Answer:
There is sufficient evidence to conclude that child booster seats meet the specific requirement.
Step-by-step explanation:
Sample: 697, 759, 1266, 621, 569, 432
Formula:
[tex]\text{Standard Deviation} = \sqrt{\displaystyle\frac{\sum (x_i -\bar{x})^2}{n-1}}[/tex]
where [tex]x_i[/tex] are data points, [tex]\bar{x}[/tex] is the mean and n is the number of observations.
[tex]Mean = \displaystyle\frac{\text{Sum of all observations}}{\text{Total number of observation}}[/tex]
[tex]Mean =\displaystyle\frac{4344}{6} = 724[/tex]
Sum of squares of differences = 415616
[tex]S.D = \sqrt{\frac{415616}{5}} = 288.31[/tex]
We are given the following in the question:
Population mean, μ = 1000 hic
Sample mean, [tex]\bar{x}[/tex] = 724
Sample size, n = 16
Alpha, α = 0.05
Sample standard deviation, s = 288.31
First, we design the null and the alternate hypothesis
[tex]H_{0}: \mu = 1000\text{ hic}\\H_A: \mu < 1000\text{ hic}[/tex]
We use one-tailed(left) t test to perform this hypothesis.
Formula:
[tex]t_{stat} = \displaystyle\frac{\bar{x} - \mu}{\frac{\sigma}{\sqrt{n}} }[/tex]
Putting all the values, we have
[tex]t_{stat} = \displaystyle\frac{724 - 1000}{\frac{288.31}{\sqrt{6}} } = -2.344[/tex]
Now, [tex]t_{critical} \text{ at 0.05 level of significance, 5 degree of freedom } = -2.015[/tex]
Calculation the p-value from table,
P-value = 0.033
Since,
Since, the p value is lower than the significance level, we fail to accept the null hypothesis and reject it. We accept the alternate hypothesis.
We conclude that the measurement is less than 1000 hic.
Thus, there is sufficient evidence to conclude that child booster seats meet the specific requirement.
Find the coordinates of the orthocenter of ABC. A(-1,0) B(0,4) C(3,0)
Answer:
[tex](0,0.75) \:or\:(0,\frac{3}{4})[/tex]
Step-by-step explanation:
Hi there!
1) Firstly, connect the points to draw a triangle.
2) From each vertex either with a pair of square or with a software trace a perpendicular line to the opposite side.
3) The concurrent point, i.e. the intersection point of these three altitudes is the orthocenter. Orthocenter means the the right center.
In equilateral triangles the Orthocenter coincides with the Centroid.
4) Finally, the coordinates of the Orthocenter found is (0,0.75)
A random sample of 16 students selected from the student body of a large university had an average age of 25 years and a population standard deviation of 2 years. We want to determine if the average age of all the students at the university is significantly more than 24. Assume the distribution of the population of ages is normal.
What is the alternative and null hypotheses?
What is the test statistic?
What is the p-value?
What is your conclusion about the stated hypotheses at a 95% confidence level?
Answer:
[tex]t=\frac{25-24}{\frac{2}{\sqrt{16}}}=2[/tex]
[tex]p_v =P(t_{15}>2)=0.0320[/tex]
If we compare the p value and the significance level given for example [tex]\alpha=0.05[/tex] we see that [tex]p_v<\alpha[/tex] so we can conclude that we reject the null hypothesis, and the true mean is significant higher than 24 years.
Step-by-step explanation:
1) Data given and notation
[tex]\bar X=25[/tex] represent the sample mean
[tex]s=2[/tex] represent the standard deviation for the sample
[tex]n=16[/tex] sample size
[tex]\mu_o =24[/tex] represent the value that we want to test
[tex]\alpha[/tex] represent the significance level for the hypothesis test.
t would represent the statistic (variable of interest)
[tex]p_v[/tex] represent the p value for the test (variable of interest)
Confidence =0.95 or 95%
[tex]\alpha=0.05[/tex]
State the null and alternative hypotheses.
We need to conduct a hypothesis in order to determine if the mean is higher than 24, the system of hypothesis would be:
Null hypothesis:[tex]\mu \leq 24[/tex]
Alternative hypothesis:[tex]\mu > 24[/tex]
We don't know the population deviation, so for this case is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:
[tex]t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}[/tex] (1)
t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".
Calculate the statistic
We can replace in formula (1) the info given like this:
[tex]t=\frac{25-24}{\frac{2}{\sqrt{16}}}=2[/tex]
Calculate the P-value
First we need to calculate the degrees of freedom given by:
[tex]df=n-1=16-1=15[/tex]
Since is a one-side upper test the p value would be:
[tex]p_v =P(t_{15}>2)=0.0320[/tex]
Conclusion
If we compare the p value and the significance level given for example [tex]\alpha=0.05[/tex] we see that [tex]p_v<\alpha[/tex] so we can conclude that we reject the null hypothesis, and the true mean is significant higher than 24 years.
Consider the equivalence relation R = {( x, y) Ix-y is an integer}.
(a) What is the equivalence class of 1 for this equivalence relation?
(b) What is the equivalence class of 1/2 for this equivalence relation?
Final answer:
The equivalence class of 1 consists of all integers plus 1, and the equivalence class of 1/2 consists of all numbers of the form 1/2 plus any integer.
Explanation:
The equivalence relation R is defined such that (x, y) is in R if and only if x - y is an integer. For any real number a, the equivalence class of a is the set of all real numbers b such that a - b is an integer.
Equivalence Class of 1
The equivalence class of 1 includes all real numbers that are an integer distance from 1. This means it contains all numbers of the form 1 + k, where k is any integer. Hence, it includes numbers like 0, 1, 2, 3, and so on, in addition to negative integers: -1, -2, -3, etc.
Equivalence Class of 1/2
Similarly, the equivalence class of 1/2 consists of all real numbers of the form 1/2 + k, where k is any integer. This set includes numbers like -1/2, 1/2, 3/2, 5/2, and so on.
A random sample of 30 varieties of cereal was selected. The average number of calories per serving for these cereals is x-120. Assuming that σ = 10, find a 95% confidence interval for the mean number of calories, μ, in a serving of cereal. a. 115.30 to 124.70 b. 116.42 to 123.58 c. 118.00 to 122.00 d. 117.00 to 123.00
Answer: confidence interval = b. ( 116.42 to 123.58)
Step-by-step explanation:
Given;
Number of samples n = 30
Standard deviation r = 10
Mean x = 120
Confidence interval of 95%
Z' = t(0.025) = 1.96
Confidence interval = x +/- Z'(r/√n)
= 120 +/- 1.96(10/√30)
= 120 +/- 3.58
= ( 116.42, 123.58)
Products produced by a machine has a 3% defective rate. (a) What is the probability that the first defective occurs in the fifth item inspected? (b) What is the probability that the first defective occurs in the first five inspections? (c) What is the expected number of inspections before the first defective occurs?
Answer:
Step-by-step explanation:
Given that the Products produced by a machine has a 3% defective rate.
Each product is independent of the other with a constant prob of being defective as 0.03
X - no of defects is binomial with p =0.03
a) the probability that the first defective occurs in the fifth item inspected
=Prob for first 4 non defective and 5th defective
=[tex](0.97)^4(0.03)^1\\=0.0266[/tex]
(b) the probability that the first defective occurs in the first five inspections
=P(X=1) in binomial with n=5
= 0.9915
c) the expected number of inspections before the first defective occurs
Expected defects in n trials = np
Expected number of inspection before the first defect = 1/p
= 33.33
=34
(c) What is the expected number of inspections before the first defective occurs?
Final answer:
The probability that the first defective product occurs on the fifth item is approximately 0.0895. The probability that a defective product occurs within the first five inspections is approximately 0.1426. The expected number of inspections before a defective product occurs is around 33 items.
