Parking at a large university has become a very big problem. University administrators are interested in determining the average parking time (e.g. the time it takes a student to find a parking spot) of its students. An administrator inconspicuously followed 280 students and carefully recorded their parking times. Which of the following graphs should not be used to display information concerning the students parking times?a. Pie chartb. Stem-and-leaf displayc. Histogramd. Box plot

Answers

Answer 1

Answer:

a. Pie chart

This one is the correct option since we use a pie chart when we have categories in the data. And for this case we don't have any category defined at the begin so for this reason the pie chart would be not useful for this case.

Step-by-step explanation:

For this case our variable of interest is the average parking time of its students. And we have 280 values for these times. So then the variable of interest is quantitative.

Which of the following graphs should not be used to display information concerning the students parking times?

a. Pie chart

This one is the correct option since we use a pie chart when we have categories in the data. And for this case we don't have any category defined at the begin so for this reason the pie chart would be not useful for this case.

b.Stem-and-leaf display

That incorrect since the Stem and leaf plot is useful when we want to plot quantitative data.

c. Histogram

That incorrect since the Histogram is ideal when we want to plot quantitative data and analyze the distribution of the data.

d. Box plot

That incorrect since the Box plot is ideal and useful when we want to plot quantitative data and see central tendency measures.

Answer 2

Final answer:

A pie chart should not be used to display the average parking times of students because it cannot effectively represent the distribution or variability of parking times, unlike a histogram, box plot, or stem-and-leaf display.

Explanation:

The question revolves around determining which graph would be inappropriate for displaying the average parking times of students at a university. The options are: a Pie chart, a Stem-and-leaf display, a Histogram, and a Box plot. To assess the suitability of each graph, we need to understand what type of data we have and what information we wish to convey with the graph.

A pie chart is typically used to show parts of a whole. However, in this scenario, we're interested in the distribution of parking times, not how a single parking time compares as a fraction of the total. Therefore, a pie chart would not effectively represent the variability or distribution of parking times.

Both a histogram and a box plot are well-suited for displaying the distribution of quantitative data such as parking times, making them good choices for this context. A stem-and-leaf display is also a valid option, especially for providing a quick visual summary that includes actual data points, which can be useful for identifying specific patterns or outliers.

In conclusion, the pie chart is the graph that should not be used to display information concerning the students' parking times. It simply doesn't align with the objective of analyzing the distribution or average of the parking times, which is best visualized through the other mentioned graphical methods.


Related Questions

A boiler has five identical relief valves. The probability that any particular valve will open on demand is 0.93. Assume independent operation of the valves. Calculate P(at least one valve opens). (Round your answer to eight decimal places.)

Answers

Answer:

There is a 99.99998% probability that at least one valve opens.

Step-by-step explanation:

For each valve there are only two possible outcomes. Either it opens on demand, or it does not. This means that we use the binomial probability distribution to solve this problem.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]

In which [tex]C_{n,x}[/tex] is the number of different combinatios of x objects from a set of n elements, given by the following formula.

[tex]C_{n,x} = \frac{n!}{x!(n-x)!}[/tex]

And p is the probability of X happening.

In this problem we have that:

[tex]n = 5, p = 0.93[/tex]

Calculate P(at least one valve opens).

This is [tex]P(X \geq 1)[/tex]

Either no valves open, or at least one does. The sum of the probabilities of these events is decimal 1. So:

[tex]P(X = 0) + P(X \geq 1) = 1[/tex]

[tex]P(X \geq 1) = 1 - P(X = 0)[/tex]

So

[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]

[tex]P(X = 0) = C_{5,0}.(0.93)^{0}.(0.07)^{5} = 0.0000016807[/tex]

Finally

[tex]P(X \geq 1) = 1 - P(X = 0) = 1 - 0.0000016807 = 0.9999983193[/tex]

There is a 99.99998% probability that at least one valve opens.

Answer:

0.930000

Step-by-step explanation:

Determine for which values of m the function variant Φ(x) = x^m is a solution to the given equation. a. 3x^2 (d^2y/dx^2) + 11x(dy/dx) - 3y = 0 b. x^2 (d^2y/dx^2) - x(dy/dx) - 5y = 0

Answers

Answer:

a)  m = -9 or m = 1

b) m =  1 + √6 or m = 1 -√6

Step-by-step explanation:

for

Φ(x) = x^m

then

dΦ/dx (x) = m*x^(m-1)

d²Φ/dx² (x) = m*(m-1)*x^(m-2)

then

for a

3x^2 (d^2y/dx^2) + 11x(dy/dx) - 3y = 0

3x^2*m*(m-1)*x^(m-2) + 11*x* m*x^(m-1) - 3*x^m = 0

3*m*(m-1)*x^m + 11*m*x^m- 3*x^m = 0

dividing by x^m

3*m*(m-1) + 11*m - 3 =0

3*m² + 8 m - 3 =0

m= [-8 ± √(64 + 4*3*3)]/2 = (-8±10)/2  

m₁ = -9 , m₂= 1

then Φ(x) = x^m is a solution for the equation a , when m = -9 or m = 1

for b)

x^2 (d^2y/dx^2) - x(dy/dx) - 5y = 0

x^2*m*(m-1)*x^(m-2) -  x* m*x^(m-1) - 5*x^m = 0

m*(m-1)*x^m -m *x^m- 5*x^m = 0

dividing by x^m

m*(m-1) -m - 5 =0

m² - 2 m - 5 =0

m= [2 ± √(4 + 4*1*5)]/2 = (2±√24)/2 = 1 ±√6

m₁ =  1 + √6 , m₂ =  1 - √6

then Φ(x) = x^m is a solution for the equation b , when m =  1 + √6 or m = 1 - √6

Answer

a) m = -3 or 1/3

b) m = 1 + root 6 or 1 - root 6

Step-by-step explanation:

The step by step calculation is as shown in the attachment.

