Answer:
So maximum when 46 of I grade and 16 of II grade are produced.
Max revenue = 2100
Step-by-step explanation:
Given that a plane delivers two types of cargo between two destinations
Crate I Crate II
Volume 9 9
Weight 187 374
Revenue 30 45
Let X be the no of crate I and y that of crate II
Then
[tex]9x+9y\leq 540\\187x+374y\leq 14212\\x\geq 2y[/tex]
Simplify these equations to get
[tex]x+y\leq 60\\x+2y\leq 76\\x\geq 2y[/tex]
Solving we get
[tex]y\leq 16\\x\leq 46 and x\geq 32\\32\leq x\leq 46[/tex]
REvenue = 30x+45y
The feasible region would have corner points as (60,0) or (32,16) or (46,16)
Revenue for (60,0) = 1800
(32,16) = 1680
(46,16)=2100
So maximum when 46 of I grade and 16 of II grade are produced.
Max revenue = 2100
To maximize revenue, we need to determine the number of crates of each cargo that should be shipped. The problem can be solved using linear programming techniques to find the optimal solution.
Explanation:To maximize revenue, we need to determine the number of crates of each cargo that should be shipped. Let's assume the number of crates of cargo I is x and the number of crates of cargo II is y.
Based on the given information, the constraints for the problem are:
Volume constraint: 9x + 9y ≤ 540Weight constraint: 187x + 374y ≤ 14,212Relationship constraint: x ≥ 2yTo find the maximum revenue, we need to maximize the objective function: Revenue = 30x + 45y.
The problem can be solved using linear programming techniques, such as graphical or simplex method, to find the optimal solution. However, since these methods require plotting and iterations, the detailed calculations are beyond the scope of this response. The optimal solution will provide the values of x and y, which can be used to determine the maximum revenue.
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For this problem, carry at least four digits after the decimal in your calculations. Answers may vary slightly due to rounding.
Santa Fe black-on-white is a type of pottery commonly found at archaeological excavations at a certain monument. At one excavation site a sample of 572 potsherds was found, of which 363 were identified as Santa Fe black-on-white.
(a) Let p represent the proportion of Santa Fe black-on-white potsherds at the excavation site. Find a point estimate for p. (Round your answer to four decimal places.)
(b) Find a 95% confidence interval for p. (Round your answers to three decimal places.)
lower limit
upper limit
Answer:
a) p = 0.6346
b) 95% confidence interval
Lower limit: 0.5951
Upper limit: 0.6741
Step-by-step explanation:
We are given the following in the question:
Sample size, n = 572
Number of Santa Fe black-on-whitepots , x = 363
a) proportion of Santa Fe black-on-white potsherds
[tex]\hat{p} = \dfrac{x}{n} = \dfrac{363}{572} = 0.6346[/tex]
b) 95% confidence interval
[tex]\hat{p}\pm z_{stat}\sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}}[/tex]
[tex]z_{critical}\text{ at}~\alpha_{0.05} = 1.96[/tex]
Putting the values, we get:
[tex]0.6346\pm 1.96(\sqrt{\frac{0.6346(1-0.6346)}{572}}) = 0.6346\pm 0.0395\\\\=(0.5951,0.6741)[/tex]
Lower limit: 0.5951
Upper limit: 0.6741
An article in Knee Surgery, Sports Traumatology, Arthroscopy, "Arthroscopic meniscal repair with an absorbable screw: results and surgical technique," (2005, Vol. 13, pp. 273-279) cites a success rate more than 90% for meniscal tears with a rim width of less than 3 mm, but only a 67% success rate for tears of 3-6 mm. If you are unlucky enough to suffer a meniscal tear of less than 3 mm on your left knee, and one of width 3-6 mm on your right knee, what are the mean and variance of the number of successful surgeries?
Answer:
Mean = 1.57
Variance=0.31
Step-by-step explanation:
To calculate the mean and the variance of the number of successful surgeries (X), we first have to enumerate the possible outcomes:
1) Both surgeries are successful (X=2).
[tex]P(e_1)=0.90*0.67=0.603[/tex]
2) Left knee unsuccessful and right knee successful (X=1).
[tex]P(e_2)=(1-0.9)*0.67=0.1*0.67=0.067[/tex]
3) Right knee unsuccessful and left knee successful (X=1).
[tex]P(e_3)=0.90*(1-0.67)=0.9*0.33=0.297[/tex]
4) Both surgeries are unsuccessful (X=0).
[tex]P(e_4)=(1-0.90)*(1-0.67)=0.1*0.33=0.033[/tex]
Then, the mean can be calculated as the expected value:
[tex]M=\sum p_iX_i=0.603*2+0.067*1+0.297*1+0.033*0\\\\M=1.206+0.067+0.297+0\\\\M=1.57[/tex]
The variance can be calculated as:
[tex]V=\sum p_i(X_i-\bar{X})^2\\\\V=0.603(2-1.57)^2+(0.067+0.297)*(1-1.57)^2+0.033*(0-1.57)^2\\\\V=0.603*0.1849+0.364*0.3249+0.033*2.4649\\\\V=0.1115+0.1183+0.0813\\\\V=0.3111[/tex]
The mean and variance of the number of successful surgeries for both knees combined are:
Mean: [tex]\({1.57}\)[/tex]
Variance: [tex]\({0.3111}\)[/tex]
The mean and variance of the number of successful surgeries for the given meniscal tears can be calculated using the information provided about the success rates.
For a meniscal tear with a rim width of less than 3 mm, the success rate is more than 90%. For simplicity, let's assume the success rate is exactly 90% (since we don't have the exact number above 90%). For a tear of 3-6 mm, the success rate is 67%.
Let's denote the success of a surgery as a random variable [tex]\( X \)[/tex], which takes the value 1 if the surgery is successful and 0 if it is not. The probability of success [tex]\( P(X = 1) \)[/tex] is the success rate, and the probability of failure [tex]\( P(X = 0) \)[/tex] is [tex]\( 1 - P(X = 1) \)[/tex].
