A citrus grower anticipates a profit of $100,000 this year if the nightly temperatures remain mild. Unfortunately, the weather forecast indicates a 10% chance that the temperature will drop below freezing during the next week. Such freezing weather will destroy 40% of the crop and reduce the profit to $60,000. However, the grower can protect the citrus fruit against the possible freezing at a cost of $5000. Should the grower spend the $5000 and thereby reduce the profit to $95,000? Hint: Compute E(X), where X is the profit the grower will get if he does nothing to protect the fruit.

Answer:

(a) The confidence level desired by the researchers is 95%.

(b) The sampling error is 0.002 microcurie per millilitre.

(c) The sample size necessary to obtain the desired estimate is 25.

Step-by-step explanation:

The mean and standard deviation of the amount of the radioactive​ element, cesium-137 present in a sample of n = 16 lichen specimen are:

[tex]\bar x=0.009\\s=0.005[/tex]

Now it is provided that the researchers want to increase the sample size in order to estimate the mean μ to within 0.002 microcurie per millilitre of its true​ value, using a​ 95% confidence interval.

The (1 - α)% confidence interval for population mean (μ) is:

[tex]CI=\bar x\pm z_{\alpha/2}\times \frac{s}{\sqrt{n}}[/tex]

(a)

The confidence level is the probability that a particular value of the parameter under study falls within a specific interval of values.

In this case the researches wants to estimate the mean using the 95% confidence interval.

Thus, the confidence level desired by the researchers is 95%.

(b)

In case of statistical analysis, during the computation of a certain statistic, to estimate the value of the parameter under study, certain error occurs which are known as the sampling error.

In case of the estimate of parameter using a confidence interval the sampling error is known as the margin of error.

In this case the margin of error is 0.002 microcurie per millilitre.

(c)

The margin of error is computed using the formula:

[tex]MOE=z_{\alpha/2}\times \frac{s}{\sqrt{n}}[/tex]

The critical value of z for 95% confidence level is:

[tex]z_{\alpha/2}=z_{0.05/2}=z_{0.025}=1.96[/tex]

*Use a z-table.

[tex]MOE=z_{\alpha/2}\times \frac{s}{\sqrt{n}}[/tex]

 [tex]0.002=1.96\times \frac{0.005}{\sqrt{n}}[/tex]

       [tex]n=[\frac{1.96\times 0.005}{0.002}]^{2}[/tex]

          [tex]=(4.9)^{2}\\=24.01\\\approx 25[/tex]

Thus, the sample size necessary to obtain the desired estimate is 25.

Answer:

Option B: Paired sample t-test, since there is a "before 20 push ups" and "after 20 push up " score for each participant.

Step-by-step explanation:

In this question, we have a case where each participant undergoes different exercises to see differences before and after 20 push ups.

Now, it's obvious that the same condition applies to all participants i.e. differences before and after 20 push ups.

Thus, this will be a case of paired sample t-test or dependent t test because conditions for all participants in both tests are dependent or same.

So, the only option that corresponds with my explanation to use paired sample is option B

Answer:B 1.5

Step-by-step explanation:

I just did it

Answer:

B 1.5

Step-by-step explanation:

Answer:

[tex]\mathrm{Number\:of\:Australians\:with\:type\:A\:Blood\:Group\:in\:2010}\:\approx\:8175987[/tex]

Step-by-step explanation:

[tex]\mathrm{Percent}:\\\\\mathrm{A\:ratio\:expressed\:as\:a\:fraction\:out\:of\:a\:hundred.}\\\\\mathrm{In\:order\:to\:convert\:a\:percent\:to\:a\:ratio,\:divide\:it\:by\:100}:\\\\38\%\:=\frac{38}{100}\\\\\mathrm{Percentage\:of\:population\:with\:A\:Blood\:Type}\:=\frac{38}{100}\times 21515754\\\\\approx8175987[/tex]

An estimated 8,175,987 Australians have Type A blood.

There were 21,515,754 in Australia in 2010. In the same year, it was estimated that 38% of people had Type A blood.

This means that the best estimate of people in Australia with Type A blood is:

= Percentage of people with type A blood x Number of people in Australia

= 38% x 21,515,754

= 8,175,986.52

= 8,175,987 people

In conclusion, approximately 8,175,987 Australians have Type A blood.

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Answer:

Step-by-step explanation:

Confidence interval gives possible estimate (range of values) that could contain the population proportion. Confidence level does not not mean probability. It only tells how confident we are that that the population proportion lies within the confidence interval.

Since we were told that the confidence interval has a lower bound of 161.9 seconds and an upper bound of 165.5 seconds, therefore, the correct option for the given situation is

A. One can be 90​% confident that the mean​ drive-through service time of this​ fast-food chain is between 161.9 seconds and 165.5 seconds.