A car manufacturer wants to assess customer satisfaction for cars sold during the previous year.(a) Describe the population involved.(b) Is the population involved hypothetical or not?

Answers

Answer:

a) The target population of interest on this case represent the "customers who bought a car during the previous year". They want to analyze all the people that satisfy this condition in order to see the satisfaction rate of these people

b) For this case the population is not hypothetical since is well defined and they have all the customers who bought a car during the last year, since that info is on the records of the manufacturer from people who bought a car. So then the population is available in order to analyze the question desired.

Step-by-step explanation:

Part a

The target population of interest on this case represent the "customers who bought a car during the previous year". They want to analyze all the people that satisfy this condition in order to see the satisfaction rate of these people

Part b

For this case the population is not hypothetical since is well defined and they have all the customers who bought a car during the last year, since that info is on the records of the manufacturer from people who bought a car. So then the population is available in order to analyze the question desired.

a. The customer who bought the car the previous year can get feedback from them so that we can evaluate the status of customer satisfaction.

b.   The customer who bought the car the previous year, so info is on the record of the manufacturer from the people who bought a car.

Data handling

the process of ensuring that research data is stored.

How to take data?

a. The customer who bought the car the previous year can get feedback from them so that we can evaluate the status of customer satisfaction.

b.   The customer who bought the car the previous year, so info is on the record of the manufacturer from the people who bought a car.

More about the data handling link is given below.

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In a sample of three people, the first person has a score 5 points above the mean, and the second person has a score 3 points above the mean.
How does the third persons score compare to the mean?

Answers

Answer:

8 points below the mean.

Step-by-step explanation:

Let 'X' be the score of the third person, and let 'M' be the mean score of the sample.

If the other two people scored 5 and 3 above the mean, in order to maintain consistency, the following expression must be true:

[tex]M+5+(M+3)+X =3M\\X=3M-2M-8\\X=M-8[/tex]

Therefore, the third person has a score 8 points below the mean.

Final answer:

Given that the other two people's scores are collectively 8 points above the mean, the third person's score must be 8 points below the mean to balance out the total deviation.

Explanation:

This question is about determining the score of the third person in a sample relative to the mean of the sample. We know that the total deviation from the mean must equal zero because the mean is the average of all the scores. Given the first person has a score 5 points above the mean, and the second person has a score 3 points above the mean, the total deviation above the mean is (5+3)=8 points. For the score deviations to balance out to zero, the third person's score must be 8 points below the mean.

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Find the domain of each function. (Enter your answer using interval notation.)(a) f(x) = 81-e^x^2/1-e^81-x^2 (b) f(x) = 7 + x/e^cos x

Answers

Answer:

a) (-∞,-9)∪(-9,9)∪(9,+∞)

b) (-∞,+∞)

Step-by-step explanation:

The domain of a real function is the largest subset og the real line in which it is defined.

a) The function [tex]f(x)=\frac{81-e^{x^2}}{1-e^{81-x^2}}[/tex] is defined for all values of x in which the denominator is not zero. The denominator is zero if [tex]0=1-e^{81-x^2}[/tex], that is,  [tex]e^{81-x^2}=1[/tex]. The exponential function is equal to one only if the exponent is equal to zero, then the values of x that nullify the denominator satisfy [tex]81-x^2=0[/tex], thus x=9 or x=-9. Then the domain of f is the set of all real numbers such that x≠9 and x≠-9, which is the set (-∞,-9)∪(-9,9)∪(9,+∞).

b) In this case the denominator is [tex]e^{\cos x}[/tex] which is always positive. Thus the denominator is not zero for all real x, then the fomain is the real line, which is the interval (-∞,+∞).

Final answer:

The domain of the function f(x) = (81-e^x^2)/(1-e^81-x^2) is (-∞, -9) ∪ (-9, 9) ∪ (9, ∞). The domain of the function f(x) = 7 + x/e^cos(x) is (-∞, ∞).

Explanation:

(a) To find the domain of the function f(x) = (81-e^x^2)/(1-e^81-x^2), we need to determine the values of x for which the function is defined. The function is defined as long as the denominator is not zero. So, we need to solve the equation 1 - e^(81-x^2) = 0 to find where the denominator equals zero. Simplifying this equation, we get e^(81-x^2) = 1. Taking the natural logarithm of both sides, we have 81-x^2 = 0. Therefore, x^2 = 81, and x = ±9. The domain of the function is (-∞, -9) ∪ (-9, 9) ∪ (9, ∞).

(b) To find the domain of the function f(x) = 7 + x/e^cos(x), we need to determine the values of x for which the function is defined. The function is defined as long as the denominator e^cos(x) is not zero. Since the exponential function is always positive, we know that e^cos(x) will never be zero. Therefore, the domain of the function is the set of all real numbers, (-∞, ∞).

Eight less than the product of a number n and 1/5 is no more than 96

Answers

Final answer:

The question translates to the inequality (1/5)n - 8 ≤ 96, which is solved by adding 8 to both sides and then multiplying by 5, resulting in n ≤ 520.