For the left knee (tear less than 3 mm):
- [tex]\( P(X = 1) = 0.90 \)[/tex] (success rate)
- [tex]\( P(X = 0) = 1 - 0.90 = 0.10 \)[/tex] (failure rate)
For the right knee (tear 3-6 mm):
- [tex]\( P(X = 1) = 0.67 \)[/tex] (success rate)
- [tex]\( P(X = 0) = 1 - 0.67 = 0.33 \)[/tex] (failure rate)
The mean (expected value) of the number of successful surgeries for each knee is calculated as follows:
For the left knee:
[tex]\[ E(X) = \sum_{i=0}^{1} x_i \cdot P(X = x_i) = 1 \cdot 0.90 + 0 \cdot 0.10 = 0.90 \][/tex]
For the right knee:
[tex]\[ E(X) = \sum_{i=0}^{1} x_i \cdot P(X = x_i) = 1 \cdot 0.67 + 0 \cdot 0.33 = 0.67 \][/tex]
The variance of the number of successful surgeries for each knee is calculated using the formula for the variance of a binary random variable:
For the left knee:
[tex]\[ \text{Var}(X) = E(X^2) - [E(X)]^2 \][/tex]
[tex]\[ E(X^2) = \sum_{i=0}^{1} x_i^2 \cdot P(X = x_i) = 1^2 \cdot 0.90 + 0^2 \cdot 0.10 = 0.90 \][/tex]
[tex]\[ \text{Var}(X) = 0.90 - (0.90)^2 = 0.90 - 0.81 = 0.09 \][/tex]
For the right knee:
[tex]\[ E(X^2) = \sum_{i=0}^{1} x_i^2 \cdot P(X = x_i) = 1^2 \cdot 0.67 + 0^2 \cdot 0.33 = 0.67 \][/tex]
[tex]\[ \text{Var}(X) = E(X^2) - [E(X)]^2 \][/tex]
[tex]\[ \text{Var}(X) = 0.67 - (0.67)^2 = 0.67 - 0.4489 = 0.2211 \][/tex]
Now, assuming the surgeries on the two knees are independent events, the mean and variance for both knees combined can be calculated as follows:
Mean for both knees:
[tex]\[ E(X_{\text{left}} + X_{\text{right}}) = E(X_{\text{left}}) + E(X_{\text{right}}) = 0.90 + 0.67 = 1.57 \][/tex]
Variance for both knees:
Since the surgeries are independent, the variance of the sum is the sum of the variances:
[tex]\[ \text{Var}(X_{\text{left}} + X_{\text{right}}) = \text{Var}(X_{\text{left}}) + \text{Var}(X_{\text{right}}) = 0.09 + 0.2211 = 0.3111 \][/tex]
Therefore, the mean and variance of the number of successful surgeries for both knees combined are:
Mean: [tex]\({1.57}\)[/tex]
Variance: [tex]\({0.3111}\)[/tex]
Based on past experience, a bank believes that 4% of the people who receive loans will not make payments on time. The bank has recently approved 300 loans. 6% of these clients did not make timely payments. What is the probability that over 6% will not make timely payments?A. 0.0721B. 0.9616C. 0.9279D. 0.0384
Answer:
D. 0.0384
Step-by-step explanation:
For each loan, there are only two possible outcomes. Either the client makes timely payments, or he does not. The probability of a client making a timely payment is independent from other clients. So we use the binomial probability distribution to solve this question.
However, our sample is big. So i am going to aproximate this binomial distribution to the normal.
Binomial probability distribution
Probability of exactly x sucesses on n repeated trials, with p probability.
Can be approximated to a normal distribution, using the expected value and the standard deviation.
The expected value of the binomial distribution is:
[tex]E(X) = np[/tex]
The standard deviation of the binomial distribution is:
[tex]\sqrt{V(X)} = \sqrt{np(1-p)}[/tex]
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.
When we are approximating a binomial distribution to a normal one, we have that [tex]\mu = E(X)[/tex], [tex]\sigma = \sqrt{V(X)}[/tex].
In this problem, we have that:
[tex]n = 300, p = 0.04[/tex]
So
[tex]\mu = E(X) = np = 300*0.04 = 12[/tex]
[tex]\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{300*0.04*0.96} = 3.39[/tex]
What is the probability that over 6% will not make timely payments?
This is 1 subtracted by the pvalue of Z when X = 0.06*300 = 18. So
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{18 - 12}{3.39}[/tex]
[tex]Z = 1.77[/tex]
[tex]Z = 1.77[/tex] has a pvalue of 0.9616
1 - 0.9616 = 0.0384
So the correct answer is:
D. 0.0384
As a freshman, suppose you had to take two of four lab science courses, one of two literature courses, two of three math courses, and one of seven physical education courses. Disregarding possible time conflicts, how many different schedules do you have to choose from?
Answer:
We have 252 different schedules.
Step-by-step explanation:
We know that as a freshman, suppose you had to take two of four lab science courses, one of two literature courses, two of three math courses, and one of seven physical education courses.
So from 4 lab science courses we choose 2:
[tex]C_2^4=\frac{4!}{2!(4-2)!}=6[/tex]
So from 2 literature courses we choose 1:
[tex]C_1^2=\frac{2!}{1!(2-1)!}=2[/tex]
So from 3 math courses we choose 2:
[tex]C_2^3=\frac{3!}{2!(3-2)!}=3\\[/tex]
So from 7 physical education courses we choose 1:
[tex]C_1^7=\frac{7!}{1!(7-1)!}=7[/tex]
We get: 6 · 2 · 3 · 7 = 252
We have 252 different schedules.
A survey among US adults of their favorite toppings on a cheese pizza reported that 43% favored pepperoni, 14% favored mushrooms, and 6% favored both pepperoni and mushrooms. What is the probability that a random adult favored pepperoni or mushrooms on their cheese pizza? Provide your answer as a whole number in the box below, i.e., .32 is 32% so you would enter 32. Round as needed.
Answer:
51% of US adults favored pepperoni or mushrooms on their cheese pizza.
Step-by-step explanation:
We are given the following in the question:
Percentage of US adults that favored pepperoni = 43%
[tex]P(P) = 0.43[/tex]
Percentage of US adults that favored mushroom = 14%
[tex]P(M) = 0.14[/tex]
Percentage of US adults that favored both pepperoni and mushroom = 6%
[tex]P(M\cap P) = 0.06[/tex]
We have to evaluate the probability that a random adult favored pepperoni or mushrooms on their cheese pizza.
Thus, we have to evaluate:
[tex]P(M\cup P) = P(M) + P(P) - P(M\cap P)\\P(M\cup P) = 0.43 + 0.14 - 0.06\\P(M\cup P) = 0.51 = 51\%[/tex]
Thus, 51% of US adults favored pepperoni or mushrooms on their cheese pizza.