Explanation:
To find the probability that the first defective occurs on the fifth item inspected (part a), we consider the first four items to be non-defective and the fifth one to be defective. The probability for one non-defective item is 97%, or 0.97. The probability we are looking for is thus the product of these probabilities:
P(non-defective, non-defective, non-defective, non-defective, defective) = (0.97)⁴* 0.03 ≈ 0.0895
For part b, we want to find the probability that a defective is found at any point in the first five inspections. This is the sum of the probabilities of finding the first defective on the first, second, third, fourth, or fifth inspection:
P(1st) + P(2nd) + P(3rd) + P(4th) + P(5th) = 0.03 + (0.97 * 0.03) + (0.97² * 0.03) + (0.97³ * 0.03) + (0.97⁴ * 0.03) ≈ 0.1426
For part c, the expected number of inspections before finding the first defective item is given by the formula E(X) = 1/p, where p is the probability of finding a defective item. Substituting the given probability:
E(X) = 1/0.03 ≈ 33.33
Therefore, we expect to inspect approximately 33 items before finding the first defective.
According to a recent study, 1 in every 9 women has been a victim of domestic abuse at some point 19) in her life. Suppose we have randomly and independently sampled twenty-five women and asked each whether she has been a victim of domestic abuse at some point in her life.
1. Find the probability that at least 2 of the women sampled have been the victim of domestic abuse. Round to six decimal places.
Answer:
[tex]P(X\geq 2)=1-P(X\leq 1)=1-[0.054294+0.167762]=0.777944[/tex]
Step-by-step explanation:
Previous concepts
A Bernoulli trial is "a random experiment with exactly two possible outcomes, "success" and "failure", in which the probability of success is the same every time the experiment is conducted". And this experiment is a particular case of the binomial experiment.
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".
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]
The complement rule is a theorem that provides a connection between the probability of an event and the probability of the complement of the event. Lat A the event of interest and A' the complement. The rule is defined by: [tex]P(A)+P(A') =1[/tex]
Find the probability that at least 2 of the women sampled have been the victim of domestic abuse.
On this case we want to find this probability
[tex]P(X\geq 2) =1-P(X<2)=1-P(X\leq 1)= 1-[P(X=0)+P(X=1)][/tex]
And we can find the individual probabilities like this:
[tex]P(X=0)=(25C0)(0.11)^0 (1-0.11)^{25-0}=0.054294[/tex]
[tex]P(X=1)=(25C1)(0.11)^1 (1-0.11)^{25-1}=0.167762[/tex]
[tex]P(X\geq 2)=1-P(X\leq 1)=1-[0.054294+0.167762]=0.777944[/tex]
Using the binomial distribution, it is found that there is a 0.287825 = 28.7825% probability that at least 2 of the women sampled have been the victim of domestic abuse.
What is the binomial distribution formula?The formula is:
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
[tex]C_{n,x} = \frac{n!}{x!(n-x)!}[/tex]
The parameters are:
x is the number of successes.n is the number of trials.p is the probability of a success on a single trial.In this problem:
1 in every 9 women has been a victim of domestic abuse at some point in her life, hence p = 1/9 = 0.1111.25 women are sampled, hence n = 25.The probability that at least 2 of the women sampled have been the victim of domestic abuse is given by:
[tex]P(X \geq 2) = 1 - P(X < 2)[/tex]
In which:
[tex]P(X < 2) = P(X = 0) + P(X = 1)[/tex]
Hence:
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
[tex]P(X = 0) = C_{25,0}.(0.1111)^{0}.(0.8889)^{25} = 0.052641[/tex]
[tex]P(X = 1) = C_{25,1}.(0.1111)^{1}.(0.8889)^{24} = 0.164484[/tex]
[tex]P(X < 2) = P(X = 0) + P(X = 1) = 0.052641 + 0.164484 = 0.217125[/tex]
[tex]P(X \geq 2) = 1 - P(X < 2) = 1 - 0.217125 = 0.782875[/tex]
0.287825 = 28.7825% probability that at least 2 of the women sampled have been the victim of domestic abuse.
More can be learned about the binomial distribution at https://brainly.com/question/24863377
Consider the following function. Without finding the inverse, evaluate the derivative of the inverse at the given point. f(x)=ln(8x+e); (1,0)
We can use the inverse function derivative theorem:
[tex]\dfrac{\textrm{d}f^{-1}}{\textrm{d}x}\Big\vert_{x=a} = \dfrac{1}{\dfrac{\textrm{d}f}{\textrm{d}x}\Big\vert_{x=f^{-1}(a)}}.[/tex]
In this case, we want to evaluate [tex]\dfrac{\textrm{d}f^{-1}}{\textrm{d}x}\Big\vert_{x=1}[/tex], so:
[tex]\dfrac{\textrm{d}f^{-1}}{\textrm{d}x}\Big\vert_{x=1} = \dfrac{1}{\dfrac{\textrm{d}f}{\textrm{d}x}\Big\vert_{x=f^{-1}(1)}}.[/tex]
The derivative is:
[tex]\dfrac{\textrm{d}f}{\textrm{d}x} = \dfrac{\textrm{d}}{\textrm{d}x}\left[\ln(8x + \textrm{e})\right] = \dfrac{1}{8x+\textrm{e}}\dfrac{\textrm{d}}{\textrm{d}x}\left(8x + \textrm{e}\right) = \dfrac{8}{8x+\textrm{e}}.[/tex]
The ordinate of the point is [tex]f^{-1}(1) = 0[/tex], so we evaluate:
[tex]\dfrac{\textrm{d}f}{\textrm{d}x}\Big\vert_{x=0} = \dfrac{8}{8 \times 0+\textrm{e}} = \dfrac{8}{\textrm{e}}.[/tex]
Finally:
[tex]\dfrac{\textrm{d}f^{-1}}{\textrm{d}x}\Big\vert_{x=1} = \dfrac{1}{\dfrac{\textrm{d}f}{\textrm{d}x}\Big\vert_{x=f^{-1}(1)}} = \dfrac{1}{\dfrac{\textrm{d}f}{\textrm{d}x}\Big\vert_{x=0}} = \dfrac{1}{\dfrac{8}{\textrm{e}}} = \dfrac{\textrm{e}}{8}.[/tex]
We can check the answer by finding the inverse:
[tex]y = \ln(8x + \textrm{e}) \implies \textrm{e}^y = 8x + \textrm{e} \iff \textrm{e}^y - \textrm{e} = 8x \iff x = \dfrac{\textrm{e}^y-\textrm{e}}{8},[/tex]
so that
[tex]f^{-1}(x) = \dfrac{\textrm{e}^x-\textrm{e}}{8}.[/tex]
Therefore:
[tex]\dfrac{\textrm{d}f^{-1}}{\textrm{d}x} = \dfrac{\textrm{e}^x}{8}.[/tex]
Which finally gives the same answer as before:
[tex]\dfrac{\textrm{d}f^{-1}}{\textrm{d}x}\Big\vert_{x=1} = \dfrac{\textrm{e}^1}{8} = \dfrac{\textrm{e}}{8}.[/tex]
Answer: [tex]\boxed{\dfrac{\textrm{d}f^{-1}}{\textrm{d}x}\Big\vert_{x=1} = \dfrac{\textrm{e}}{8}}.[/tex]
"The Munchies Cereal Company makes a cereal from several ingredients. Two of the ingredients, oats and rice, provide vitamins A and B. The company wants to know how many ounces of oats and rice it should include in each box of cereal to meet the minimum requirements of 48 milligrams of vitamin A and 12 milligrams of vitamin B while minimizing cost. An ounce of oats contributes 8 milligrams of vitamin A and 1 milligram of vitamin B, whereas an ounce of rice contributes 6 milligrams of A and 2 milligrams of B. An ounce of oats costs $0.05, and an ounce of rice costs $0.03. a. Formulate a linear programming model for this problem. b. Solve this model by using graphical analysis."
Answer:
8x + 6y >/= 48 ......1
x + 2y >/= 12 .......2
The cost function is given as;
C = 0.05x + 0.03y .........3
The minimum cost is $0.24 at (0,8)
That is 0 ounces of oats and 8 ounces of rice.