Explanation:

The question involves translating a word problem into a mathematical inequality. The phrase 'Eight less than the product of a number n and 1/5' can be written as (1/5)n - 8. When it states that this is 'no more than 96', it implies that the expression should be less than or equal to 96. Therefore, the inequality we need to solve is (1/5)n - 8 ≤ 96.

Now, let's solve this inequality step-by-step:

Add 8 to both sides of the inequality: (1/5)n ≤ 104.

Multiply both sides by 5 to solve for n: n ≤ 520.

This gives us the solution to the inequality, indicating that the number n can be any value less than or equal to 520 to satisfy the initial condition.

A professor gives a statistics exam. The exam has 100 possible points. The scores for the students in the second classroom are as follows: 88 84 84 52 100 84 92 84 88 Calculate the sample size, n, and the sample mean, M.

Answers

Answer: Sample size n is 9.

The sample mean M is 84.

Step-by-step explanation:

Given : A professor gives a statistics exam.

The exam has 100 possible points.

The scores for the students in the second classroom are as follows:

88 84 84 52 100 84 92 84 88

Since , the number of scores of different students= 9

∴ Sample size is n= 9.

Now , Sample mean = [tex]\dfrac{\text{Sum of all data values}}{\text{Sample size}}[/tex]

[tex]=\dfrac{88+ 84 +84 +52 +100+ 84+ 92+ 84+ 88 }{9}\\\\=\dfrac{756}{9} =84[/tex]

Hence, the sample mean M is 84.

We have that the sample size, n, and the sample mean, M. is mathematically given as

n=9M_n=88

From the question we are told

A professor gives a statistics exam.

The exam has 100 possible points. The scores for the students in the second classroom are as follows: 88 84 84 52 100 84 92 84 88 Calculate the sample size, n, and the sample mean, M.Mean

Generally the equation for the  Mean is mathematically given as

[tex]M=\sum/n\\\\\Therefore\\\\M=\frac{88 +84 +84+ 52+ 100+ 84+ 92+ 84+ 88 }{9}\\\\M=84[/tex]

Therefore

New mean

M_n=84+4

M_n=88

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Students in a mathematics class were given an exam and then tested monthly with an equivalent exam. The average scores for the class are given by the human memory model f(t) = 76 − 18 log10(t + 1), 0 ≤ t ≤ 12 where t is the time in months. Verify your answers in parts (a), (b), and (c) using a graphing utility.

(a) What was the average score on the original exam (t = 0)? f(0) =
(b) What was the average score after 2 months? (Round your answer to one decimal place.) f(2) =
(c) What was the average score after 11 months? (Round your answer to one decimal place.) f(11) =

Answers

Answer:

a) [tex] f(t) = 76-18 log_{10} (0+1)= 76 - 18 log_{10} 1= 76[/tex]

b) [tex] f(t) = 76-18 log_{10} (2+1)= 76 - 18 log_{10} 3= 76-8.588=67.41[/tex]

c) [tex] f(t) = 76-18 log_{10} (11+1)= 76 - 18 log_{10} 12= 76-19.425=56.574[/tex]

Step-by-step explanation:

For this case we know that the average scores for the class are given by the following model:

[tex] f(t) = 76-18 log_{10} (t+1) , 0 \leq t \leq 12[/tex]

Where t is in months. The graph attached illustrate the function for this case

And for this case we can answer the questions like this:

Part a

We just need to replace t =0 into the model and we got:

[tex] f(t) = 76-18 log_{10} (0+1)= 76 - 18 log_{10} 1= 76[/tex]

Part b

We just need to replace t =2 into the model and we got:

[tex] f(t) = 76-18 log_{10} (2+1)= 76 - 18 log_{10} 3= 76-8.588=67.41[/tex]

Part c

We just need to replace t =11 into the model and we got:

[tex] f(t) = 76-18 log_{10} (11+1)= 76 - 18 log_{10} 12= 76-19.425=56.574[/tex]

12 ounces is roughly the same

Answers

Answer:

Roughly the same as what?

Step-by-step explanation:

Answer:

A. 400 GRAMS

B. 120 GRAMS

C. 356 GRAMS

D. 340 GRAMS

PENN FOSTER

ANSWER IS D 340 GRAMS

Step-by-step explanation:1 ounce = 28.3495 12 ounces = 28.3495 * 12 = 340.194 grams

P (A or B ). P(A or B) indicates​ _______. Choose the correct answer below. A. the probability that A and B both occur in the same trial. B. the probability that in a single​ trial, event A​ occurs, event B​ occurs, or they both occur. C. the probability that event A occurs in one trial followed by event B in another trial. D. the probability that event A or event B does not occur in a single trial.

Answers

Answer:

Option B is right

Step-by-step explanation:

P(A or B)

In set theory if A and B are two sets we have either A occurs or B occurs or both as

P(A or B) = P(AUB)

This implies that A occurs or B occurs

Probability is calculated as P(A)+P(B)-P(AB)

P(AB) here represents both occurring and this is subtracted as this was added two times

Option A is wrong because this is AB

Option C is wrong because it is not necessary A occurs first

D is wrong because this is A'UB' where A' represents the complement of A

So correct answer out of four options is

B. the probability that in a single​ trial, event A​ occurs, event B​ occurs, or they both occur.

Final answer:

The notation P(A or B) in probability specifies the probability that in a single trial, either event A happens, event B happens, or they both happen.

Explanation:

The notation P(A or B) in probability theory denotes the probability that event A occurs, event B occurs, or both events occur in a single trial. If we look at the four options provided, option B fits this definition. Hence, P(A or B) indicates the probability that in a single trial, event A occurs, event B occurs, or they both occur. It does not imply that both events A and B have to occur concurrently in the same trial (option A), nor that event A has to occur in one trial followed by event B in another trial (option C), and certainly not the probability that event A or event B does not occur in a single trial (option D).