Each year, taxpayers are able to contribute money to various charities via their IRS tax forms. The following list contains the amounts of money (in dollars) donated via IRS tax forms by Each year, taxpayers are able to contribute money taxpayers:
2 , 22 , 27 , 31 , 36 , 51 , 57 , 57 , 60 , 62 , 62 , 62 , 73 , 77 , 83 , 95 , 99 , 104 , 105 , 127 , 153 , 162 , 197
(a) For these data, which measures of central tendency take more than one value? Choose all that apply.
Mean
Median
Mode
None of these measures
(b) Suppose that the measurement 197 (the largest measurement in the data set) were replaced by 246. Which measures of central tendency would be affected by the change? Choose all that apply.
Mean
Median
Mode
None of these measures
(c) Suppose that, starting with the original data set, the largest measurement were removed. Which measures of central tendency would be changed from those of the original data set? Choose all that apply.
Mean
Median
Mode
None of these measures
(d) Which of the following best describes the distribution of the original data? Choose only one.
Negatively skewed
Positively skewed
Roughly symmetrical
Answer:
(a) None of these measures
(b) Mean
(c) Mean and Median
(d) Roughly Symmetrical
Step-by-step explanation:
(a)
Mean
Total number in the set = 23
Summation of the set = 2+22+27+31+36+51+57+57+60+62+62+62+73+77+83+95+99+104+105+127+153+162+197 = 1804
Mean = Sum of set / total no of set
1804/23 = 78.435
Median is the middle number in the set after it had been arranged from lowest to highest
2 , 22 , 27 , 31 , 36 , 51 , 57 , 57 , 60 , 62 , 62 , 62 , 73 , 77 , 83 , 95 , 99 , 104 , 105 , 127 , 153 , 162 , 197
The Median is 62
Mode the value that appear most
Mode is 62
None of them takes more than one value
(b) If 197 is replaced by 246, the set becomes
2 , 22 , 27 , 31 , 36 , 51 , 57 , 57 , 60 , 62 , 62 , 62 , 73 , 77 , 83 , 95 , 99 , 104 , 105 , 127 , 153 , 162 , 246
The mean becomes
Total number in the set = 23
Summation of the set = 2+22+27+31+36+51+57+57+60+62+62+62+73+77+83+95+99+104+105+127+153+162+246= 1853
Mean = Sum of set / total no of set
1853/23 = 80.565
The Median and Mode remains the same.
(c) When the largest measurements are removed, the number of values in the set reduces and this affects the Mean and the Median. The mode will still remain unchanges since it is a small number and appears the most.
The circumference of a sphere was measured to be 74 cm with a possible error of 0.5 cm. (a) Use differentials to estimate the maximum error in the calculated surface area. (Round your answer to the nearest integer.) cm2 What is the relative error?
Using differentials, the estimated maximum error in the calculated surface area of a sphere with a measured circumference of 74 cm and a possible error of 0.5 cm is 24 cm². The relative error is approximately 5%.
Explanation:The subject concerns the application of differentials in estimating the maximum error in the calculated surface area of a sphere. Given the circumference C = 74 cm with a possible error δC = 0.5 cm, we can calculate the radius r = C / (2π). With the surface area formula of a sphere A = 4πr², differentiating this equation gives dA = 8πr dr. By substituting the values, the maximum error in calculated surface area δA = dA = 8πr δr = 8π(C/2π) (δC/2π) = 2C δC / π. Plugging the values of C = 74 cm and δC = 0.5 cm, we get δA ≈ 24 cm² which is the maximum error in the calculated surface area. For the relative error, it is the absolute error divided by the actual measurement, hence, the relative error is δA/A = δA / 4πr² = (2C δC / π) / 4π(C/2π)² ≈ 0.05 or 5%.
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To find a formula for the moose population, calculate the rate of change and use it in the formula P = 190t + 4360. The model predicts the moose population to be 7710 in 2003.
Explanation:To find a formula for the moose population, we need to determine the rate of change in the population. We can do this by finding the slope of the line that represents the change in population from 1991 to 1999. First, we calculate the change in population: 5880 - 4360 = 1520. Then, we calculate the change in time: 1999 - 1991 = 8. Next, we divide the change in population by the change in time to find the rate of change: 1520/8 = 190. So, the formula for the moose population, P, is P = 190t + 4360, where t represents the years after 1991.
To predict the moose population in 2003, we substitute t = 12 (since 2003 is 12 years after 1991) into the formula: P = 190(12) + 4360 = 7710. Therefore, the model predicts the moose population to be 7710 in 2003.
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A reasonable estimate of the moment of inertia of an ice skater spinning with her arms at her sides can be made by modeling most of her body as a uniform cylinder. Suppose the skater has a mass of 64 kg . One eighth of that mass is in her arms, which are 60 cm long and 20 cm from the vertical axis about which she rotates. The rest of her mass is approximately in the form of a 20-cm-radius cylinder.
Answer:
Step-by-step explanation:
Given data:
Mass of the one arm of the skater, m = (1/16) x 64 = 4 kg
Rest mass of the skater in the form of cylinder, M = (7 / 8) x 64 kg = 56 kg
Radius of the cylinder, R = 20 cm = 0.20 m
The parallel axis theorem:
The population of mosquitoes in a certain area increases at a rate proportional to the current pop-ulation, and in the absence of other factors, the population doubles each week. There are 200,000mosquitoes in the area initially, and predators (birds, bats, and so forth) eat 20,000 mosquitoes perday. Set up a differential equation for the population of mosquitoes and make sure to solve for theproportionality constant using the information given. Determine the population of mosquitoes in thearea at any time.
Final answer:
To model the mosquito population considering both exponential growth and daily predation, a differential equation was formulated and solved, revealing how the population changes over time.
Explanation:
To determine the population of mosquitoes in the area at any time, given that the population doubles each week and predators eat 20,000 mosquitoes per day, we can set up a differential equation. To start, we know the initial population is 200,000 mosquitoes. Given the population increases proportionally, we use the formula P(t) = P_0e^{rt}, where P(t) is the population at time t, P_0 is the initial population, r is the rate of growth, and t represents time in weeks.
To find r, we use the fact that the population doubles each week. So, when t = 1, P(t) = 2P_0, leading to 2P_0 = P_0e^{r(1)}, simplifying to 2 = e^r, which gives r = ln(2).