Step-by-step explanation:
let x represent the number of ounces of oats
And y represent the number of ounces of rice
For Vitamin A
Minimum requirements = 48mg
x ounces of oats contribute 8mg × x
y ounces of Rice contribute 6mg × y
Therefore, we have;
8x + 6y >/= 48 ......1
For Vitamin B
Minimum requirements = 12mg
x ounces of oats contribute 1mg × x
y ounces of Rice contribute 2mg × y
Therefore, we have;
x + 2y >/= 12 .......2
The cost function is given as;
C = 0.05x + 0.03y .........3
Attached is the graphical representation.
The feasible points are (x,y) = (0,8),(2.4,4.8),(12,0)
The minimum cost is determined by substituting each point into the cost function
For (0,8)
C= 0.05(0) + 0.03(8)
C = $0.24
For (12,0)
C= 0.60
For (2.4,4.8)
C= $0.264
The minimum cost is $0.24 at (0,8)
The problem can be modelled with a system of linear inequalities to represent the constraints of the cereal company. You graph these constraints and find the feasible region. After graphing the objective function, move this line towards the origin until it just leaves the feasible region. This point gives the optimal solution.
Explanation:In this problem, we are dealing with linear equations and inequalities. The goal of the Munchies Cereal Company is to determine the amount of oats and rice, measured in ounces, to include in its cereal mix so as to meet the minimum requirements of 48 milligrams of vitamin A and 12 milligrams of vitamin B while at the same time minimizing the cost.
Let's denote the amount of oats as 'x' and the amount of rice as 'y'. The nutrition constraints can be formulated as:
8x + 6y >= 48 (to meet the requirement for vitamin A)x + 2y >= 12 (to meet the requirement for Vitamin B)And since quantities cannot be negative, we also have the constraints: x >= 0 and y >= 0. The objective is to minimize the cost, which can be expressed as C = 0.05x + 0.03y.
To solve this problem graphically, you would plot the constraint lines and see the feasible region (the area that satisfies all constraints). The cost line (C = 0.05x + 0.03y) is then drawn and moved towards the origin until the last point of the feasible region is touched. That point gives the optimal solution.
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As risk management officer at your firm, you are in charge of analyzing the data on personal injury claims filed against your firm. Some summary statistics for a random sample ofthe costs of 100 claims filed in the recent past are below.
Mean = $1,040.47 1st Quartile = $989.72
Median $1,039.71 3rd Quartile $1,088.18 .
Standard deviation = $89.50 !)o.:,
1. Which statement is correct?
The middle 50% ofthe costs are between $989.72 and $1,088.18."
Answer
The answer and procedures of the exercise are attached in the following archives.
Step-by-step explanation:
You will find the procedures, formulas or necessary explanations in the archive attached below. If you have any question ask and I will aclare your doubts kindly.
The correct statement is that the middle 50% of personal injury claim costs fall between $989.72 and $1,088.18, which represents the interquartile range. This range and the standard deviation are key in evaluating the distribution of claim costs.
The statement that the middle 50% of the costs are between $989.72 and $1,088.18 is correct in reference to the provided summary statistics of personal injury claims. This range is defined by the first and third quartiles, also known as the interquartile range (IQR). The IQR is a measure of variability and represents the span between the 25th percentile (first quartile) and the 75th percentile (third quartile), which indeed encompasses the middle 50% of data in a given sample.
In the context of personal injury claims costs at your firm, this means that half of the claim costs fall within that range, with fewer costs being less than $989.72 (the lower 25%) and fewer costs being more than $1,088.18 (the upper 25%). This can be useful information for assessing claims costs and preparing for future claims expenses. The provided standard deviation of $89.50 indicates the average amount that claim costs vary from the mean ($1,040.47).
The probabilities that two students will not show up for class on a beautiful spring day are 0.3 and 0.4, respectively. It is a beautiful spring day. Find the following probabilities. (See Example 6. Round your answers to two decimal places.) (a) Neither will show up for class. (b) Both will show up for class. (c) Exactly one will show up for class.
Answer:
(a) 0.12
(b) 0.42
(c) 0.46
Step-by-step explanation:
Probability of student A not showing up = 0.3
Probability of student B not showing up = 0.4
(a) Neither will show up for class
[tex]P(A\ and\ B) = 0.3*0.4 = 0.12[/tex]
(b) Both will show up for class
[tex]P (A\ nor\ B) = (1-0.3)*(1-0.4) = 0.42[/tex]
(c) Exactly one will show up for class
[tex]P(A\ or\ B) = 1 -P(A\ and\ B) - P(A\ nor\ B)\\P(A\ or\ B) = 1 -0.12-0.42 = 0.46[/tex]
Life after college. We are interested in estimating the proportion of graduates at a mid-sized university who found a job within one year of completing their undergraduate degree. Suppose we conduct a survey and find out that 348 of the 400 randomly sampled graduates found jobs. The graduating class under consideration included over 4500 students.(a) Describe the population parameter of interest. What is the value of the point estimate of this parameter?(b) Check if the conditions for constructing a confidence interval based on these data are met.(c) Calculate a 95% confidence interval for the proportion of graduates who found a job within one year of completing their undergraduate degree at this university, and interpret it in the context of the data.(d) What does "95% confidence" mean?(e) Now calculate a 99% confidence interval for the same parameter and interpret it in the context of the data.(f) Compare the widths of the 95% and 99% confidence intervals. Which one is wider? Explain.(Please show work for all problems, thank you)
Answer:
a) The parameter of interest is p who represent the proportion of graduates from this university who found a job within one year after graduating, and the estimated value is:
[tex]\hat p=\frac{348}{400}=0.87[/tex]
b) [tex]np=400*0.87=348>10[/tex]
[tex]n(1-p)=400(1-0.87)=52>10[/tex]
So both conditions are satisifed and we can construct the confidence interval.
c) The 95% confidence interval would be given (0.837;0.903).
We are confident (95%) that that the true proportion of graduates that found jobs is between 0.837 and 0.903
d) On this case that the 95% of the selected random samples will produce a 95% confidence interval that includes the true proportion of interest.
e) The 99% confidence interval would be given (0.827;0.913).
We are confident (99%) that that the true proportion of graduates that found jobs is between 0.827 and 0.913
f) The width for the 95% interval is 0.903-0.837=0.066, and for the 99% interval 0.913-0.827=0.086. And we see that the 99% is wider since we have more confidence that the true parameter of interest would be on the range provided.
Step-by-step explanation:
Data given and notation
n=400 represent the random sample taken
X=348 represent the number of graduates that found jobs in the sample
[tex]\hat p=\frac{348}{400}=0.87[/tex] estimated proportion of graduates that found jobs
[tex]\alpha[/tex] represent the significance level
z would represent the statistic to calculate the confidence interval
p= population proportion of graduates that found jobs
(a) Describe the population parameter of interest. What is the value of the point estimate of this parameter?
The parameter of interest is p who represent the proportion of graduates from this university who found a job within one year after graduating, and the estimated value is:
[tex]\hat p=\frac{348}{400}=0.87[/tex]
(b) Check if the conditions for constructing a confidence interval based on these data are met.
[tex]np=400*0.87=348>10[/tex]
[tex]n(1-p)=400(1-0.87)=52>10[/tex]
So both conditions are satisifed and we can construct the confidence interval.
(c) Calculate a 95% confidence interval for the proportion of graduates who found a job within one year of completing their undergraduate degree at this university, and interpret it in the context of the data.
The confidence interval would be given by this formula
[tex]\hat p \pm z_{\alpha/2} \sqrt{\frac{\hat p(1-\hat p)}{n}}[/tex]
For the 95% confidence interval the value of [tex]\alpha=1-0.95=0.05[/tex] and [tex]\alpha/2=0.025[/tex], with that value we can find the quantile required for the interval in the normal standard distribution.
