(3 points) In class we described the anti-malarial drug artemisinin (structure given below). What is the active functional group on the drug that is most responsible for its potency?
Answer: QINGHAOSU
Step-by-step explanation:Artemisinin is an antimalarial (Drug used to cure Malaria) drug whose active ingredient QINGHAOSU was isolated in the 1970s from the Plant called Artemisia annua by a Chinese Physician.
Artemisinin has been in use in ancient Chinese communities since the 4th century to cure Diseases. Artemisinin is potent in killing several forms of the Plasmodium specie and has been used to derive other antimalarial Drugs in use today like Artemeter and Artesunate.
David's gasoline station offers 4 cents off per gallon if the customer pays in cash. Past evidence indicates that 40% of all customers pay in cash. During a one-hour period, 15 customers buy gasoline at this station. What is the probability that more than 8 and less than 12 customers pay in cash?
Answer:
The probability that more than 8 and less than 12 customers pay in cash is 0.0931.
Step-by-step explanation:
Let X = a customer at David's gasoline station pay in cash.
The probability of a customer paying in cash is, P (X) = p = 0.40
The number of customers at the gasoline station during a 1-hour period is,
n = 15.
Then the random variable X follows a binomial distribution, Bin (15, 0.40).
The probability function for a Binomial distribution is:
[tex]P(X=x)={n\choose x}p^{x}(1-p)^{n-x}[/tex]
Compute the probability that more than 8 and less than 12 customers pay in cash as follows:
[tex]P(8< X< 12)=P(X<12)-P(X<8)\\=P(X=9)+P(X=10)+P(X=11)\\=[{15\choose 9}(0.40)^{9}(1-0.40)^{15-9}]+[{15\choose 10}(0.40)^{10}(1-0.40)^{15-10}]\\+[{15\choose 11}(0.40)^{11}(1-0.40)^{15-11}]\\=0.0612+0.0245+0.0074\\=0.0931[/tex]
Thus, the probability that more than 8 and less than 12 customers pay in cash is 0.0931.
The problem comes to calculating binomial probabilities for when 9, 10, and 11 customers pay in cash and adding them together. This scenario applies to binomial distribution, where we have a success (paying in cash) happening with a probability of 40%.
Explanation:This question is a problem of the binomial distribution. The binomial distribution is used when there are exactly two mutually exclusive outcomes of a trial (often referred to as success and failure). In this case, the success is the customer paying in cash, which happens 40% of the time according to past evidence.
The formula for the binomial distribution is:
P(X = k) = C(n, k) * (p^k) * ((1-p)^(n-k))
where
P(X = k) is the probability we are trying to calculateC(n, k) is the number of combinations of n items taken k at a timep is the probability of success on an individual trial (0.4 or 40% for pay in cash)n is the number of trials (15 customers)k is the number of successes we want (more than 8 and less than 12, so we calculate for 9, 10, and 11 separately and then add them together)Carry out this calculation for k=9, 10, 11, and then add these probabilities together to get the probability that more than 8 and less than 12 customers pay in cash.
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Given the following data for resting heartrate among college students, what is the IQR? 57, 59, 60, 62, 62, 63, 64, 68, 70, 70, 71, 71, 73, 79, 87, 89, 90
Answer:
14
Step-by-step explanation:
The inter-quartile range (IQR) is the difference between the third and first quartile. The data gathered is:
57, 59, 60, 62, 62, 63, 64, 68, 70, 70, 71, 71, 73, 79, 87, 89, 90
The data set has 17 values, the first quartile is the average between the 4th and 5th values, while the third quartile is the average between the 13th and 14th values:
[tex]Q_1 = \frac{62+62}{2}=62\\Q_3 = \frac{73+79}{2}=76[/tex]
The IQR is:
[tex]IQR = Q_3 - Q_1 = 76 -62\\IQR = 14[/tex]
The Interquartile Range (IQR) of the given data for resting heart rate among college students is 11, calculated by finding the difference between the upper (third) and lower (first) quartiles.
Explanation:Given the following data for resting heartrate among college students: 57, 59, 60, 62, 62, 63, 64, 68, 70, 70, 71, 71, 73, 79, 87, 89, 90, we are to find the Interquartile Range (IQR). The IQR is a statistical term that measures the statistical spread, or variability, of data points. It is calculated as the difference between the 75th percentile (Q3) and the 25th percentile (Q1).
First, we need to arrange the data points in ascending order, which has already been done for us in this case. Next, let's divide the data into two halves: the lower half and the upper half. For our data, the lower half is 57 to 68 and the upper half is 70 to 90. Find the median (middle value) of each half. The lower half median (Q1) is 62, and upper half median (Q3) is 73.