Including the effect of predators, the amended differential equation becomes dP/dt = rP - 20,000. Substituting r with ln(2) and solving this equation gives us the mosquito population at any time, accounting for both natural growth and predation.
A sprint duathlon consists of a 5 km run, a 20 km bike ride, followed by another 5 km run. The mean finish time of all participants in a recent large duathlon was 1.67 hours with a standard deviation of 0.25 hours. Suppose a random sample of 30 participants was taken and the mean finishing time was found to be 1.59 hours with a standard deviation of 0.30 hours. What is the standard error for the mean finish time of 30 randomly selected participants
Answer:
The standard error is 0.0456 for the mean finish time of 30 randomly selected participants.
Step-by-step explanation:
We are given the following in the question:
Population mean, [tex]\mu[/tex] = 1.67 hours
Population standard deviation, [tex]\sigma[/tex] = 0.25 hours
Sample mean, [tex]\bar{x}[/tex] = 1.59 hours
Sample standard deviation, s = 0.30 hours
Sample size, n = 30
We have to find the standard error for the mean finish time of 30 randomly selected participants.
Formula:
[tex]\text{Standard error} = \dfrac{\sigma}{\sqrt{n}} = \dfrac{0.25}{\swqrt{30}} = 0.0456[/tex]
Thus, the standard error is 0.0456 for the mean finish time of 30 randomly selected participants.
The standard error for the mean finish time of 30 randomly selected participants is 0.0549 hours.
Explanation:The standard error for the mean finish time of 30 randomly selected participants can be calculated using the formula:
Standard Error = Standard Deviation / √(Sample Size)
Plugging in the given values, the standard error would be:
Standard Error = 0.30 / √(30) = 0.0549 hours
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Use the information given to find the appropriate minimum sample size. (Round your answer up to the nearest whole number.)Estimating μ correct to within 3 with probability 0.99. Prior experience suggests that the measurements will range from 8 to 40.
The minimum sample size required to estimate μ within 3 with a confidence level of 0.99, given a standard deviation of 8, is approximately 48. This was determined by plugging the values into the sample size formula and rounding up to the nearest whole number.
Explanation:To find the minimum sample size, we need to use the formula for sample size n, = (Z_α/2 * σ / E)^2. In this problem, you want to estimate μ correct to within 3 with a probability of 0.99. In other words, you want the error E to be 3 and the confidence level to be 0.99.
The Z value corresponding to a confidence level of 0.99 is approximately 2.576 (you can find this value from a standard Z-table). The measurements range from 8 to 40, so we can estimate the standard deviation σ as (40 - 8) / 4 = 8.
Plugging these values into the formula, we get n = (2.576 * 8 / 3)^2 = 47.36. This number must be rounded up to the nearest whole number because the sample size cannot be a fraction. So, the minimum sample size required is 48.
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Please help me find the answer.
Answer:
b/a = c/b
if a = b, then b = c
Answer: the second one (b/a = c/b) and the last one (if a = b then b = c) are the only ones that are true
Step-by-step explanation:
1 point) Consider the following game of chance based on the spinner below: Each spin costs $2. If the spinner lands on A the player wins a quarter, if the spinner stops on D the player wins $9 otherwise the player wins nothing. Calculate the players expected winnings. Express your answer to at least three decimal places in dollar form. .
The game of chance discussed is a question about probability and expected value in mathematics. To calculate the expected winnings of the game, we use given game information and probabilities. If the probabilities are not given, the question usually assumes a fair spinner, i.e., all outcomes are equally likely.
Explanation:The subject at hand deals with probability and expected value, which are mathematical concepts typically covered in a high school math curriculum. The game described illustrates these concepts. Each possible outcome of the game (A or D, otherwise lose) corresponds to an event that has a certain probability. These probabilities are all added together to determine the expected value of the game in dollars.
Suppose the probabilities of landing on A and D are p(A) and p(D), and the probability of not landing on either A or D is 1 - p(A) - p(D), then the expected value of the game is: Expected Value = $2 * [p(A)*0.25 + p(D)*9 + (1 - p(A) - p(D))*0] .
To find the expected value, we would need to know the probabilities of landing on each of these segments on the spinner. If these probabilities are not given in the problem, it can be assumed that the spinner is fair (i.e., all outcomes are equally likely). If there are n total segments on the spinner, then p(A) = p(D) = 1/n, and the probability of not landing on A or D would be (n-2)/n. Substitute these probabilities into the expected value equation can give the answer.
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Write the equation of the line that passes through (3, 4) and (2, −1) in slope-intercept form. (2 points) a y = 3x − 7 b y = 3x − 5 c y = 5x − 11 d y = 5x − 9
Answer: y = 5x − 11
Step-by-step explanation:
The equation of a straight line can be represented in the slope-intercept form, y = mx + c
Where c = intercept
Slope, m =change in value of y on the vertical axis / change in value of x on the horizontal axis represent
change in the value of y = y2 - y1
Change in value of x = x2 -x1
y2 = final value of y
y 1 = initial value of y
x2 = final value of x
x1 = initial value of x
The line passes through (3,4) and (2, -1),
y2 = - 1
y1 = 4
x2 = 2
x1 = 3
Slope,m = (- 1 - 4)/(2 - 3) = - 5/- 1 = 5
To determine the y intercept, we would substitute x = 3, y = 4 and m= 5 into
y = mx + c. It becomes
4 = 5 × 3 + c
4 = 15 + c
c = 4 - 15 = - 11
The equation becomes
y = 5x - 11
The proportion of high school seniors who are married is 0.02. Suppose we take a random sample of 300 high school seniors; a.) Find the mean and standard deviation of the sample count X who are married. b.) What is the probability that, in our sample of 300, we find that 8 of the seniors are married? c.) What is the probability that we find less than 4 of the seniors are married? d.) What is the probability that we find at least 1 of the seniors are married?
Answer:
a) Mean 6, standard deviation 2.42
b) 10.40% probability that, in our sample of 300, we find that 8 of the seniors are married.
c) 14.85% probability that we find less than 4 of the seniors are married.
d) 99.77% probability that we find at least 1 of the seniors are married
Step-by-step explanation:
For each high school senior, there are only two possible outcomes. Either they are married, or they are not. The probability of a high school senior being married is independent from other high school seniors. So we use the binomial probability distribution to solve this question.