The correct interpretation of the 90% confidence interval for the fast-food chain's drive-through service times is that it likely contains the true mean service time (A. One can be 90% confident that the mean drive-through service time is between 161.9 seconds and 165.5 seconds).

The correct answer to the question is A. One can be 90% confident that the mean drive-through service time of this fast-food chain is between 161.9 seconds and 165.5 seconds. This statement is a proper interpretation of a confidence interval in statistics. Confidence intervals provide a range of values, derived from the data sample, that likely contain the population parameter of interest. In this case, the parameter of interest is the mean drive-through service time. Option B is incorrect because it overly simplifies the range into a single value. Option C is misleading because it attributes a probability to the mean's location, which is not accurate in the context of confidence intervals. Lastly, option D incorrectly suggests that the mean service time falls within a specific range 90% of the time, rather than conveying the level of confidence in the interval containing the true mean.

Answer:

(a) [tex]\bar x\sim N(\mu_{\bar x}=50,\ \sigma_{\bar x}=0.5)[/tex]

(b) The z-score for the sample mean [tex]\bar x[/tex] = 51 is 2.

(c) The value of [tex]P(\bar X < 51)[/tex] is 0.9773.

Step-by-step explanation:

The random variable X has mean, μ = 50 and standard deviation, σ = 4.

A random sample of size n = 64 is selected.

(a)

According to the Central Limit Theorem if we have a population with mean μ and standard deviation σ and appropriately huge random samples (n > 30) are selected from the population with replacement, then the distribution of the sample mean will be approximately normally distributed.

Then, the mean of the  distribution of sample mean is given by,

[tex]\mu_{\bar x}=\mu[/tex]

And the standard deviation of the  distribution of sample mean is given by,

[tex]\sigma_{\bar x}=\frac{\sigma}{\sqrt{n}}[/tex]

The sample of X selected is, n = 64 > 30.

So, the Central limit theorem can be applied to approximate the distribution of sample mean ([tex]\bar x[/tex]).

[tex]\bar x\sim N(\mu_{\bar x}=50,\ \sigma_{\bar x}=0.5)[/tex]

(b)

The z-score for the sample mean [tex]\bar x[/tex] is given as follows:

[tex]z=\frac{\bar x-\mu_{\bar x}}{\sigma_{\bar x}}[/tex]

Compute the z-score for [tex]\bar x[/tex] = 51 as follows:

[tex]z=\frac{\bar x-\mu_{\bar x}}{\sigma_{\bar x}}[/tex]

  [tex]=\frac{51-50}{0.5}\\[/tex]

  [tex]=2[/tex]

Thus, the z-score for the sample mean [tex]\bar x[/tex] = 51 is 2.

(c)

Compute the value of [tex]P(\bar X < 51)[/tex] as follows:

[tex]P(\bar X < 51)=P(\frac{\bar x-\mu_{\bar x}}{\sigma_{\bar x}}<\frac{51-50}{0.5})[/tex]

                  [tex]=P(Z<2)\\=0.97725\\\approx 0.9773[/tex]

*Use a z-table for the probability.

Thus, the value of [tex]P(\bar X < 51)[/tex] is 0.9773.

The x distribution is normally distributed with a mean of 50 and a standard deviation of 4. The z-value corresponding to x = 51 is 0.25. The probability of x being less than 51 is approximately 0.5987.

a) The x distribution is normally distributed with a mean of 50 and a standard deviation of 4.

The mean of the distribution is μx = 50 and the standard deviation is σx = 4.

b) To find the z-value corresponding to x = 51, we can use the formula z = (x - μ) / σ. Plugging in the values, we get z = (51 - 50) / 4 = 0.25.

c) To find P(x < 51), we can use the standard normal distribution table or a calculator to find the corresponding cumulative probability. The value is approximately 0.5987, rounded to four decimal places.

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Answer:

g-x=3

Step-by-step explanation:

Answer:

Cov (X,Y) = 6

Step-by-step explanation:

hello,

Cov(X,Y) = E(XY) - E(X)E(Y)

we must  first find E(XY), E(X), and E(Y).

since X is uniformly distributed on the interval (0,12), then E(X) = 6.

next we find the joint density f(x,y) using the formula

[tex]f(x,y) = g(y|x)f_{X}(x)[/tex]

[tex]f_{X} (x) = \frac{1}{12} \ $for$\ 0<x<12[/tex] this is because f is uniformly distributed on the the interval (0,12)

also since the conditional probability density of Y given X=x, is  uniformly distributed on the interval [0,x], then

[tex]g(y|x)=\frac{1}{x}[/tex]  for 0≤y≤x≤12

thus

[tex]f(x,y)=\frac{1}{12x}[/tex].