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An automobile manufacturer claims that its van has a 39.0 miles/gallon (MPG) rating. An independent testing firm has been contracted to test the MPG for this van since it is believed that the van has an incorrect manufacturer's MPG rating. After testing 120 vans, they found a mean MPG of 38.8. Assume the standard deviation is known to be 2.2. A level of significance of 0.02 will be used. Find the value of the test statistic. Round your answer to 2 decimal places. Enter the value of the test statistic.

Answers

Final answer:

To evaluate the manufacturer's claim, a one-sample t-test can be conducted. The test statistic can be calculated using the given information.

Explanation:

To conduct a hypothesis test to evaluate the manufacturer's claim, we can use a one-sample t-test. The null hypothesis (H0) is that the mean MPG is equal to 39.0, and the alternative hypothesis (Ha) is that the mean MPG is not equal to 39.0. We can calculate the test statistic using the formula:

test statistic = (sample mean - population mean) / (standard deviation / sqrt(sample size))

Using the given values:

sample mean = 38.8

population mean = 39.0

standard deviation = 2.2

sample size = 120

Substituting these values, the test statistic is:

test statistic = (38.8 - 39.0) / (2.2 / sqrt(120))

drag each expression to show whether it can be used to find the surface area, volume or neither​

Answers

Final answer:

The expressions can be used to calculate volume if they incorporate height, length, and width. They could calculate surface area if they involve length and width or the sum of all faces of a 3D object, without height. If these measurements are not present, they may not calculate surface area or volume.

Explanation:

The original question appears to be a task related to a mathematical exercise instead of a traditional query. It's asking whether certain expressions can be used to calculate volume or surface area, or if they're not meant to calculate either.

If the expression used includes measurements for length, width, and height and involves their multiplication, it's likely being used to calculate the volume of a 3D object. For a cube, volume is calculated as length x width x height, for example.

If the expression incorporates the multiplication of length and width, without incorporating height, it's being used to calculate surface area. The expression might also calculate the sum of all the faces of a 3D object to find total surface area.

If there aren't any apparent calculations for length, width, or height, it could be that these expressions aren't being used for calculating surface area or volume.

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n insurance company pays hospital claims. The number of claims that include emergency room or operating room charges is 85% of the total number of claims. The number of claims that do not include emergency room charges is 25% of the total number of claims. The occurrence of emergency room charges is independent of the occurrence of operating room charges on hospital claims. Calculate the probability that a claim submitted to the insurance company includes operating room charges.

Answers

Answer:

[tex] 0.85 = P(C) + 0.75 -0.75 P(C)[/tex]

[tex]0.1 = 0.25 P(C)[/tex]

[tex] P(C) = 0.4[/tex]

Step-by-step explanation:

First we can define some notation useful:

C ="represent the event of incurring in operating charges"

R= represent the event of emergency rooms charges"

For this case we are interested on P(C) since they want "the probability that a claim submitted to the insurance company includes operating room charges."

We have some probabilities given:

[tex] P(R') = 0.25 , P(C \cup R) =0.85[/tex]

Solution to the problem

By the complement rule we have this:

[tex] P(R') = 0.25 =1-P(R)[/tex]

[tex] P(R) = 1-0.25 = 0.75[/tex]

Since the two events C and R are considered independent we have this:

[tex]P(C \cap R) = P(C) *P(R)[/tex]

Now we can use the total probability rule like this:

[tex] P(C \cup R) = P(C) + P(R) - P(R)*P(C)[/tex]

And if we replace we got:

[tex] 0.85 = P(C) + 0.75 -0.75 P(C)[/tex]

[tex]0.1 = 0.25 P(C)[/tex]

[tex] P(C) = 0.4[/tex]

The domain of discourse are the students in a class. Define the predicates: S(x): x studied for the test A(x): x received an A on the test What is the logical expression that is equivalent to?

Answers

Answer:

[tex] (\forall x) S(x)\rightarrow A(x) [/tex]

[tex] (\exists x)\neg S(x)\wedge A(x) [/tex]

Step-by-step explanation:

Everyone who studied for the test received an A on the test.

That means- if you studied for the test you will recived an A, and it is hold for everyone, so we will use quantifire [tex]\forall[/tex].

[tex] (\forall x) S(x)\rightarrow A(x) [/tex]

It means: for every student holds- If it is correct that student x studied then  the student got an A.

Someone who did not study for the test received an A on the test.

It means that, there is at least one student hwo didn't studie but student got an A. So we have conjuction of two sentences (negation of the S(x) and A(x) for some student- for that we use existential quantifie).

[tex] (\exists x)\neg S(x)\wedge A(x) [/tex]

Final answer:

The logical expression involving predicates S(x) and A(x) can be represented by ∀x(S(x) → A(x)), stating that all students who studied for the test received an A.

Explanation:

The question involving the predicates S(x): x studied for the test and A(x): x received an A on the test revolves around predicate logic, where we aim to understand and analyze the logical relations of sentences with subjects and predicates within a specified domain of discourse.

To define the logical expression that describes a relationship between studying for a test and receiving an A would depend on the specific relationship we want to express. For instance, a possible logical expression could be ∀x(S(x) → A(x)), which translates to 'For all students x in the class, if x studied for the test, then x received an A on the test.'