Binomial probability distribution
The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
In which [tex]C_{n,x}[/tex] is the number of different combinations of x objects from a set of n elements, given by the following formula.
[tex]C_{n,x} = \frac{n!}{x!(n-x)!}[/tex]
And p is the probability of X happening.
The expected value of the binomial distribution is:
[tex]E(X) = np[/tex]
The standard deviation of the binomial distribution is:
[tex]\sqrt{V(X)} = \sqrt{np(1-p)}[/tex]
In this problem, we have that:
[tex]n = 300, p = 0.02[/tex]
a.) Find the mean and standard deviation of the sample count X who are married.
Mean
[tex]E(X) = np = 300*0.02 = 6[/tex]
Standard deviation
[tex]\sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{300*0.02*0.98} = 2.42[/tex]
b.) What is the probability that, in our sample of 300, we find that 8 of the seniors are married?
This is P(X = 8).
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
[tex]P(X = 8) = C_{300,8}.(0.02)^{8}.(0.98)^{292} = 0.1040[/tex]
10.40% probability that, in our sample of 300, we find that 8 of the seniors are married.
c.) What is the probability that we find less than 4 of the seniors are married?
[tex]P(X < 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)[/tex]
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
[tex]P(X = 0) = C_{300,0}.(0.02)^{0}.(0.98)^{300} = 0.0023[/tex]
[tex]P(X = 1) = C_{300,1}.(0.02)^{1}.(0.98)^{299} = 0.0143[/tex]
[tex]P(X = 2) = C_{300,2}.(0.02)^{2}.(0.98)^{298} = 0.0436[/tex]
[tex]P(X = 3) = C_{300,3}.(0.02)^{3}.(0.98)^{297} = 0.0883[/tex]
[tex]P(X < 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) = 0.0023 + 0.0143 + 0.0436 + 0.0883 = 0.1485[/tex]
14.85% probability that we find less than 4 of the seniors are married.
d.) What is the probability that we find at least 1 of the seniors are married?
Either no seniors are married, or at least 1 one is. The sum of the probabilities of these events is decimal 1. So
[tex]P(X = 0) + P(X \geq 1) = 1[/tex]
From c), we have that [tex]P(X = 0) = 0.0023[/tex]. So
[tex]0.0023 + P(X \geq 1) = 1[/tex]
[tex]P(X \geq 1) = 0.9977[/tex]
99.77% probability that we find at least 1 of the seniors are married
In this problem, the mean and standard deviation of a binomial distribution, with probability of success 0.02 and sample size 300, are found. Subsequently, the probabilities that 8, less than 4, and at least 1 of the seniors are married are computed using the binomial formula.
Explanation:This problem deals with the Binomial distributions in statistics. Since we know the proportion of high school seniors who are married is 0.02, and the sample size is 300, we can use these values to calculate the mean and the standard deviation.
a.) The mean (mean = np) of the sample count X who are married is 0.02*300=6, and the standard deviation would be sqrt(n*p*(1-p)) = sqrt(300*0.02*0.98) = √5.88≈2.43.
b.) The probability that, in our sample of 300, we find that 8 of the seniors are married is given by the binomial formula P(X=k) = C(n,k)(p^k)(1-p)^(n-k). Plugging n=300, k=8, p=0.02 into the formula, we get the desired probability.
c.) The probability that we find less than 4 of the seniors are married is sum of P(X=k) from k=0 to 3. This could be computed using the aforementioned binomial formula. Remember, you're summing the probabilities for each k.
d.) The probability that we find at least 1 of the seniors are married can be found by subtracting the probability that none of the seniors are married from 1 (i.e., P(X >=1) = 1 - P(X=0)).
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A computer virus is trying to corrupt two files. The first file will be corrupted with probability 0.4. Independently of it, the second file will be corrupted with probability 0.3. (a) Compute the probability mass function (pmf) of X, the number of corrupted files.
Answer:
[tex]P(X = 0) = 0.42[/tex]
[tex]P(X = 1) = 0.46[/tex]
[tex]P(X = 2) = 0.12[/tex]
Step-by-step explanation:
We have these following probabilities:
40% probability that the first file is corrupted. So 60% probability that the first file is not corrupted.
30% probability that the second file is corrupted. So 70% probability that the second file is not corrupted.
Probability mass function
Probability of each outcome(0, 1 and 2 files corrupted).
No files corrupted:
60% probability that the first file is not corrupted.
70% probability that the second file is not corrupted.
So
[tex]P(X = 0) = 0.6*0.7 = 0.42[/tex]
One file corrupted:
First one corrupted, second no.
40% probability that the first file is corrupted.
70% probability that the second file is not corrupted.
First one ok, second one corrupted.
60% probability that the first file is not corrupted.
30% probability that the second file is corrupted.
[tex]P(X = 1) = 0.4*0.7 + 0.6*0.3 = 0.46[/tex]
Two files corrupted:
40% probability that the first file is corrupted.
30% probability that the second file is corrupted.
[tex]P(X = 2) = 0.4*0.3 = 0.12[/tex]
A group of students bakes 100 cookies to sell at the school bake sale. The students want to ensure that the price of each cookie offsets the cost of the ingredients. If all the cookies are sold for $0.10 each, the net result will be a loss of $4. If all the cookies are sold for $0.50 each. The students will make a $36 profit. First, write the linear function p(x) that represents the net profit from selling all the cookies, where x is the price of each cookie. Then, determine how much profit the students will make if they sell the cookies for $0.60 each. Explain. Tell how your answer is reasonable.
Answer:
46
Step-by-step explanation:
-Let b be the constant in the linear equation.
#The linear equation can be expressed as:
[tex]p(x)=100x+b[/tex]
Substitute the values in the equation to find b:
[tex]p(x)=100x+b\\\\-4=100(0.1)+b\\\\b=-14\\\\\#or\\\\36=100(0.5)+b\\\\b=-14[/tex]
We know have the constant value b=-14, substitute the values of b and x in the p(x) function:
[tex]p(x)=100x+b\\\\p(x)=100(0.6)-14\\\\p(x)=60-14\\\\p(x)=46[/tex]
Hence, the profit when selling price is $0.60 is $46
#From our calculations, it's evident that the cookies production has a very high fixed cost which can only be offset by raisng the selling price or the number of units sold at any given time.
If the students sell the cookies for $0.60 each, they will make a profit of $46.