[tex]z_{\alpha/2}=1.96[/tex]
And replacing into the confidence interval formula we got:
[tex]0.87 - 1.96 \sqrt{\frac{0.87(1-0.87)}{400}}=0.837[/tex]
[tex]0.87 + 1.96 \sqrt{\frac{0.87(1-0.87)}{400}}=0.903[/tex]
And the 95% confidence interval would be given (0.837;0.903).
We are confident (95%) that that the true proportion of graduates that found jobs is between 0.837 and 0.903
(d) What does "95% confidence" mean?
On this case that the 95% of the selected random samples will produce a 95% confidence interval that includes the true proportion of interest.
(e) Now calculate a 99% confidence interval for the same parameter and interpret it in the context of the data
For the 99% confidence interval the value of [tex]\alpha=1-0.99=0.01[/tex] and [tex]\alpha/2=0.005[/tex], with that value we can find the quantile required for the interval in the normal standard distribution.
[tex]z_{\alpha/2}=2.58[/tex]
And replacing into the confidence interval formula we got:
[tex]0.87 - 2.58 \sqrt{\frac{0.87(1-0.87)}{400}}=0.827[/tex]
[tex]0.87 + 2.58 \sqrt{\frac{0.87(1-0.87)}{400}}=0.913[/tex]
And the 99% confidence interval would be given (0.827;0.913).
We are confident (99%) that that the true proportion of graduates that found jobs is between 0.827 and 0.913
(f) Compare the widths of the 95% and 99% confidence intervals. Which one is wider?
The width for the 95% interval is 0.903-0.837=0.066, and for the 99% interval 0.913-0.827=0.086. And we see that the 99% is wider since we have more confidence that the true parameter of interest would be on the range provided.
Suppose that the average number of airline crashes in a country is 2 per month. (a) What is the probability that there will be at least 3 accidents in the next month? Probability = 0.3233 (b) What is the probability that there will be at least 6 accidents in the next two months? Probability = 1-(643/(15e^4)) (c) What is the probability that there will be at most 6 accidents in the next three months? Probability =
Answer:
a) [tex]P(X\geq 3)=1-P(X<3)=1-P(X\leq 1)=1-[0.1353+0.2707+0.2707]=0.3234[/tex]
b) [tex]P(X\geq 6)=1-P(X<6)=1-P(X\leq 5)=1-[0.0183+0.0733+0.1465+0.1954+0.1954+0.1563]=0.2148[/tex]
c) [tex]P(X\leq 6)=0.00248+0.0149+0.0446+0.0892+0.1339+0.1606+0.1606=0.6063[/tex]
Step-by-step explanation:
Let X the random variable that represent the number of airline crashes in a country. We know that [tex]X \sim Poisson(\lambda=2)[/tex]
The probability mass function for the random variable is given by:
[tex]f(x)=\frac{e^{-\lambda} \lambda^x}{x!} , x=0,1,2,3,4,...[/tex]
And f(x)=0 for other case.
For this distribution the expected value is the same parameter [tex]\lambda[/tex]
[tex]E(X)=\mu =\lambda[/tex]
(a) What is the probability that there will be at least 3 accidents in the next month?
On this case we are interested on the probability of having at least three accidents in the next month, and using the complement rule we have this:
[tex]P(X\geq 3)=1-P(X<3)=1-P(X\leq 1)=1-[P(X=0)+P(X=1)+P(X=2)][/tex]
Using the pmf we can find the individual probabilities like this:
[tex]P(X=0)=\frac{e^{-2} 2^0}{0!}=0.1353[/tex]
[tex]P(X=1)=\frac{e^{-2} 2^1}{1!}=0.2707[/tex]
[tex]P(X=2)=\frac{e^{-2} 2^2}{2!}=0.2707[/tex]
And replacing we have this:
[tex]P(X\geq 3)=1-P(X<3)=1-P(X\leq 1)=1-[0.1353+0.2707+0.2707]=0.3234[/tex]
(b) What is the probability that there will be at least 6 accidents in the next two months?
For this case since we want the amount in the next two months the rate changes [tex]\lambda=2x2= 4[/tex] accidents per 2 months.
[tex]P(X\geq 6)=1-P(X<6)=1-P(X\leq 5)=1-[P(X=0)+P(X=1)+P(X=2)+P(X=3)+P(X=4)+P(X=5)][/tex]
Using the pmf we can find the individual probabilities like this:
[tex]P(X=0)=\frac{e^{-4} 4^0}{0!}=0.0183[/tex]
[tex]P(X=1)=\frac{e^{-4} 4^1}{1!}=0.0733[/tex]
[tex]P(X=2)=\frac{e^{-4} 4^2}{2!}=0.1465[/tex]
[tex]P(X=3)=\frac{e^{-4} 4^3}{3!}=0.1954[/tex]
[tex]P(X=4)=\frac{e^{-4} 4^4}{4!}=0.1954[/tex]
[tex]P(X=5)=\frac{e^{-4} 4^5}{5!}=0.1563[/tex]
Replacing we got:
[tex]P(X\geq 6)=1-P(X<6)=1-P(X\leq 5)=1-[0.0183+0.0733+0.1465+0.1954+0.1954+0.1563]=0.2148[/tex]
(c) What is the probability that there will be at most 6 accidents in the next three months?
For this case since we want the amount in the next two months the rate changes [tex]\lambda=2x3= 6[/tex] accidents per 3 months.
[tex]P(X\leq 6)=P(X=0)+P(X=1)+P(X=2)+P(X=3)+P(X=4)+P(X=5)+P(X=6)[/tex]
Using the pmf we can find the individual probabilities like this:
[tex]P(X=0)=\frac{e^{-6} 6^0}{0!}=0.00248[/tex]
[tex]P(X=1)=\frac{e^{-6} 6^1}{1!}=0.0149[/tex]
[tex]P(X=2)=\frac{e^{-6} 6^2}{2!}=0.0446[/tex]
[tex]P(X=3)=\frac{e^{-6} 6^3}{3!}=0.0892[/tex]
[tex]P(X=4)=\frac{e^{-6} 6^4}{4!}=0.1339[/tex]
[tex]P(X=5)=\frac{e^{-6} 6^5}{5!}=0.1606[/tex]
[tex]P(X=6)=\frac{e^{-6} 6^6}{6!}=0.1606[/tex]
[tex]P(X\leq 6)=0.00248+0.0149+0.0446+0.0892+0.1339+0.1606+0.1606=0.6063[/tex]
The formula Upper A equals 23.1 e Superscript 0.0152 tA=23.1e0.0152t models the population of a US state, A, in millions, t years after 2000.
a. What was the population of the state in 2000?
b. When will the population of the state reach 28.328.3 million?
Answer:
a) [tex]A(t=0)= 23.1 e^{0.0152(0)}=23.1e^0 =23.1[/tex]
b) [tex]t = \frac{ln(\frac{28.3}{23.1})}{0.0152}=13.357 years[/tex]
So for this case the answer would be 13.347 years, the population will be 28.3 million and ye year would be 2000+13.347 and that would be approximately in 2014
Step-by-step explanation:
For this case we assume the following model:
[tex]A(t)= 23.1 e^{0.0152 t}[/tex]
Where t is the number of years after 2000/
Part a
For this case we want the population for 2000 and on this case the value of t=0 since we have 0 years after 2000. If we rpelace into the model we got:
[tex]A(t=0)= 23.1 e^{0.0152(0)}=23.1e^0 =23.1[/tex]
So then the initial population at year 2000 is 23.1 million of people.
Part b
For this case we want to find the time t whn the population is 28.3 million.