In a 2-sample z-test for two proportions, you find the following: sub(hat(p),1)

Answers

Answer:

[tex]z=\frac{0.32-0.36}{\sqrt{0.34(1-0.34)(\frac{1}{50}+\frac{1}{50})}}=-0.422[/tex]

So on this case the only option that satisfy the calculated statistic is:

z=-0.42

Step-by-step explanation:

Assuming this complete problem: "In a 2-sample z-test for two proportions, you find the following:

^P1 = 0.32, (n,1)=50

^P,2= 0.36, (n,2)=50

Find the test statistic you will use while executing this test:

z=-0.67 , z=±1.64 , z=-1.96 , z=0.34 , z=-0.42"

Solution to the problem

Data given and notation  

[tex]n_{1}=50[/tex] sample 1 selected

[tex]n_{2}=50[/tex] sample 2 selected

[tex]p_{1}=0.32[/tex] represent the sample proportion for 1

[tex]p_{2}=0.36[/tex] represent the sample proportion for 2  

z would represent the statistic (variable of interest)  

[tex]p_v[/tex] represent the value for the test (variable of interest)

We need to apply a z test to compare proportions, and the statistic is given by:  

[tex]z=\frac{\hat p_{1}-\hat p_{2}}{\sqrt{\hat p (1-\hat p)(\frac{1}{n_{1}}+\frac{1}{n_{2}})}}[/tex]   (1)

Or equivalently:

[tex]z=\frac{\hat p_{2}-\hat p_{1}}{\sqrt{\hat p (1-\hat p)(\frac{1}{n_{1}}+\frac{1}{n_{2}})}}[/tex]   (1)

Where [tex]\hat p=\frac{X_{1}+X_{1}}{n_{1}+n_{1}}=\frac{\hat p_1 +\hat p_2}{2}=\frac{0.32+0.36}{2}=0.34[/tex]

Calculate the statistic

Replacing in formula (1) the values obtained we got this:  

[tex]z=\frac{0.32-0.36}{\sqrt{0.34(1-0.34)(\frac{1}{50}+\frac{1}{50})}}=-0.422[/tex]

So on this case the only option that satisfy the calculated statistic is:

z=-0.42

Suppose that there are 2 million inhabitants of a country in which 1 the mean gross (before tax) income is $20,000 per year. Suppose that the mean individual pays taxes on $3000 per year. What is the total disposable income (gross income minus taxes) per year for the country as a whole?

Answers

Answer:

$34 billion

Step-by-step explanation:

Data provided in the question:

Total number of inhabitants of a country = 2 million

1 mean gross income = $20,000 per year

Tax paid by mean individual = $3000 per year

Now,

Mean disposable income

= Mean gross income - Mean tax

= $20,000 - $3,000

= $17,000 per year

Therefore,

Total disposable income per year

= Mean disposable income × Total number of inhabitants of a country

= $17,000 × 2 million

= $34 billion

55. If a test statistic falls in the critical region the null hypothesis is _______________. a) Rejected b) Not Rejected c) It depends d) accepted

Answers

Answer:

Option a) Rejected              

Step-by-step explanation:

We define critical region as:

It is also known as the rejection region.It is the region where the null hypothesis is rejected.Thus, if our calculated test statistic lies in the critical region, we fail to accept the null hypothesis and reject it.For right tailed test: The critical region is the values greater than the critical value.For left tailed test: The critical region is the values less than the critical value.The critical region for two tailed test is the value apart from the range of critical values.

Thus,

If a test statistic falls in the critical region the null hypothesis is rejected.

What dose the equation P =?
I think $8.95

Answers

Answer: Yes, you are correct it's $8.95 because like they stated in the word problem p= the price of each bowl and like they said in the word problem they charge $8.95 a bowl. So, $8.95 is the answer.

Step-by-step explanation:

that is correct. answer is 8.95

After having a​ sonogram, a pregnant woman learns that she will have twins. What is the probability that she will have identical​ twins?

Answers

Answer:

[tex]\frac{1}{7}[/tex]

Step-by-step explanation:

The table containing information about the sexes of twins and and the type of twins are given in the table below with corresponding weightage.

Now, since we know that she is expecting twins, the probability of the event

                [tex]A - '\text{the woman will have identical twins }'[/tex]

is calculated as follows

[tex]Pr(A) = \dfrac{\text{total number of boy/boy, boy/girl, girl/boy and girl/girl identical twins}}{\text{total number of both identical and frathernal twins}}[/tex]

By using the informations from the given table, we obtain

                                        [tex]Pr(A) = \frac{2 + 0 + 0 +2}{28} = \frac{4}{28} = \frac{1}{7}[/tex]

Find the exact value of cos theta​, given that sin thetaequalsStartFraction 15 Over 17 EndFraction and theta is in quadrant II. Rationalize denominators when applicable.

Answers

Answer:

[tex] cos \theta = -\frac{8}{17}[/tex]

Step-by-step explanation:

For this case we know that:

[tex] sin \theta = \frac{15}{17}[/tex]

And we want to find the value for [tex] cos \theta[/tex], so then we can use the following basic identity:

[tex] cos^2 \theta + sin^2 \theta =1 [/tex]

And if we solve for [tex] cos \theta [/tex] we got:

[tex] cos^2 \theta = 1- sin^2 \theta[/tex]

[tex] cos \theta =\pm \sqrt{1-sin^2 \theta}[/tex]

And if we replace the value given we got:

[tex] cos \theta =\pm \sqrt{1- (\frac{15}{17})^2}=\sqrt{\frac{64}{289}}=\frac{\sqrt{64}}{\sqrt{289}}=\frac{8}{17}[/tex]

For our case we know that the angle is on the II quadrant, and on this quadrant we know that the sine is positive but the cosine is negative so then the correct answer for this case would be:

[tex] cos \theta = -\frac{8}{17}[/tex]

Answer:

It is D

Step-by-step explanation:

EDGE 2021

An athlete ran 200 meters in 19.19 seconds. Suppose that he ran the first half of that race (around a curve) in 10.75 seconds. How long did it take him to run the second half of the race (on a straight track)?