To solve this problem, let's first define the variables and set up the linear function p(x) that represents the net profit based on the selling price x per cookie.
Given information:
- Selling each cookie for $0.10 results in a net loss of $4.
- Selling each cookie for $0.50 results in a net profit of $36.
From this information, we can set up two equations based on the net profit:
1. When selling each cookie for $0.10:
[tex]\[ R = 100 \cdot 0.10 = 10 \] \[ P(0.10) = R - C = 10 - C = -4 \] \[ C = 10 + 4 = 14 \][/tex]
(Total cost of ingredients)
2. When selling each cookie for $0.50:
[tex]\[ R = 100 \cdot 0.50 = 50 \] \[ P(0.50) = R - C = 50 - C = 36 \] \[ C = 50 - 36 = 14 \][/tex]
Total cost of ingredients)
So, the total cost of ingredients C is $14 regardless of the selling price, since it's consistent in both scenarios.
Now, let's define the linear function P(x) :
[tex]\[ P(x) = R - C \][/tex]
Where ( R = 100x ) (total revenue from selling 100 cookies at x dollars each), and ( C = 14 ) (total cost of ingredients).
Therefore,
[tex]\[ P(x) = 100x - 14 \][/tex]
This function P(x) gives us the net profit when each cookie is sold for x dollars.
Now, to find out how much profit the students will make if they sell the cookies for $0.60 each:
[tex]\[ x = 0.60 \]\[ P(0.60) = 100 \cdot 0.60 - 14 \]\[ P(0.60) = 60 - 14 \]\[ P(0.60) = 46 \][/tex]
So, if the students sell each cookie for $0.60, they will make a profit of $46.
Explanation of Reasonableness:
The function [tex]\( P(x) = 100x - 14 \)[/tex] is a linear function that accurately represents the relationship between the selling price x and the net profit ( P(x) ). The function is derived from the given conditions where selling at $0.10 results in a loss and selling at $0.50 results in a profit, confirming the slope and intercept of the function.
You wish to estimate the average weight of a mouse. You obtain 10 mice, sampled uniformly at random and with replacement from the mouse population. Their weights are 21; 23; 27; 19; 17; 18; 20; 15; 17; 22 grams respectively. (a) What is the best estimate for the average weight of a mouse, from this data
Answer:
The best estimate for the average weight of a mouse, from this data is 19.9 grams.
Step-by-step explanation:
The best estime for the weight of a mouse from this data is the sum of all these weights divided by the number of mices.
10 mices
Their weights are 21; 23; 27; 19; 17; 18; 20; 15; 17; 22 grams
So
[tex]M = \frac{21+23+27+19+17+18+20+15+17+22}{10} = 19.9[/tex]
The best estimate for the average weight of a mouse, from this data is 19.9 grams.
Your DVD membership costs $16 per month for 10 DVD rentals. Each additional DVD rental is $2. a. Write an equation in two variables that represents the monthly cost of your DVD rentals. b. Identify the independent and dependent variables. c. How much does it cost to rent 15 DVDs in one month?
C(15) = $26
Step-by-step explanation:
According to a 2013 study by the Pew Research Center, 15% of adults in the United States do not use the Internet (Pew Research Center website, December, 15, 2014). Suppose that 10 adults in the United States are selected randomly.
a. Is the selection of the 10 adults a binomial experiment? Explain.
b. What is the probability that none of the adults use the Internet (to 4 decimals)?
c. What is the probability that 3 of the adults use the Internet (to 4 decimals)? If you calculate the binomial probabilities manually, make sure to carry at least 4 decimal digits in your calculations.
d. What is the probability that at least 1 of the adults uses the Internet (to 4 decimals)?
Answer:
a) For this case we can use the binomial model since we assume independent events and the same probability for each trial is the same p =0.15
b) [tex]P(X=0)=(10C0)(0.15)^0 (1-0.15)^{10-0}=0.1969[/tex]
c) [tex]P(X=3)=(10C3)(0.15)^3 (1-0.15)^{10-3}=0.1298[/tex]
d) [tex] P(X \geq 1)= 1-P(X <1) = 1-P(X=0)[/tex]
And using the result from part a we got:
[tex] P(X \geq 1)= 1-P(X <1) = 1-P(X=0)= 1-0.1969 =0.8031[/tex]
Step-by-step explanation:
Previous concepts
The binomial distribution is a "DISCRETE probability distribution that summarizes the probability that a value will take one of two independent values under a given set of parameters. The assumptions for the binomial distribution are that there is only one outcome for each trial, each trial has the same probability of success, and each trial is mutually exclusive, or independent of each other".
Let X the random variable of interest, on this case we now that:
[tex]X \sim Binom(n p)[/tex]
The probability mass function for the Binomial distribution is given as:
[tex]P(X)=(nCx)(p)^x (1-p)^{n-x}[/tex]
Where (nCx) means combinatory and it's given by this formula:
[tex]nCx=\frac{n!}{(n-x)! x!}[/tex]
Solution to the problem
Part a
For this case we can use the binomial model since we assume independent events and the same probability for each trial is the same p =0.15
Part b
For this case we want this probability:
[tex] P(X=0)[/tex]
And replacing we got:
[tex]P(X=0)=(10C0)(0.15)^0 (1-0.15)^{10-0}=0.1969[/tex]
Part c
For this case we want this probability:
[tex] P(X=3)[/tex]
And replacing we got:
[tex]P(X=3)=(10C3)(0.15)^3 (1-0.15)^{10-3}=0.1298[/tex]
Part d
For this cae we want thi probability:
[tex] P(X \geq 1)[/tex]
And we can use the complment rule and we got:
[tex] P(X \geq 1)= 1-P(X <1) = 1-P(X=0)[/tex]
And using the result from part a we got:
[tex] P(X \geq 1)= 1-P(X <1) = 1-P(X=0)= 1-0.1969 =0.8031[/tex]
Use inverse trigonometric functions to solve the following equations. If there is more than one solution, enter all solutions as a comma-separated list (like "1, 3"). If an equation has no solutions, enter "DNE".solve tan ( θ ) = 1 tan(θ)=1 for θ θ (where 0 ≤ θ < 2 π 0≤θ< 2π).