Now, subtract Q1 from Q3 to find the IQR: 73 - 62 = 11. So, the IQR of the resting heart rates of the given college students data is 11.
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A parcel delivery company delivered 103,000 packages last year, when its average employment was 84 drivers. This year the firm handled 112,000 deliveries with 96 drivers. What was the percentage change in productivity over the past two years
Answer: 4.85%
Step-by-step explanation: productivity = parcel delivered / number of drivers.
At when parcel delivered was 103, 000 the average number of drivers was 84 thus making the productivity to be
103,000/ 84 = 1226.19
At when the parcel delivered was 112, 800 the average number of drivers was 96 thus making the productivity to be
112, 000/ 96 = 1166.67
Thus the change in productivity = 1226.19 - 1166.67 = 59.52.
Initial productivity = 1226.19
final productivity = 1166.67
% change in productivity = change in productivity / initial productivity * 100
% change in productivity = 59.52 /1226 * 100
% change in productivity = 4.85%
Luna Company accepted credit cards in payment for $7,950 of services performed during July 2018. The credit card company charged Luna a 1.55 percent service fee; it paid Luna as soon as it received the invoices. Required Based on this information alone, what is the amount of net income earned during the month of July? (Do not round intermediate values. Round final answer to 2 decimal places.)
Answer:
$7826.78
Step-by-step explanation:
Total income = 7950
With a service fee charge of 1.55%, the service fee charge = [tex]1.55\%\times7950 = \dfrac{1.55}{100}\times7950 = 123.225[/tex]
Net income = 7950 - 123.225 = 7826.775 = $7826.78
An alternative solution:
Service charge = 1.55%
Net income = (100 - 1.55)% × 7950
= 98.45% × 7950 = [tex]\dfrac{98.45}{100}\times7950 = 7826.775=7826.78[/tex]
Answer:
Step-by-step explanation:
Accepted credit= $ 7950
Charges for services =1.55% of the total
7950 - 1.55% of 7950
7950 - (1.55/100)*7950
7950 - 123.225
7826.775
7826,78 two decimal places
This is the net income earned during the month of July
Denzel bought headphones two months ago, Solo2 Beats by Dre, for $130. He gives them to his little brother and goes online to buy another for himself but they are now $160.
What is the percentage change in the headphone’s price?
a) 23% b) 81% c) 19% d) 21%
Answer:
a) 23%
Step-by-step explanation:
To find the price change as percentage :
Multiply 100 by 160 then divide it by 130
100 × 160 ÷ 130 = 123 approximately which means there's an additional 23% to the 100% price
Which of the following best describes galena’s cleavage (see animation in placemark)? a. three directions at ~90° b. three directions not at ~90° c. two directions at ~90° d. two directions not at ~90°
Answer:
a)
Step-by-step explanation:
Given problem:
Which of the following best describes galena’s cleavage (see animation in placemark)?
Options:
a)three directions at ~90
b) three directions not at ~90°
c) two directions at ~90°
d) two directions not at ~90°
Solution:
- As per the animation where the place-mark is observed as a point. We can see that corner point is a point of intersection of three planes, left, right and bottom. We can also see that these three planes are at right angles @ 90 degrees to each other. Hence, option A.
An urn contains 6 red balls and 3 blue balls. One ball is selected at random and is replaced by a ball of the other color. A second ball is then chosen. What is the conditional probability that the first ball selected is red, given that the second ball was red?
Answer:
0.5882 or 58.82%
Step-by-step explanation:
The probability that both balls were red (A) is:
[tex]P(A)=\frac{6}{9}*\frac{5}{9}=0.3704[/tex]
The probability that the first ball was blue and the second ball was red (B) is:
[tex]P(B) = \frac{3}{9}*\frac{7}{9}=0.2593[/tex]
The conditional probability that the first ball selected is red, given that the second ball was red is:
[tex]P = \frac{P(A)}{P(A)+P(B)}=\frac{0.3704}{0.3704+0.2593} =0.5882[/tex]
The University of Michigan's business school claims it has the highest average GPA in the Big 10 among its business students. The business school claims that the business student average GPA is 3.5. Your friend believes that Michigan's claims are falsely inflated. In an effort to prove whether the grades are falsely inflated, your friend collects a random sample of 100 business students from Michigan and gets an average GPA of 3.31 with a standard deviation of 0.3.