So we need to solve this equation:
[tex]28.3= 23.1 e^{0.0152(t)}[/tex]
We can divide both sides by 23.1 and we got:
[tex]\frac{28.3}{23.1}= e^{0.0152t}[/tex]
Now we can apply natural log on both sides and we got:
[tex]ln(\frac{28.3}{23.1})= 0.0152 t[/tex]
And then for t we got:
[tex]t = \frac{ln(\frac{28.3}{23.1})}{0.0152}=13.357 years[/tex]
So for this case the answer would be 13.347 years, the population will be 28.3 million and ye year would be 2000+13.347 and that would be approximately in 2014
Tara wants to weigh her three stuffed animals. They will only fit on the scale two at a time. Together Addie and Missy weight 18 ounces. Missy and Corky weigh 22 ounces, and Addie and Corky weigh 12 ounces. How much does each animal weigh?
Answer: Addie weighs 4 ounces
Missy weighs 14 ounces
Corky weighs 8 ounces
Step-by-step explanation:
Let a represent the weight of Addie.
Let m represent the weight of Missy.
Let c represent the weight of Corky.
Together Addie and Missy weigh 18 ounces. This means that
a + m = 18 - - - - - - - - - 1
Missy and Corky weigh 22 ounces. This means that
m + c = 22
m = 22 - c - - - - - - - - - - 2
Addie and Corky weigh 12 ounces. This means that
a + c = 12
a = 12 - c - - - - - - - - - - - 3
Substituting equation 2 and equation 3 into equation 1, it becomes
22 - c + 12 - c = 18
34 - 2c = 18
- 2c = 18 - 34 = - 16
c = - 16/ - 2 = 8
Substituting c = 8 into equation 2, it becomes
m = 22 - 8
m = 14
Substituting c = 8 into equation 3, it becomes
a = 12 - 8
a = 4
find x from the picture
Answer: x = 120 degrees
Step-by-step explanation:
The diagram is that of a polygon with 5 sides. This means that it is a Pentagon. The sum of the interior angles in a polygon is expressed as
180(n -2)
Where n represents the number of sides that the polygon has.
Since the given polygon has 5 sides, then the sum of the interior angles would be
180(5 - 2) = 180 × 3 = 540 degrees.
Therefore,
x + x + x + 90 + 90 = 540
3x + 180 = 540
3x = 540 - 180 = 360
x = 360/3 = 120 degrees
Solve the inequality. Graph the solution. 4(n-3) -6>18
Answer:
n>9
Step-by-step explanation:
4(n-3)-6>18
4n-12-6>18
4n-18>18
4n>18+18
4n>36
n>36/4
n>9
The weekly salary paid to employees of a small company that supplies part-time laborers averages $750 with a standard deviation of $450. (a) If the weekly salaries are normally distributed, estimate the fraction of employees that make more than $300 per week. (b) If every employee receives a year-end bonus that adds $100 to the paycheck in the final week, how does this change the normal model for that week? (c) If every employee receives a 5% salary increase for the next year, how does the normal model change? (d) If the lowest salary is $300 and the median salary is $525, does a normal model appear appropriate? (a) If the weekly salaries are normally distributed, the fraction of employees that make more than $300 per week is approximately nothing. (Type an integer or a fraction.)
Answer:
(a) The fraction of employees is 0.84.
(b)
[tex]\mu=850\\\\\sigma=450[/tex]
(c)
[tex]\mu=787.5\\\\\sigma=472.5[/tex]
(d) No. The left part of the distribution would be truncated too much.
Step-by-step explanation:
(a) If the weekly salaries are normally distributed, estimate the fraction of employees that make more than $300 per week.
We have to calculate the z-value and compute the probability
[tex]z=\frac{X-\mu}{\sigma}= \frac{300-750}{450}=\frac{-450}{450}=-1\\\\P(X>300)=P(z>-1)=0.84[/tex]
(b) If every employee receives a year-end bonus that adds $100 to the paycheck in the final week, how does this change the normal model for that week?
The mean of the salaries grows $100.
[tex]\mu_{new}=E(x+C)=E(x)+E(C)=\mu+C=750+100=850[/tex]
The standard deviation stays the same ($450)
[tex]\sigma_{new}=\sqrt{\frac{1}{N} \sum{[(x+C)-(\mu+C)]^2} } =\sqrt{\frac{1}{N} \sum{(x+C-\mu-C)^2} }\\\\ \sigma_{new}=\sqrt{\frac{1}{N} \sum{(x-\mu)^2} } =\sigma[/tex]
(c) If every employee receives a 5% salary increase for the next year, how does the normal model change?
The increases means a salary X is multiplied by 1.05 (1.05X)
The mean of the salaries grows 5%, to $787.5.
[tex]\mu_{new}=E(ax)=a*E(x)=a*\mu=1.05*750=487.5[/tex]
The standard deviation increases by a 5% ($472.5)
[tex]\sigma_{new}=\sqrt{\frac{1}{N} \sum{[(ax)-(a\mu)]^2} } =\sqrt{\frac{1}{N} \sum{a^2(x-\mu)^2} }\\\\ \sigma_{new}=\sqrt{a^2}\sqrt{\frac{1}{N} \sum{(x-\mu)^2}}=a*\sigma=1.05*450=472.5[/tex]
(d) If the lowest salary is $300 and the median salary is $525, does a normal model appear appropriate?
No. The left part of the distribution would be truncated too much.
Normal distribution has its mean, median and mode coincident on single point. The solutions to the given problems are specified as:
a) P(X > 300) = 0.8413b) The normal model shifts to the right, 100 units, but its structure stays same.c) If salary is increased 5%, then the normal model gets scaled .d) If lowest salary = $300, and median = $525, then it isn't a normal model as median ≠ mean = $750. It is going to be negatively skewed.How to get the z scores?If we've got a normal distribution, then we can convert it to standard normal distribution and its values will give us the z score.
If we have
[tex]X \sim N(\mu, \sigma)[/tex]
(X is following normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex])
then it can be converted to standard normal distribution as
[tex]Z = \dfrac{X - \mu}{\sigma}, \\\\Z \sim N(0,1)[/tex]
(Know the fact that in continuous distribution, probability of a single point is 0, so we can write
[tex]P(Z \leq z) = P(Z < z) )[/tex]
Also, know that if we look for Z = z in z tables, the p value we get is
[tex]P(Z \leq z) = \rm p \: value[/tex]
For the given case, if we take X = salary of employees weekly of the considered company, then:
[tex]X \sim N(750, 450)[/tex]
The fraction of employees that make more than $300 per week is
P(X > 300).
Using the standard normal distribution, we can estimate this fraction of employees that make more than $300 per week as:
[tex]P(X > 300 ) = 1 - P(X \leq 300) = 1 - P(Z = \dfrac{X - \mu}{\sigma} \leq \dfrac{300 - 750}{450} =-1)\\ \\P(X > 300) = 1 - P(Z \leq -1)\\[/tex]
Using the z-table, the p-value for Z = -1 is: 0.1587
Thus, [tex]P(X > 300) = 1 - P(Z \leq -1) = 1 - 0.1587 = 0.8413[/tex]
When we add $100 to each value of X, it doesn't change the structure of the graph of normal distribution.
When we increase salary by 5%, it means, new salary [tex]Y = X + 5\% \text{of X} = \dfrac{21X}{20}[/tex]
The random variable is scaled by 21/20 = 1.05, so the graph will stay on same origin but it will be stretched a bit in thickness and height.
For the 4th case(d), the median is specified to be $525, but as it was known to us that mean is $750, so mean and median aren't coinciding, and specially, the median is in left of mean, showing that the graph is leaning on right side(negatively skewed). (we deduce it when median < mean, as median shows that mid value is reached, but mean shows that probability is still not reached, so its being late, and reaches later, showing that there is tail in the left of the graph, so being negatively skewed).
Thus, The solutions to the given problems are specified as:
a) P(X > 300) = 0.8413b) The normal model shifts to the right, 100 units, but its structure stays same.c) If salary is increased 5%, then the normal model gets scaled .d) If lowest salary = $300, and median = $525, then it isn't a normal model as median ≠ mean = $750. It is going to be negatively skewed.Learn more about standard normal distribution here:
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True or False? Tell whether the pair of ratios form a proportion. 4/5 and 5/6 Please explain why you chose what you chose
Answer:
False. The products from cross multiplication are different.