The solutions to the equation tan(θ) = 1 within the specified range 0 ≤ θ < 2π: θ = 0.7854, 3.9270
Apply the inverse tangent function:
We begin by applying the inverse tangent function (arctan) to both sides of the equation: arctan(tan(θ)) = arctan(1)
Since arctan is the inverse of tangent, they cancel each other out on the left side, leaving us with: θ = arctan(1)
Determine the reference angle:
arctan(1) = π/4, which is the reference angle in the first quadrant where tangent is 1.
Find solutions in other quadrants:
The tangent function has a period of π, meaning it repeats its values every π radians.
Since tangent is also positive in the third quadrant, we add π to the reference angle to find the solution in that quadrant: θ = π/4 + π = 5π/4
Consider the specified range:
We're given the range 0 ≤ θ < 2π. Both π/4 and 5π/4 fall within this range, so they are the valid solutions.
Therefore, the solutions to the equation tan(θ) = 1 within the specified range are θ = 0.7854 (π/4) and θ = 3.9270 (5π/4).
Final answer:
To solve the equation tan(θ) = 1 for θ, we need to use the inverse tangent function. The solution to the equation is θ = π/4.
Explanation:
To solve the equation tan(θ) = 1 for θ, we need to use the inverse trigonometric function. In this case, we will use the inverse tangent function, also known as arctan or atan.
Applying the inverse tangent function to both sides of the equation, we get θ = atan(1).
Using a calculator, we find that atan(1) = π/4. Therefore, the solution to the equation is θ = π/4.
y=−7x+3 y=−x−3
Find the solution to the system of equations.
Answer:
(x,y)=(1,-4)
Step-by-step explanation:
y=−7x+3
y=−x−3
(y=) −7x+3=−x−3
-7x+x=-3-3
-6x=-6
x=-6/(-6)
x=1
y=-7*1+3=-7+3=-4
(x,y)=(1,-4)
Answer:
[tex](x,y)= (1,-4)\\[/tex]
Step-by-step explanation:
We will solve it using the substitution method
Using Substitution method
Let [tex]y = -7x + 3[/tex] be equation 1 and [tex]y = -x - 3[/tex] be equation 2
putting value of y from equation 1 in equation 2 and further simplifying:
we get
[tex]-7x +3 = -x - 3\\-7x + x = -3 -3\\-6x =-6\\\\6x=6x\\x= 1[/tex]
Now put value of x i.e. [tex]x=1[/tex] in equation 1 and by further simplifying
[tex]y = -7x + 3\\y= -7(1) +3\\y= -7+3\\y=-4[/tex]
So the solution to the system is written as\[tex](x,y)= (1,-4)[/tex]
1. A manufacturer of a printer determines that the mean number of days before a cartridge runs out of ink is 75 days, with a standard deviation of 6 days. Assuming a normal distribution, what is the probability that the number of days will be less than 67.5 days?
Answer:
[tex]P(X<67.5)=P(\frac{X-\mu}{\sigma}<\frac{67.5-\mu}{\sigma})=P(Z<\frac{67.5-75}{6})=P(z<-1.25)[/tex]
And we can find this probability using the normal standard table or excel:
[tex]P(z<-1.25)=0.106[/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 number of days before cartridge runs out of a population, and for this case we know the distribution for X is given by:
[tex]X \sim N(75,6)[/tex]
Where [tex]\mu=75[/tex] and [tex]\sigma=6[/tex]
We are interested on this probability
[tex]P(X<67.5)[/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<67.5)=P(\frac{X-\mu}{\sigma}<\frac{67.5-\mu}{\sigma})=P(Z<\frac{67.5-75}{6})=P(z<-1.25)[/tex]
And we can find this probability using the normal standard table or excel:
[tex]P(z<-1.25)=0.106[/tex]
Following Exercise 3.5.9, let p1, . . . , pk be a pairwise relatively prime set of naturals, each greater than 1. Let X be the set {0, 1, . . . , p1 −1}× . . . ×{0, 1, . . . , pk −1}. Define a function f from {0, 1, . . . , p1p2 . . . pk − 1} to X by the rule f(x) = x%p1, . . . , x%pk. Prove that f is a subject
Answer: see the pictures attached
Step-by-step explanation:
You can now sell 80 cups of lemonade per week at 40¢ per cup, but demand is dropping at a rate of 4 cups per week each week. Assuming that raising the price does not affect demand, how fast do you have to raise your price if you want to keep your weekly revenue constant? HINT [Revenue = Price × Quantity.]
Final answer:
To keep the weekly revenue constant while demand drops, we can set up an equation using the revenue formula. By equating the original revenue with the new revenue, we can find the rate at which the price needs to be raised. Taking the derivative, we can determine the rate of change of the price.
Explanation:
To keep the weekly revenue constant, we need to find the rate at which the price has to be raised to offset the drop in demand. Currently, the price is 40¢ per cup and demand is dropping at a rate of 4 cups per week. Since revenue is equal to price times quantity, we can set up the equation:
Revenue = Price × Quantity.
Initially, we have 80 cups of lemonade sold at 40¢ per cup, resulting in a revenue of $32 (80 x 40¢). As demand drops by 4 cups per week each week, the new quantity sold can be represented by 80 - 4t, where t represents the number of weeks. Let P be the new price per cup that needs to be raised. The new revenue equation can be written as:
Revenue = P(80 - 4t).
To find the value of P, we equate the original revenue ($32) with the new revenue:
$32 = P(80 - 4t).
Simplifying the equation, we get:
32 = 80P - 4Pt.
Moving the terms around, we have:
4Pt = 80P - 32.
Dividing both sides by 4P, we get:
t = (80P - 32)/(4P).
So, the rate at which the price needs to be raised to keep the weekly revenue constant is given by the derivative of t with respect to P. Taking the derivative, we get:
t' = (4(80P - 32) - 4P(80))/(4P)^2.
Simplifying further, we have:
t' = (320P - 128 - 320P)/(4P)^2.
Simplifying again, we get:
t' = -128/(4P)^2.
Thus, the rate of change of t with respect to P is given by -128/(4P)^2. This represents the rate at which the price needs to be raised in order to keep the weekly revenue constant.
The following data on average daily hotel room rate and amount spent on entertainment (The Wall Street Journal, August 18, 2011) lead to the estimated regression equation ŷ = 17.49 + 1.0334x. For these data SSE = 1541.4.
City Room Rate ($) Entertainment ($)
Boston 148 161
Denver 96 105
Na.shville 91 101
New Orleans 110 142
Phoenix 90 100
San Diego 102 120
San Francisco 136 167
San Jose 90 140
Tampa 82 98
(a) Predict the amount spent on entertainment for a particular city that has a daily room rate of $89 (to 2 decimals).