Interpret a 5% chance of a type I error occurring:
A. an alpha level of .05 means that 5% of the time, the null hypothesis is rejected when it is actually correct.
B. an alpha level of .05 means that 5% of the time, the null hypothesis is rejected when it is actually incorrect.
C. an alpha level of .05 means that 5% of the time, the null hypothesis is not rejected when it is actually correct.
D. an alpha level of .05 means that 5% of the time, the null hypothesis is not rejected when it is actually incorrect.
Answer:
A
Step-by-step explanation:
The type I error arises when we wrongfully reject the null hypothesis. The probability of occurrence of type I error is denoted as α. Thus, α=0.05 means that there is 5% probability that we reject the null hypothesis when it is true. So, we can say that the α=0.05, means that 5% of the time, we reject the null hypothesis when it is correct.
Answer:
A
Step-by-step explanation:
According to DeMorgan 's theorem, the complement of W · X + Y · Z is W' + X' · Y' + Z'. Yet both functions are 1 for WXYZ = 1110. How can both a function and its complement be 1 for the same input combination? What's wrong here?
Answer:
The parenthesis need to be kept intact while applying the DeMorgan's theorem on the original equation to find the compliment because otherwise it will introduce an error in the answer.
Step-by-step explanation:
According to DeMorgan's Theorem:
(W.X + Y.Z)'
(W.X)' . (Y.Z)'
(W'+X') . (Y' + Z')
Note that it is important to keep the parenthesis intact while applying the DeMorgan's theorem.
For the original function:
(W . X + Y . Z)'
= (1 . 1 + 1 . 0)
= (1 + 0) = 1
For the compliment:
(W' + X') . (Y' + Z')
=(1' + 1') . (1' + 0')
=(0 + 0) . (0 + 1)
=0 . 1 = 0
Both functions are not 1 for the same input if we solve while keeping the parenthesis intact because that allows us to solve the operation inside the parenthesis first and then move on to the operator outside it.
Without the parenthesis the compliment equation looks like this:
W' + X' . Y' + Z'
1' + 1' . 1' + 0'
0 + 0 . 0 + 1
Here, the 'AND' operation will be considered first before the 'OR', resulting in 1 as the final answer.
Therefore, it is important to keep the parenthesis intact while applying DeMorgan's Theorem on the original equation or else it would produce an erroneous result.
The error originates from an incorrect application of DeMorgan's theorem. The correct complement of (W · X) + (Y · Z) is (W' + X') · (Y' + Z'). This corrects the discrepancy seen for WXYZ = 1110.
Explanation:The confusion here likely originates from a mistake in the DeMorgan's theorem application. According to DeMorgan's Theorem, the complement of (W · X) + (Y · Z) is given by (W' + X') · (Y' + Z'), not W' + X' · Y' + Z'.
So, if we have WXYZ = 1110, the given function (W · X) + (Y · Z) equals 1 because we have (1 · 1) + (1 · 0) = 1 + 0 = 1. Whereas, the correct complement function, (W' + X') · (Y' + Z') equals 0 because we have (0 + 0) · (0 + 1) = 0 · 1 = 0.
This explains why we were seeing both original function and the incorrectly applied complement function evaluating to 1 for the same input combination.
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Assume that T is a linear transformation. Find the standard matrix of T.
T: set of real numbers R^2 →R^2 first rotates points through ( -pi/6) radians (clockwise) and then reflects points through the horizontal x1-axis.
Answer:
The Matrix of T is
[tex]\left[\begin{array}{cc}cos(\pi/6)&-sin(\pi/6)\\-sin(\pi/6)&-cos(\pi/6)\end{array}\right][/tex]
Step-by-step explanation:
Rotate -pi/6 Clockwise is the same as rotating pi/6 anticlockwise. The matrix of that rotation is
[tex]\left[\begin{array}{cc}cos(\pi/6)&-sin(\pi/6)\\sin(\pi/6)&cos(\pi/6)\end{array}\right][/tex]
The matrix of the reflection through the x1-axis is
[tex]\left[\begin{array}{cc}1&0\\0&-1\end{array}\right][/tex]
Therefore, the composition is the product of both matrices is the matrix of T
[tex]MT = \left[\begin{array}{cc}1&0\\0&-1\end{array}\right] * \left[\begin{array}{cc}cos(\pi/6)&-sin(\pi/6)\\sin(\pi/6)&cos(\pi/6)\end{array}\right] = \left[\begin{array}{cc}cos(\pi/6)&-sin(\pi/6)\\-sin(\pi/6)&-cos(\pi/6)\end{array}\right][/tex]
I hope that works for you!
To find the standard matrix of the linear transformation [tex]\( T \)[/tex], we need to perform two operations in sequence: a rotation through [tex]\( -\frac{\pi}{6} \)[/tex] radians (clockwise) and a reflection through the horizontal [tex]\( x_1 \)-axis[/tex]. We will find the matrices for each of these transformations and then multiply them to get the standard matrix for [tex]\( T \)[/tex].