Step-by-step explanation:
To know if a pair of ratios form a proportion, cross multiply. If the products are equal, they are a proportion.
Write like this to see top (numerator) and bottom (denominator) clearly.
[tex]\frac{4}{5} =\frac{5}{6}[/tex]
Multiply each numerator with the other side's denominator:
4 X 6 = 24
5 X 5 = 25
Are they equal? No. 24 ≠ 25
Therefore it's not a proportion.
In a fund-raising game for your school, you bet $1 to roll two dice. If your total is 8,9,10 or 11 you win $2. If your total is 12, you win $6. If your total is 7 or less, you lose the dollar you bet. How much, on average do you expect to win or lose with each dollar bet?
A. You will lose 56 cents
B. You will 5.6 cents.
C. On average, you will break even.
D. You will win 2/36
E. You will lose 5.6 cents
Answer: E , You will lose 5.6 cents
Step-by-step explanation:
Because with two dice, there are 36 possible outcomes, 21 are 7 or less, 14 are 8 through 11, and 1 is twelve.
Also when you have a total of 8,9,10 or 11, you gain $1 deducing the $1 you bet. The same with when you have 12 you gain $5.
Average $ to gain when total is 8,9,10 or 11 = P(8,9,10,11)
P(8,9,10,11) = ($2-$1)(14/36) = $14/36 gain
P(12) = ($6-$1)(1/36) = $5/36 gain
P(7 or less) = (0-$1)(21/36) = -$21/36 loss
P(loss or gain)= P(8,9,10,11) + P(12) + P(7 or less)
P(loss or gain) = $( 14/36 + 5/36 - 21/36) = -$2/36
P(loss or gain) = -$0.056 = -5.6 cents loss
Therefore, For every $1 bet you will lose 5.6 cents.
The manager of a grocery store has taken a random sample of 100 customers. The average length of time it took the customers in the sample to check out was 3.11 minutes with a standard deviation of 0.5 minutes. We want to test to determine whether or not the mean waiting time of all customers is significantly more than 3 minutes. If the test statistic is 2.2, what is the p-value associated with this hypothesis test?(Round your answer to three decimal places.)
Answer:
At 0.05 significance level, the p-value is 0.014
Step-by-step explanation:
We are given the following in the question:
Population mean, μ = 3 minutes
Sample mean, [tex]\bar{x}[/tex] = 3.11 minutes
Sample size, n = 100
Alpha, α = 0.05
Sample standard deviation, σ = 0.5 minutes
First, we design the null and the alternate hypothesis
[tex]H_{0}: \mu = 3\text{ minutes}\\H_A: \mu > 3\text{ minutes}[/tex]
We use one-tailed z test to perform this hypothesis.
Formula:
[tex]z_{stat} = \displaystyle\frac{\bar{x} - \mu}{\frac{\sigma}{\sqrt{n}} }[/tex]
[tex]z_{stat} = 2.2[/tex]
Now, we calculate the p-value from the normal standard z-table.
P-value = 0.014
At 0.05 significance level, the p-value is 0.014
Location is known to affect the number, of a particular item, sold by HEB Pantry. Two different locations, A and B, are selected on an experimental basis. Location A was observed for 18 days and location B was observed for 13 days. The number of the particular items sold per day was recorded for each location. On average, location A sold 39 of these items with a sample standard deviation of 8 and location B sold 49 of these items with a sample standard deviation of 4. Select a 99% confidence interval for the difference in the true means of items sold at location A and B. a) O [-1242,-7582]
b) O132.76, 45.24]
c)。8 1.76, 94.24]
d) 0-1 6.03,-3.97]
e)。[42.76, 55.24]
F. None of the above
Answer:
d) [-16.03,-3.97]
[tex]-16.03 \leq \mu_A -\mu_B \leq -3.97[/tex].
Step-by-step explanation:
Notation and previous concepts
A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".
The margin of error is the range of values below and above the sample statistic in a confidence interval.
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".
[tex]n_A=18[/tex] represent the sample of A
[tex]n_B =13[/tex] represent the sample of B
[tex]\bar x_A =39[/tex] represent the mean sample for A
[tex]\bar x_B =49[/tex] represent the mean sample for B
[tex]s_A =8[/tex] represent the sample deviation for A
[tex]s_B =4[/tex] represent the sample deviation for B
[tex]\alpha=0.01[/tex] represent the significance level
Confidence =99% or 0.99
The confidence interval for the difference of means is given by the following formula:
[tex](\bar X_A -\bar X_B) \pm t_{\alpha/2}\sqrt{(\frac{s^2_A}{n_A}+\frac{s^2_B}{n_B})}[/tex] (1)
The point of estimate for [tex]\mu_A -\mu_B[/tex] is just given by:
[tex]\bar X_A -\bar X_B =39-49=-10[/tex]
The appropiate degrees of freedom are [tex]df=n_1+ n_2 -2=18+13-2=29[/tex]
Since the Confidence is 0.99 or 99%, the value of [tex]\alpha=0.01[/tex] and [tex]\alpha/2 =0.005[/tex], and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.005,29)".And we see that [tex]t_{\alpha/2}=2.756[/tex]
The standard error is given by the following formula:
[tex]SE=\sqrt{(\frac{s^2_A}{n_A}+\frac{s^2_B}{n_B})}[/tex]
And replacing we have:
[tex]SE=\sqrt{(\frac{8^2}{18}+\frac{4^2}{13})}=2.188[/tex]
Confidence interval
Now we have everything in order to replace into formula (1):
[tex]-10-2.756\sqrt{(\frac{8^2}{18}+\frac{4^2}{13})}=-16.03[/tex]
[tex]-10+2.756\sqrt{(\frac{8^2}{18}+\frac{4^2}{13})}=-3.97[/tex]
So on this case the 99% confidence interval for the differences of means would be given by [tex]-16.03 \leq \mu_A -\mu_B \leq -3.97[/tex].
d) [-16.03,-3.97]
The 99% confidence interval for the difference in the true means of items sold at location A and B is [-15.64, -4.36], therefore the correct answer is F. None of the above.
Explanation:This question is about computing a confidence interval for the difference of two sample means. The formula for the 99% confidence interval for the difference between two means is:
(X1 - X2) ± Z * sqrt [s1^2/n1 + s2^2/n2]
Where X1 and X2 are the sample means, s1 and s2 are the sample standard deviations, n1 and n2 are the sample sizes, and Z is the Z-score for the desired confidence level. For a 99% confidence level, the Z-score is approximately 2.576. We plug the given values into the equation to calculate:
(39 - 49) ± 2.576 * sqrt [(8^2 / 18) + (4^2 / 13)] => -10 ± 2.576 * sqrt [3.56 + 1.23] => -10 ± 2.576 * sqrt [4.79] => -10 ± 2.576 * 2.19 => -10 ± 5.64
This means the 99% confidence interval for the difference in the true means of items sold at location A and B is [-15.64, -4.36], which is not among the given options, so the correct answer is F. None of the above.
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A media research company conducted an online survey of adults in a certain country to determine their favorite sport. A friend believes professional football (with revenue of about $13 billion per year) is the favorite sport of 53% of adults in the country. Specify the null and alternative hypotheses for testing this belief. Be sure to identify the parameter of interest.
Answer:
Null hypothesis: The favorite sport of 53% of adults in the country is professional football ( with revenue of about $13 billion per year)
Alternate hypothesis: The adults in the country whose favorite sport is not professional football (with revenue of about $13 billion per year) is less than 53%
Step-by-step explanation:
The given claim is that the favorite sport of 53% of adults in the country is professional football (with revenue of about $13 per year) which is the null hypothesis. This means that the favorite sport of the remaining 47%(less than 53%) of adults in the country is not professional football (with revenue of about $13 billion per year) which is the alternate hypothesis
Teachers in a medium-sized suburban school district have an average salary of $47,500 per year, with a standard deviation of $4,600. After negotiating with the school district, teachers recieve a 5% raise and a one-time $500 bonus. What are the new mean and standard deviation of the teacher's salaries during the year in which they recieve bonus?A. $50,125; $4,960B. $49,875; $4,830C. $49,875; $5,330D. $50,375; $4,830E. $50,375; $5,330
Answer: $49,875; $4,830C.