hence,

[tex]E(X,Y)= \int\limits^{12}_{x=0} \int\limits^x_{y=o} xy\frac{1}{12x} \,dy dx[/tex]

[tex]E(X,Y)=\frac{!}{24} \int\limits^{12}_{x=0} x^2 \, dx = 24[/tex]

also,

[tex]E(Y) = \int\limits^{12}_{x=0} \int\limits^x_{y=0} y\frac{1}{12x} \, dydx[/tex]

[tex]E(Y)=\frac{1}{24}\int\limits^{12}_{x=0} {x} \, dx =3[/tex]

thus Cov(X,Y) = E(XY) - E(X)E(Y)

                      =  24 - (6)(3)

                      =     6

Final answer:

To find the volume of a cone with a radius of 3 inches and a height of 10 inches, use the formula V = [tex]\frac{1}{3}[/tex] * π * r² * h. Substitute the values and calculate to find a volume of B) 94.26 cubic inches.

Explanation:

The volume of the cone can be calculated using the formula V = [tex]\frac{1}{3}[/tex] * π * r² * h.

Substitute the values for the radius (3 inches) and height (10 inches) into the formula to find the volume:

V = [tex]\frac{1}{3}[/tex] * 3.142 * 3² * 10

V = [tex]\frac{1}{3}[/tex] * 3.142 * 9 * 10

V = 94.26 cubic inches

Therefore, the cone's volume is 94.26 cubic inches.

Answer:

The answer is Option E; both B and C are correct

Step-by-step explanation:

Both (b) and (c) are correct. Simpson's paradox is a paradox in which a trend that appears in different groups of data disappears when these groups are combined, and the reverse trend appears for the aggregate data. This result is often encountered in social-science and medical-science statistics, and is particularly confounding when frequency data are unduly given causal interpretations. Simpson's Paradox disappears when causal relations are brought into consideration.

Now, this question is an example of Simpsons paradox because the groups of collected data over a period of time from five major cities showed a trend that StatsAir does better overall, but this trend is reversed when the groups are studied separately to show that air median does better.

So, option B is correct.

Also, City is a variable that influences both the dependent variable and independent variable, causing a spurious association. That is it is the cause of why the 2 results are biased. Thus, city is a lurking variable.

So, option C is also correct

Answer:

360

Step-by-step explanation:

formula is L x B x H so first layer would be 6 rows of 20 cubes which is 120cubes

H is 3cm so 120x3 = 360

or find out the volume of the box, 20x6x3 = 360

Answer:

See explanation

Step-by-step explanation:

Solution:-

- A survey was conducted among the College students for their motivations of using credit cards two years ago. A randomly selected group of sample size n = 425 college students were selected.

- The results of the survey test taken 2 years ago and recent study are as follows:

                                           Old Survey ( % )            New survey ( Frequency )

                  Reward                 27                                              112

                  Low rate               23                                              96

                  Cash back           21                                              109

                  Discount              9                                               48

                  Others                  20                                             60

- We are to test the claim for any changes in the expected distribution.

We will state the hypothesis accordingly:

Null hypothesis: The expected distribution obtained 2 years ago for the motivation behind the use of credit cards are as follows: Rewards = 27% , Low rate = 23%, Cash back = 21%, Discount = 9%, Others = 20%

Alternate Hypothesis: Any changes observed in the expected distribution of proportion of reasons for the use of credit cards by college students.

( We are to test this claim - Ha )

We apply the chi-square test for independence.

- A chi-square test for independence compares two variables in a contingency table to see if they are related. In a more general sense, it tests to see whether distributions of categorical variables differ from each other.

- We will compute the chi-square test statistics ( X^2 ) according to the following formula:

                                [tex]X^2 = Sum [ \frac{(O_i - E_i)^2}{Ei} ][/tex]

Where,

 O_i : The observed value for ith data point

 E_i : The expected value for ith data point.

- We have 5 data points.

So, Oi :Rewards = 27% , Low rate = 23%, Cash back = 21%, Discount = 9%, Others = 20% from a group of n = 425.

     Ei : Rewards = 112 , Low rate = 96, Cash back = 109, Discount = 48, Others = 60.

Therefore,

                     [tex]X^2 = [ \frac{(112 - 425*0.27)^2}{425*0.27} + \frac{(96 - 425*0.23)^2}{425*0.23} + \frac{(109 - 425*0.21)^2}{425*0.21} + \frac{(48 - 425*0.09)^2}{425*0.09} + \frac{(60 - 425*0.20)^2}{425*0.20}]\\\\X^2 = [ 0.06590 + 0.03132 + 4.37044 + 2.48529 + 7.35294]\\\\X^2 = 14.30589[/tex]

- Then we determine the chi-square critical value ( X^2- critical ). The two parameters for evaluating the X^2- critical are:

                     Significance Level ( α ) = 0.10

                     Degree of freedom ( v ) = Data points - 1 = 5 - 1 = 4  

Therefore,

                     X^2-critical = X^2_α,v = X^2_0.1,4

                    X^2-critical = 7.779

- We see that X^2 test value = 14.30589 is greater than the X^2-critical value = 7.779. The test statistics value lies in the rejection region. Hence, the Null hypothesis is rejected.