(b) Develop a 95% confidence interval for the mean amount spent on entertainment for all cities that haye a daily room rate of $89 (to 2 decimals).
(c) The average room rata in Chicago is $128. Develop a 95% prediction interval for the amount spent on entertainment in Chicago (to 2 decimals).
Answer:
a. Predicted Amount = $109.46
b. Confidence Interval = (94.84,124.08)
c. Interval = (110.6883,188.8517)
Step-by-step explanation:
Given
ŷ = 17.49 + 1.0334x.
SSE = 1541.4
a.
ŷ = 17.49 + 1.0334(89)
ŷ = 109.4626
ŷ = 109.46 --- Approximated
Predicted Amount = $109.46
b.
ŷ = 17.49 + 1.0334(89)
ŷ = 109.4626
ŷ = 109.46
First we calculate the standard deviation
variance = SSE/(n-2)
v = 1541.4/(9-2)
v = 1541.4/7
v = 220.2
s = √v
s = √220.2
s = 14.839
Then we calculate mean(x) and ∑(x - (mean(x))²
X --- Y -- Mean(x) --- ∑(x - (mean(x))²
148 -- 161 -- 43-- 1849
96 || 105|| -9 || 81
91 ||101 || -14 || 196
110 || 142 || 5 || 25
90 || 100 || -15 || 225
102 || ||120 ||-3|| 9
136 || 167 ||31 ||961
90 || 140 ||-15 ||225
82 || 98 ||-23 || 529
Sum 945 || 1134|| 0 ||4100
Mean (x) = 945/9 = 105
∑(x - (mean(x))² = 4100
α = 1 - 95% = 5%
α/2 = 2.5% = 0.025
tα,df = n − 2 = t0.025,7 =2.365
Confidence interval = 109.46 ± 2.365 * 14.839 √((1/9)+ (89-105)²/4100
Confidence Interval = (109.46 ± 14.62)
Confidence Interval = (94.84,124.08)
c.
ŷ = 17.49 + 1.0334(128)
ŷ = 149.7652
ŷ = 149.77
Interval = 149.77 ± 2.365 * 14.839 √((1/9)+ (128-105)²/4100
Interval = 149.77 ± 39.0817
Interval = (110.6883,188.8517)
Given the regression equation ŷ = 17.49 + 1.0334x, we can predict the amount spent on entertainment in cities based on their daily room rate. For instance, a city with a daily room rate of $89 is estimated to spend about $109.67 on entertainment. However, we don't have enough information to calculate the 95% confidence interval or the 95% prediction interval.
Explanation:To solve these questions, we use the provided regression equation, which is ŷ = 17.49 + 1.0334x. The variable 'x' represents the daily room rate, and 'ŷ' represents the predicted amount spent on entertainment.
(a) To predict the amount spent on entertainment for a city that has a daily room rate of $89, substitute x with 89 in the equation: ŷ = 17.49 + 1.0334 * 89. The computed prediction is $109.67.
(b) To develop a 95% confidence interval for the mean amount spent on entertainment for all cities with a daily room rate of $89, we would need additional statistical data such as the standard error or the number of data points. There isn't sufficient information in the question to accurately compute this.
(c) To find the 95% prediction interval for the amount spent on entertainment in Chicago with an average room rate of $128, we would also need additional statistical data like the standard error, degrees of freedom, or the number of observations. Again, the question does not provide sufficient details to calculate this.
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Draw a rectangle that shows 8 equal parts . Shade more than 3/8 of the rectangle but less than 5/8 .what fraction did you model? Use multiplication and division to write two equivalent fractions for your model.
Answer:
4/8 more than 3/ but less than 5/8
Answer: I modeled 4/8 because it is greater than 3 less than 5 2 equivalent fractions are 8/16 12/24
Step-by-step explanation:
Brian is filling a conic container with water. He has the container half full. The radius of the container is 5 inches and the height is 20 inches. What is the current volume of the water?
The current volume of the water is 261.66 square inches.
Solution:
The container is in cone shape.
Radius of the container = 5 inch
Height of the container = 20 inch
Volume of the container = [tex]\frac{1}{3} \pi r^2 h[/tex]
[tex]$=\frac{1}{3}\times 3.14 \times 5^2 \times 20[/tex]
Volume of the container = 523.33 square inch
Current volume of the water = Half of the volume of container
[tex]$=\frac{1}{2}\times523.33[/tex]
= 261.66 square inch
The current volume of the water is 261.66 square inches.
In order to estimate the height of all students at your university, let's assume you have measured the height of all psychology majors at the university. The resulting raw scores are called _________. constants data coefficients statistics
Answer:
Data
Step-by-step explanation:
We are given the following in the question:
We want to measure height of all psychology majors at the university.
Thus, the resulting raw scores of each individual are called the data.
Data point:
Height of each psychology majors at the university
Data:
Collection of all heights of all psychology majors at the university
These value are constants but comprises a data.
They are neither coefficients nor statistic because they do not describe a sample.
Thus, the correct answer is
Data
A company determines that its marginal cost, in dollars, for producing x units of a product is given by Upper C prime (x )equals4500 x Superscript negative 1.9, where xgreater than or equals1.11. Suppose that it were possible for the company to make infinitely many units of this product. What would the total cost be?
Answer:
Total Cost = Fixed Cost as x --> ∞
Step-by-step explanation:
C'(x) = 4500 x⁻¹•⁹ where x ≥ 1
Marginal Cost = C'(x) = (dC/dx)
C(x) = ∫ (marginal cost) dx
C(x) = ∫ (4500 x⁻¹•⁹)
C(x) = (-5000 x⁻⁰•⁹) + k
where k = constant of integration or in economics term, K = Fixed Cost.
C(x) = [-5000/(x⁰•⁹)] + Fixed Cost
The company wants to make infinitely many units, that is, x --> ∞
C(x --> ∞) = [-5000/(∞⁰•⁹)] + Fixed Cost
(∞⁰•⁹) = ∞
C(x --> ∞) = [-5000/(∞)] + Fixed Cost
But mathematically, any number divide by infinity = 0;
(-5000/∞) = 0
C(x --> ∞) = 0 + Fixed Cost = Fixed Cost.
Total Cost of producing infinite number of units for this cost function is totally the Fixed Cost.