Step-by-step explanation:
The average salary of Teachers in a medium-sized suburban school district is $47,500 per year.
The standard deviation is $4,600
After negotiating with the school district, teachers recieve a 5% raise and a one-time $500 bonus. The bonus of $500 will not alter the mean and standard deviation because equal amount is added for each teacher.
5% increase in each teacher's salary would differ. Therefore, it will affect the mean and standard deviation by 5%. Therefore, the new mean would be
47500 + (5/100 × 47500) = $49875
The new standard deviation would be
4600 + (5/100 × 4600) = $4830
Answer:
$50,375; $4,830 (I just answered it on one of my quizzes)
Step-by-step explanation:
Gravel is being dumped from a conveyor belt at a rate of 20 ft3/min, and its coarseness is such that it forms a pile in the shape of a cone whose base diameter and height are always equal. How fast is the height of the pile increasing when the pile is 14 ft high? (Round your answer to two decimal places.)
Answer:
The height of the pile is increasing [tex]\frac{20}{49\pi}[/tex] a minute when the pile is 14ft high.
Step-by-step explanation:
The volume of a cone is given by the following formula:
[tex]V = \frac{\pi r^{2}h}{3}[/tex]
We have that the diameter and the height are equal, so [tex]r = \frac{h}{2}[/tex]
So
[tex]V = \frac{\pi h^{3}}{12}[/tex]
Let's derivate this equation, using implicit derivatives.
[tex]\frac{dV}{dt} = \frac{\pi h^{2}}{4}\frac{dh}{dt}[/tex]
In this problem, we have to:
Find [tex]\frac{dh}{dt}[/tex], when [tex]\frac{dV}{dt} = 20, h = 14[/tex]. So
[tex]\frac{dV}{dt} = \frac{\pi h^{2}}{4}\frac{dh}{dt}[/tex]
[tex]20 = \frac{196\pi}{4}\frac{dh}{dt}[/tex]
[tex]\frac{dh}{dt} = \frac{20}{49\pi}[/tex]
The height of the pile is increasing [tex]\frac{20}{49\pi}[/tex] a minute when the pile is 14ft high.
This involves relationship between rates using Calculus.
dh/dt = 0.13 ft/min
We are given;Volumetric rate; dv/dt = 20 ft³/min
height of pile; h = 14 ft
We are not given the diameter here but as we are dealing with a right circular cone, we will assume that the diameter is equal to the height.
Thus; diameter; d = 14 ft
radius; r = h/2 = d/2 = 14/2
radius; r= 7 ft
Formula for volume of a cone is; V = ¹/₃πr²h We want to find how fast the height is increasing and this is dh/dt. Thus, we will need to express r in the volume formula in terms of h; V = ¹/₃π(h/2)²h V = ¹/₃π(h²/4)h V = ¹/₁₂πh³ differentiating both sides with respect to time t gives; dV/dt = 3(¹/₁₂πh²)dh/dt dV/dt = ¹/₄πh²(dh/dt)Plugging in the relevant values, we have;
20 = ¹/₄π × 14² × (dh/dt)
dh/dt = (20 × 4)/(π × 14²)
dh/dt = 0.13 ft/min
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Keitaro walks at a pace of 3 miles per hour and runs at a pace of 6 miles per hour. Each month, he wants to complete at least 36 miles but not more than 90 miles. The system of inequalities represents the number of hours he can walk, w, and the number of hours he can run, r, to reach his goal.3w + 6r ≥ 363w + 6r ≤ 90Which combination of hours can Keitaro walk and run in a month to reach his goal?A. 2 hours walking; 12 hours runningB. 4 hours walking; 3 hours runningC. 9 hours walking; 12 hours runningD. 12 hours walking; 10 hours running
Answer:
A. 2 hours walking; 12 hours running
Step-by-step explanation:
The combination of hours walking and running has to respect both these inequalities:
[tex]3w + 6r \geq 36[/tex]
[tex]3w + 6r \leq 90[/tex]
A. 2 hours walking; 12 hours running
3w + 6r = 3*2 + 6*12 = 6+72 = 78.
Ok, it is larger than 35 and smaller than 91.
B. 4 hours walking; 3 hours running
3w + 6r = 3*4 + 6*3 = 12 + 18 = 30.
Invalid. Lesser than 36.
C. 9 hours walking; 12 hours running
3w + 6r = 3*9 + 6*12 = 27 + 72 = 99
Larger than 90. Invalid
D. 12 hours walking; 10 hours running
3w + 6r = 3*12 + 6*10 = 96
Larger than 90. Invalid
The combination of hours Keitaro can walk and run in a month to reach his goal is 2 hours walking; 12 hours running
3w + 6r ≥ 36. (1)
3w + 6r ≤ 90. (2)
substitute each option into the equation
A. 2 hours walking; 12 hours running
3w + 6r ≥ 36
3(2) + 6(12) ≥ 36
6 + 72 ≥ 36
78 ≥ 36
True
3w + 6r ≤ 90
3(2) + 6(12) ≤ 90
6 + 72 ≤ 90
78 ≤ 90
True
B. 4 hours walking; 3 hours running
3w + 6r ≤ 90
3(4) + 6(3) ≤ 90
12 + 18 ≤ 90
30 ≤ 90
True
B. 4 hours walking; 3 hours running
3w + 6r ≥ 36
3(4) + 6(3) ≥ 36
12 + 18 ≥ 36
30 ≥ 36
False
C. 9 hours walking; 12 hours running
3w + 6r ≥ 36
3(9) + 6(12) ≥ 36
27 + 72 ≥ 36
99 ≥ 36
True
3w + 6r ≤ 90
3(9) + 6(12) ≤ 90
27 + 72 ≤ 90
99 ≤ 90
False
D. 12 hours walking; 10 hours running
3w + 6r ≥ 36
3(12) + 6(10) ≥ 36
36 + 60 ≥ 36
96 ≥ 36
True
3w + 6r ≤ 90
3(12) + 6(10) ≤ 90
36 + 60 ≤ 90
96 ≤ 90
False.
Therefore, the combination of hours Keitaro can walk and run in a month to reach his goal is 2 hours walking; 12 hours running
Learn more about inequality:
https://brainly.com/question/15816805
Find the area of the surface generated when the given curve is revolved about the x-axis y=4x+5 [0,2 ]a. 36√17.xb. 36πc. 36π/√17d. 32√17.π
Answer:
The area of this revolted surface is 36π
Step-by-step explanation:
To obtain the area of a revolted surface, you have to define:
1) which is the axis on which the surface is revolted: this defines the limits on that axis or hight of the surface. In this case x∈[0;2]
2) which is the expression of the radius of the revolted surface and its dependence with the hight. In this case, the radius expression could be Y=4x+5
3) Define the angular variable: If this is a fully revolted surface, the angular variable will go from 0 to 2π
Now we can obtain the area with a double integral:
[tex]A=\int\limits^{2}_0 { \int\limits^{2\pi}_0 {r} \, d \varphi } \, dx =\int\limits^{2}_0 { \int\limits^{2\pi}_0 {4x+5} \, d \varphi } \, dx =\int\limits^{2}_0 { (2\pi)(4x+5)} \, dx=36\pi[/tex]
There are 4 suits (heart, diamond, clover, and spade) in a 52-card deck, and each suit has 13 cards. Suppose your experiment is to draw one card from a deck and observe what suit it is. Express the probability in fraction format. (Show all work. Just the answer, without supporting work, will receive no credit.)