Answers

The time needed to run the second half of the race is 8.44 s

Step-by-step explanation:

The total distance covered by the athlete during the race is

d = 200 m

And the total time taken to cover this distance is

T = 19.19 s

We also know that the time the athlete needs to cover the first half of the race is

[tex]t_1 = 10.75 s[/tex]

Also, we know that

[tex]T=t_1 + t_2[/tex]

where [tex]t_2[/tex] is the time the athlete takes to cover the second half of the race.

Re-arranging this equation and susbtituting the values, we find the value of t2:

[tex]t_2 = T-t_1 = 19.19-10.75=8.44 s[/tex]

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Find the missing lengths:
OK=1 and OL=3, find KH and LH.

Answers

Answer:

KH = 2  and  LH = 2√3

Step-by-step explanation:

Using Euclidean theorem for the right triangle.

∵ ΔLHK is a right triangle at H

OK = 1 , OL = 3

KL = KO + OL = 1 + 3 = 4

KH² = KO * KL = 1 * 4 = 4

KH = √4 = 2

And LH² = LO * LK = 3 * 4 = 12

∴ LH = √12 = 2√3

Select the most likely answer for the coefficient of linear correlation for the following two variables: x = the number of hours spent studying for a test, and y = the number of points earned on the test
a. r = 1.20b. r = 0.70c. r = - 0.85d. r = 0.05

Answers

Answer:

Option b) r = 0.70      

Step-by-step explanation:

We are given the following in he question:

Variables:

x = the number of hours spent studying for a test

y = the number of points earned on the test

Correlation is a technique that help us to find or define a relationship between two variables.  A positive correlation means that an increase in one quantity leads to an increase in another quantityA negative correlation means with increase in one quantity the other quantity decreases.+1 tells about a a perfect positive linear relationship and −1 indicates a perfect negative linear relationship.Values between 0 and 0.3 tells about a weak positive linear relationship, values between 0.3 and 0.7 shows a moderate positive correlation and a correlation of 0.7 and 1.0 states a strong positive linear relationship.Values between 0 and -0.3 tells about a weak negative linear relationship, values between -0.3 and -0.7 shows a moderate negative correlation and a correlation value of of -0.7 and -1.0 states a strong negative linear relationship.

As the number of hours increases, the number of points earned on the test increases. Thus, the two variables are positively correlated.

Thus, the coefficient correlation between two variables can be given by r = 0.70, that shows a moderate positive correlation.

Option b) r = 0.70

Fertilizer is sold in 100 pound bags labelled with the amount of nitrogen ( N), phosphoric acid ( P2O5), and potash ( K2O) present. The mixture of these nutrients varies from one type of fertilizer to the next. For example, a 100 pound bag of Vigoro Ultra Turf fertilizer contains 29 pounds of nitrogen, 3 pounds of phosphoric acid, and 4 pounds of potash. Another type of fertilizer, Parkerâ's Premium Starter, has 18 pounds of nitrogen, 25 pounds of phosphoric acid, and 6 pounds of potash per 100 pounds. Determine the amount of each type required to yield a mixture containing the 134 pounds of nitrogen, 37 pounds of phosphoric acid, and 22 pounds of potash.

Answers

Answer:

4 pounds of of Vigoro Ultra Turf fertilizer and 1 pound of Parkera's Premium Starter will yield the required mixture.

Step-by-step explanation:

In bag of Vigoro Ultra Turf fertilizer:

29 pounds of nitrogen, 3 pounds of phosphoric acid, and 4 pounds of potash

In bag of Parkera's Premium Starter:

18 pounds of nitrogen, 25 pounds of phosphoric acid, and 6 pounds of potash

Let the mass of Vigoro Ultra Turf fertilizer = x

Amount of nitrogen in x amount = 29 x

Amount of phosphoric acid  in  x amount = 3x

Let the mass of Parkera's Premium Starter= y

Amount of nitrogen in y amount = 18 y

Amount of phosphoric acid  in y  amount = 25 y

Amount of nitrogen in desired mixture formed by combing x and y fertilizers bag : 134 pounds

29 x + 18 y = 134 ..[1]

Amount of phosphoric acid  in desired mixture formed by combing x and y fertilizers bag : 37 pounds

3x + 25 y = 37 ..[2]

Solving [1] anf [2] by substitution method :

[tex]x=\frac{134-18y}{29}[/tex]

[tex]3\times \frac{134-18y}{29}+25y=37[/tex]

y = 1

[tex]x=\frac{134-18y}{29}=4[/tex]

4 pounds of of Vigoro Ultra Turf fertilizer and 1 pound of Parkera's Premium Starter will yield the required mixture.

To create a mixture containing 134 pounds of nitrogen, 37 pounds of phosphoric acid, and 22 pounds of potash, 400 pounds of Vigoro Ultra Turf and 100 pounds of Parker's Premium Starter are required, based on solving a system of linear equations.