1. Rotation Matrix [tex]\( R \)[/tex]:
The matrix that represents a rotation through an angle [tex]\( \theta \)[/tex] in the counterclockwise direction is given by:
[tex]\[ R(\theta) = \begin{bmatrix} \cos(\theta) -\sin(\theta) \\ \sin(\theta) \cos(\theta) \end{bmatrix} \][/tex]
For a clockwise rotation, we use a negative angle, so for [tex]\( -\frac{\pi}{6} \)[/tex] radians, the rotation matrix is:
[tex]\[ R\left(-\frac{\pi}{6}\right) = \begin{bmatrix} \cos\left(-\frac{\pi}{6}\right) -\sin\left(-\frac{\pi}{6}\right) \\ \sin\left(-\frac{\pi}{6}\right) \cos\left(-\frac{\pi}{6}\right) \end{bmatrix} = \begin{bmatrix} \frac{\sqrt{3}}{2} \frac{1}{2} \\ -\frac{1}{2} \frac{\sqrt{3}}{2} \end{bmatrix} \][/tex]
2. Reflection Matrix [tex]\( M \)[/tex]:
The matrix that represents a reflection through the horizontal[tex]\( x_1 \)[/tex]-axis is given by:
[tex]\[ M = \begin{bmatrix} 1 0 \\ 0 -1 \end{bmatrix} \][/tex]
3. Standard Matrix of [tex]\( T \)[/tex]
To find the standard matrix of[tex]\( T \)[/tex], we multiply the rotation matrix [tex]\( R \)[/tex] by the reflection matrix [tex]\( M \)[/tex]:
[tex]1 0 \\ 0 -1[/tex]
[tex]\end{bmatrix} \begin{bmatrix} \frac{\sqrt{3}}{2} \frac{1}{2} \\ -\frac{1}{2} \frac{\sqrt{3}}{2} \end{bmatrix} \] \[ T = \begin{bmatrix} \frac{\sqrt{3}}{2} \frac{1}{2} \\ \frac{1}{2} -\frac{\sqrt{3}}{2} \end{bmatrix} \][/tex]
Therefore, the standard matrix of the linear transformation [tex]\( T \)[/tex] is:
[tex]\[ \boxed{T = \begin{bmatrix} \frac{\sqrt{3}}{2} \frac{1}{2} \\ \frac{1}{2} -\frac{\sqrt{3}}{2} \end{bmatrix}} \][/tex]
This matrix represents the transformation that first rotates points through [tex]\( -\frac{\pi}{6} \)[/tex] radians (clockwise) and then reflects them through the horizontal [tex]\( x_1 \)-axis.[/tex]
A crystal growth furnace is used in research to determine how best to manufacture crystals used in electronic components for the space shuttle. For proper growth of the crystal, the temperature must be controlled accurately by adjusting the input power. Suppose the relationship is given by
T(w)=0.1w2+2.155w+20
where T is the temperature in degrees Celsius and w is the power input in watts.
Answer: The question is incomplete ad some details are missing.
it says Suppose the relationship is given by T(w)=0.1w2+2.155w+20
where T is the temperature in degrees Celsius and w is the power input in watts.
a) How many watts of power are needed to maintain the temperature at exactly 200degree celsius
= 33watts of power are needed
Step-by-step explanation:
The detailed steps and appropriate substitution is as shown in the attachment
A measurement of the circumference of a disk has an uncertainty of 1 . 5 mm. How many measurements must be made so that the diameter can be estimated with an uncertainty of only 0 . 5 mm
Answer:
How many measurements must be made = 9
Step-by-step explanation:
The steps are as shown in the attachment.
To reduce the uncertainty in the diameter of a disk from 1.5mm to 0.5mm, three times more measurements of the circumference would need to be made. This is due to the relationship between the circumference and diameter, and how the uncertainty propagates through this relationship.
Explanation:This question pertains to the areas of accuracy, precision, and uncertainty in measurements. Understanding these concepts is vital in the field of physics. The circumference of a disk and its diameter are related by the constant π (Pi): Diameter = Circumference / π.
The question states that there is an uncertainty of 1.5 mm in measuring the circumference. Given that the diameter and circumference are directly connected, when you reduce the uncertainty in the measurement of the circumference (e.g., by taking more measurements), you also reduce the uncertainty in the diameter. However, the relationship is not linear. Through propagation of uncertainty principles, the uncertainty in diameter would be the uncertainty in the circumference divided by π. To reduce this to 0.5 mm, you would require three times more measurements.
The precision of a measurement system is closely linked to the size of its measurement increments. The smaller the measurement increment, the more precise the tool. Various factors can contribute to the uncertainty of a measurement, including the smallest division on a given tool, the ability of the person making the measurement, irregularities in the object being measured, and unforeseen circumstances that affect the outcome.
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