Answer:
The probability of drawing a heart or diamond is 1/2 or 0.5
The probability that the card is not a spade is 3/4 or 0.75
Step-by-step explanation:
Consider the provided information.
Part (a) Find the probability of drawing a heart or diamond.
There are 13 cards of heart and 13 cards of diamond.
We need to find the probability of drawing a heart or diamond.
[tex]P(\text{Heart or Diamond})=P(\text{Heart card Drawn})+P(\text{Diamond card Drawn})[/tex]
[tex]P(\text{Heart or Diamond})=\frac{13}{52}+\frac{13}{52}[/tex]
[tex]P(\text{Heart or Diamond})=\frac{26}{52}=\frac{1}{2}=0.5[/tex]
Hence, the probability of drawing a heart or diamond is 1/2 or 0.5
(b) Find the probability that the card is not a spade.
Out of 52 cards 13 are spade,
That means 52 - 13 = 39 cards are not a spade.
[tex]P(\text{Not spade})=\frac{39}{52}=\frac{3}{4}=0.75[/tex]
Hence, the probability that the card is not a spade is 3/4 or 0.75
Devise the exponential growth function that fits the given data, then answer the accompanying question. Be sure to identify the refernce point (t=0) and units of time.Between 2003 and 2008, the average rate of inflation in a certain country was about 4% per year. If a cart of groceries cost $120 in 2003, what will it cost in 2013 assuming the rate of inflation remains constant?
Answer:
[tex]A(t=10) = 120 e^{ln(1.04)10}=177.629[/tex]
And that would be the approximately cost for 2013.
Step-by-step explanation:
For this case we need to define some notation first.
A= population , t= represent the years after 2003, C= constant for the exponential model.
The starting point t=0 correspond to the year of 2003.
On this case we are assuming the following exponential model:
[tex]A(t) = A_o e^{Ct}[/tex]
The initial value on this case is for t=0 A(t=0)= 120 and if we replace we got this:
[tex]120=A_o e^{C(0)}=A_o e^0 = A_o[/tex]
And then the model is:
[tex]A(t) =120 e^{Ct}[/tex]
Now we need to determine the value for C. Since we know that inflation increase 4% per year we have that after one year we have 1.04 times the value of the original value, and we have this equation:
[tex]1.04 A_o= A_o e^{C(1)}= A_o e^C[/tex]
And we got this:
[tex]1.04= e^C [/tex]
Applying ln on both sides we got:
[tex]ln(1.04)= C=0.0392207[/tex]
So then our model is given by:
[tex]A(t) = 120 e^{ln(1.04)t}[/tex]
For 2013 we have that t=10 since 2013-2003 = 10 after 2003, if we replace t=10 we got this:
[tex]A(t=10) = 120 e^{ln(1.04)10}=177.629[/tex]
And that would be the approximately cost for 2013.
The cost of groceries in 2013, after applying an annual inflation rate of 4% for 10 years, will be approximately $177.63.
Explanation:To calculate the annual rate of inflation and the cost in 2013, you can use the exponential growth function, where cost =[tex]initial_{cost} * (1 + rate)^{time[/tex]. In this case, the initial cost in 2003 (t=0) is $120 and the annual rate of inflation is 4%, or 0.04. To calculate the cost in 2013, which is 10 years after 2003, apply the exponential growth function:
Cost in 2013 = $120 * (1 + 0.04)¹⁰
This calculation yields:
Cost in 2013 = $120 * (1.04)¹⁰
Cost in 2013 = $120 * 1.48024
Cost in 2013 = $177.63
Therefore, in 2013, assuming the rate of inflation remains constant, the cart of groceries will cost approximately $177.63.
Explain how to solve 3^(x − 4) = 6 using the change of base formula log base b of y equals log y over log b. Include the solution for x in your answer. Round your answer to the nearest thousandth.
Answer:
x = 4 + (log 6 / log 3)
x ≈ 5.631
Step-by-step explanation:
3^(x − 4) = 6
Take log base 3 of both sides.
log₃ 3^(x − 4) = log₃ 6
x − 4 = log₃ 6
Use change of base formula.
x − 4 = log 6 / log 3
Solve for x.
x = 4 + (log 6 / log 3)
x ≈ 5.631
Answer:
5.631
Step-by-step explanation:
Using the change of base formula log base b of y equals log y over log b
Log y (base b) = log y /log b
3^(x − 4) = 6
Taking the log of both sides
log 3^(x − 4) = log 6
using the logarithm law that states that
log a ^ x = x log a
x - 4 log 3 = log 6
x - 4 = log 6 / log 3
x - 4 = 1.630929754
x = 5.630929754
≈ 5.631
A scientist measured the speed of light. His values are in km/sec and have 299,000 subtracted from them. He reported the results of 25 trials with a mean of 756.22 and a standard deviation of 100.89.
(a) Find a 90% confidence interval for the true speed of light from these statistics.
(b) State in words what this interval means. Keep in mind that the speed of light is a physical constant that, as far as we know, has a value that is true throughout the universe.
(c) What assumptions must you make in order to use your method?
Answer:
a) The 90% confidence interval would be given by (721.716;790.724)
b) We are 90% confident that the true mean for the true speed of light is between (721.716;790.724)
c) We assume the following conditions:
RandomizationIndependenceDeviation unknown [tex]\sigma[/tex]Step-by-step explanation:
A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".
The margin of error is the range of values below and above the sample statistic in a confidence interval.
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".
Part a
[tex]\bar X=756.22[/tex] represent the sample mean
[tex]\mu[/tex] population mean (variable of interest)
[tex]s=100.89[/tex] represent the sample standard deviation
n=25 represent the sample size
90% confidence interval
The confidence interval for the mean is given by the following formula:
[tex]\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}[/tex] (1)
The degrees of freedom are given by:
[tex]df=n-1=25-1=24[/tex]
Since the Confidence is 0.90 or 90%, the value of [tex]\alpha=0.1[/tex] and [tex]\alpha/2 =0.05[/tex], and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.05,24)".And we see that [tex]t_{\alpha/2}=1.71[/tex]
Now we have everything in order to replace into formula (1):
[tex]756.22-1.71\frac{100.89}{\sqrt{25}}=721.716[/tex]
[tex]756.22+1.71\frac{100.89}{\sqrt{25}}=790.724[/tex]
So on this case the 90% confidence interval would be given by (721.716;790.724)
Part b
We are 90% confident that the true mean for the true speed of light is between (721.716;790.724)
Part c
We assume the following conditions:
RandomizationIndependenceDeviation unknown [tex]\sigma[/tex]Final answer:
The 90% confidence interval for the true speed of light is between 299,723.02683 km/sec and 299,789.41317 km/sec. This interval suggests we can be 90% confident that the constant speed of light falls within this range, with the understanding that the true speed of light is approximately 299,792,458 m/s.
Explanation:
The sample mean is 756.22, and the standard deviation is 100.89 with 25 trials.
First, add 299,000 km/sec to the sample mean to revert to the actual speed of light. Adjusted mean = 756.22 + 299,000 = 299,756.22 km/sec.
Since the sample size (25) is greater than 30, we use the z-score for a 90% confidence interval, which is 1.645.
To find the margin of error (ME), use the formula ME = z * (σ/√n), where σ is the standard deviation and n is the sample size. ME = 1.645 * (100.89/√25)= 1.645 * 20.178 = 33.19317 km/sec.
The confidence interval is then mean ± ME. That gives us the interval: [299,756.22 - 33.19317, 299,756.22 + 33.19317] or [299,723.02683, 299,789.41317] km/sec.
Interpretation: We are 90% confident that the true speed of light lies within the interval of 299,723.02683 to 299,789.41317 km/sec.
The samples are independent and randomly selected.
The data reported is accurate and measured without systematic errors.
The data is normally distributed or the sample size is large enough for the Central Limit Theorem to apply.
These results align with the known fact that the speed of light is a constant at approximately 299,792,458 meters/second, and any deviations observed in the experiment are likely due to measurement error or experimental uncertainties.