We are tasked with finding the amount of two different fertilizers needed to obtain a mixture with 134 pounds of nitrogen, 37 pounds of phosphoric acid, and 22 pounds of potash. The two fertilizers are Vigoro Ultra Turf and Parkerâ's Premium Starter. We'll use a system of linear equations to solve this problem.

Let V represent the amount (in pounds) of Vigoro Ultra Turf and P represent the amount (in pounds) of Parkerâ's Premium Starter. The equations based on the given information are:

0.29V + 0.18P = 134 (nitrogen content)0.03V + 0.25P = 37 (phosphoric acid content)0.04V + 0.06P = 22 (potash content)

Now we solve these equations simultaneously to determine V and P. After solving, we find that V = 400 and P = 100. Thus, we need 400 pounds of Vigoro Ultra Turf and 100 pounds of Parkerâ's Premium Starter to obtain the desired mixture.

If Ms. P wants to withdraw $900 from an account earning 4% average annual interest rate at the start of each year for 7 years, how much must she have in the account today?

Answers

Answer:

Amount he must have in his account today is  $5,617.92

Step-by-step explanation:

Data provided in the question:

Regular withdraw amount = $900

Average annual interest rate, i = 4% = 0.04

Time, n = 7 years

Now,

Present Value = [tex]C \times\left[ \frac{1-(1+i)^{-n}}{i} \right] \times(1 + i)[/tex]

here,

C = Regular withdraw amount

Thus,

Present Value = [tex]C \times\left[ \frac{1-(1+i)^{-n}}{i} \right] \times(1 + i)[/tex]

Present Value = [tex]900 \times\left[ \frac{1-(1+0.04)^{-7}}{ 0.04 } \right] \times(1 + 0.04)[/tex]

Present Value = [tex]936 \times\left[ \frac{1 - 1.04^{-7}}{ 0.04} \right][/tex]

Present Value = [tex]936 \times\left[ \frac{1 - 0.759918}{ 0.04} \right][/tex]

Present Value = 936 × 6.00205

or

Present Value = $5,617.92

Hence,

Amount he must have in his account today is  $5,617.92

A die is thrown twice. Let X1 and X2 denote the outcomes, and define random variable X to be the minimum of X1 and X2. Determine the distribution of X.

Answers

Answer:

Step-by-step explanation:

Given that a die is thrown twice.  

X1, X2 are the outcomes in 2 throws

X1 = minimum of two

X1 can be either 1 or 2...6

Total outcomes are 36.

For x1 =1, fav ourable outcomes are (1,1) (1,2)...(1,6) (6,1)...(2/,1)= 11

P(X1=1) = [tex]\frac{11}{36}[/tex]\

P(X1=2) =Prob for one die showing two and other die showing 2 to 6

= [tex]\frac{9}{36}[/tex]

P(x1=3) = Prob for one die showing three and other die showing 3 to 6

=[tex]\frac{7}{36}[/tex]

thus we find that probability is reducing by 2 in the numerator

P(x1=4) = 5/36 followed by 3/36 for 5 and 1/36 for 6

Final answer:

The random variable X represents the minimum outcome when rolling a die twice. The values of X range from 1 to 6 and its distribution is given.

Explanation:

The random variable X represents the minimum outcome when rolling a die twice. In other words, X is the smaller of the two outcomes.

X can take on values from 1 to 6 since the outcomes of rolling a die are integers between 1 and 6. The minimum value of X is 1, which occurs when both dice show a 1.

The distribution of X is as follows:

p(X = 1) = 1/36p(X = 2) = 3/36p(X = 3) = 5/36p(X = 4) = 7/36p(X = 5) = 9/36p(X = 6) = 11/36

Geno wants to purchase gym membership. He has no more than y dollars to spend. Total Fitness charges an initial fee of $100 plus $30 per month. Gymania charges initial fee of $25 plus $50 per month. Write a system of equations that can be used to determine which company offers the better deal.

Answers

Answer:

Gymania is a better deal if the membership is for 3 months and below.

Total Fitness is a better if the membership is for 4 months and above.

Step-by-step explanation:

Let the number of months be 'x'.

Given:

Money Geno has = 'y' dollars.

Total Fitness charges:

Monthly fee = $30

Initial fee = $100

Gymania charges:

Monthly fee = $50

Initial fee = $25

Total charges is equal to the sum of initial fee and monthly fee multiplied by number of months.

So, for 'x' months, monthly fee charged by Total Fitness = [tex]30x[/tex]

For 'x' months, monthly fee charged by Gymania = [tex]50x[/tex]

Now, total charge by Total Fitness = Initial fee + Fee for 'x' months

Total charge by Total Fitness = [tex]100+30x[/tex]

Now, total charge by Gymania = Initial fee + Fee for 'x' months

Total charge by Gymania = [tex]25+50x[/tex]

Now, Geno has only 'y' dollars to spend. So, 'y' must be less than or equal to the total charge.

Therefore, the total charge for each membership is:

[tex]y=30x+100\\\\y= 50x+25[/tex]

Now, we graph both the equations. The graph is shown below.

From the graph, it is clear that, the total cost for Gymania (blue line) is less than that of Total Fitness (red line) till number of months equals 3.75 or 3 months. After 3.75 months, the graph of Gymania is above Total Fitness. So, if the membership is 4 months or above, then Total Fitness is more efficient.

Therefore, Gymania is a better deal if the membership is for 3 months and below.

Total Fitness is a better if the membership is for 4 months and above.

Final answer:

To determine which company offers the better deal for a gym membership, set up a system of equations for the two companies' costs and compare. Total Fitness charges $30 per month, while Gymania charges $50 per month. The equation is solved to find the break-even point where their costs are equal.

Explanation:

To determine which company offers the better deal, we can set up a system of equations based on the given information:

Let x be the total number of months for the gym membership.

Total Fitness charges an initial fee of $100 plus $30 per month, so the total cost can be represented by the equation: y = 30x + 100.

Gymania charges an initial fee of $25 plus $50 per month, so the total cost can be represented by the equation: y = 50x + 25.

To compare the two deals, we need to find the values of x where the total cost is the same for both companies. We can set up the following equation:

30x + 100 = 50x + 25.

Simplifying, we get:

20x = 75.

Dividing both sides by 20, we find that x = 3.75.

Since x represents the number of months, it cannot be a decimal, so we round up to the nearest whole number. Therefore, Geno should join Total Fitness if he plans to have the membership for 4 or more months, and Gymania if he plans to have the membership for 3 or fewer months.

In each of Problems 5 through 10, verify that each given function is a solution of the differential equation.
y''-y=0, y1(t)=e^t , y2(t)=cosht

Answers

Answer:

For First Solution: [tex]y_1(t)=e^t[/tex]

[tex]y_1(t)=e^t[/tex] is the solution of equation y''-y=0.

For 2nd Solution:[tex]y_2(t)=cosht[/tex]

[tex]y_2(t)=cosht[/tex]  is the solution of equation y''-y=0.

Step-by-step explanation:

For First Solution: [tex]y_1(t)=e^t[/tex]

In order to prove whether it is a solution or not we have to put it into the equation and check. For this we have to take derivatives.

[tex]y_1(t)=e^t[/tex]

First order derivative:

[tex]y'_1(t)=e^t[/tex]

2nd order Derivative:

[tex]y''_1(t)=e^t[/tex]

Put Them in equation y''-y=0

e^t-e^t=0

0=0

Hence [tex]y_1(t)=e^t[/tex] is the solution of equation y''-y=0.

For 2nd Solution:

[tex]y_2(t)=cosht[/tex]

In order to prove whether it is a solution or not we have to put it into the equation and check. For this we have to take derivatives.

[tex]y_2(t)=cosht[/tex]

First order derivative:

[tex]y'_2(t)=sinht[/tex]

2nd order Derivative:

[tex]y''_2(t)=cosht[/tex]

Put Them in equation y''-y=0

cosht-cosht=0

0=0

Hence [tex]y_2(t)=cosht[/tex]  is the solution of equation y''-y=0.

The functions [tex]y(t)=e^t[/tex] and [tex]y(t)=cosht[/tex] are solutions of given differential equation.

We have to prove that given function is a solution of the differential equation.

Given function,   [tex]y(t)=e^t[/tex]

Given differential equation are, [tex]y''-y=0[/tex]

  So that,  [tex]y'(t)=e^t , y''(t)=e^t[/tex]

Substitute values in given differential equation.

                    [tex]e^t - e^t=0[/tex]

Thus, function [tex]y(t)=e^t[/tex] is solution of given differential equation.

Another function is,  [tex]y(t)=cosht[/tex]

So that,  [tex]y'(t)=sinht , y''(t)=cosht[/tex]

Substitute values in given differential equation.

                    [tex]cosht-cosht=0[/tex]

Thus, function [tex]y(t)=cosht[/tex] is solution of given differential equation.

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Scientists at the Hopkins Memorial Forest have been collecting environmental data for more than 100 years. Sulfate content in water samples from Birch Brook is known to be 7.38mg/L with a standard deviation of 1.60mg/L. If 10 students measure the sulfate in their samples to together estimate the mean, what is the probability that they get a sampling error of greater than 1.0 mg/L? What assumptions do we need to make to calculate that value?

Answers

Answer:

The probability that they get a sampling error of greater than 1.0 mg/L is 0.04.

To calculate this value, I assume that the samples are randomly collected

Step-by-step explanation:

Sampling error can be calculated using the formula

[tex]\frac{t*s}{\sqrt{n}}[/tex]  where

t is the statistic of the probability getting the standard errors is the standard deviation (1.60mg/L)n is the sample size (10)

For the sampling error of 1.0mg/L we have

[tex]1=\frac{t*1.60}{\sqrt{10}}[/tex]    

Solving for t we have t≈1.976

Then the probability that they get a sampling error of greater than 1.0 mg/L is

P(t>1.976) ≈ 0.04.

In other words, we are 96% confident that the sampling error is within 1.0 mg/L.

To calculate this value, I assume that the samples are randomly collected.

Ace Truck leases its 10-ft box truck at $30/day and $0.50/mi, whereas Acme Truck leases a similar truck at $25/day and $0.55/mi.

(a) Find the daily cost of leasing from each company as a function of the number of miles driven.

Answers

Answer:

Ace Truck

[tex]C(m) = 30 + 0.5*m[/tex]

Acme Truck

[tex]C(m) = 25 + 0.55*m[/tex]

Step-by-step explanation:

The cost function to lease a box truck from a company has the following format:

[tex]C(m) = F + a*m[/tex]

In which F is the fixed cost and a is the cost per mile m.

(a) Find the daily cost of leasing from each company as a function of the number of miles driven.

Ace Truck

$30/day and $0.50/mi. This means that [tex]F = 30, a = 0.50[/tex]. So

[tex]C(m) = 30 + 0.5*m[/tex]

Acme Truck

$25/day and $0.55/mi. This means that [tex]F = 25 a = 0.55[/tex]. So

[tex]C(m) = 25 + 0.55*m[/tex]

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