Conclusion:-

This provides us enough evidence to conclude that there as been a change in the claimed/expected distribution of the motivations of college students to use credit cards.

Final answer:

To test for changes in credit card usage motivations among college students, calculate the expected frequencies from the old survey percentages, apply the chi-square goodness-of-fit test, and compare the statistic to the chi-square critical value at alpha = 0.10.

Explanation:

Understanding the Hypothesis Test for Changed Distribution

To determine whether there has been a shift in the distribution of college students' motivations for using credit cards since the previous survey, we conduct a hypothesis test for goodness of fit. The test will compare the observed frequencies from the new survey with the expected frequencies based on the old survey percentages.

First, calculate the expected frequency for each category using the formula: Expected Frequency = (Old Survey Percentage / 100) × Total Number of Students. Then, employ the chi-square goodness-of-fit test to analyze the data:

Reward: Expected = 0.27 × 425 = 114.75

Low rate: Expected = 0.23 × 425 = 97.75

Cash back: Expected = 0.21 × 425 = 89.25

Discount: Expected = 0.09 × 425 = 38.25

Others: Expected = 0.20 × 425 = 85.00

With the expected frequencies calculated, the chi-square statistic will be computed:

χ² = Σ((Observed - Expected) ² / Expected)

This statistic will be compared to a critical value from the chi-square distribution table with (n - 1) degrees of freedom, where n is the number of categories, at alpha = 0.10 to determine whether to reject the null hypothesis (no change in distribution).

Hey there!

"A number" is referred to an unknown number so we can say it is labled as

[tex]x[/tex]

"Increased" means you're going up/ adding

ten = 10

150 stays the same

"Ten times a number" =

[tex] \bf{10x}[/tex]

"Increased by 150" =

[tex] \bf{ + 150}[/tex]

Thus your answer should look like this:

[tex] \bf{10x + 150}[/tex]

Good luck on your assignment and enjoy your day!

~

[tex] \frak{loveyourselffirst }[/tex]

To solve this problem, we can use the algebraic expression 10x + 150, where 'x' represents the number.

To solve the problem, we can translate the given phrase into an algebraic expression. Let's assume the number is represented by the variable 'x'. 'Ten times a number increased by 150' can be written as 10x + 150. This expression represents ten times the number 'x' plus 150.

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The remaining part of Question:

4) what can we conclude about sampling variability in the sample proportion calculated in the sample at John Hopkins as compared to that calculated in the sample at Ohio State.

5) The number of undergraduates at Johns Hopkins is approximately 2000 while the number at Ohio State is approximately 40000, suppose instead that at both schools, a simple random sample of about 3% of the undergraduates Will be taken.

Answer:

4) The sample proportion from Johns Hopkins will have about the same sampling variability as that from Ohio State

5) The sample proportion from John Hopkins will have more sampling variability than that from Ohio State

Step-by-step explanation:

Note: The sampling variability in the sample proportion decreases with increase in the sample size.

4) since the sample size at both Johns Hopkins and Ohio State is the same (i.e. n = 50), the sample variability of the sample proportion will be the same for both cases.

5) 3% of the population are selected for the observation in both cases.

At Johns Hopkins, sample size, n = 3% * 2000

n = 60

At Ohio State, sample size, n = 3% * 40000

n = 1200

Since sampling variability in the sample proportion decreases with increase in the sample size, the sampling variability in sample proportion will be higher at Johns Hopkins than at Ohio State.

Answer:

27 hours

Step-by-step explanation:

So t is going to represent time. $8t is equivalent to how much she makes for t amount of time. Being she is already down $16 for the gas, we have to take that into consideration that she must earn that back. Set up:

$8t - $16 = $200

Add like terms to get:

$8t = $216

Now solve for t by dividing both sides by $8.

t = 27 hours

Answer:

-3 (x-3) (x+1)

Step-by-step explanation:

− 3 x ^2 + 6 x + 9

Factor out -3

-3( x^2 -2x-3)

The terms inside the parentheses can be factored

What 2 numbers multiplies to -3 and adds to -2

-3*1 = -3

-3+1 =-2

-3 (x-3) (x+1)

Answer:

h ≥ -12

Step-by-step explanation:

at least means greater than or equal to

h ≥ -12

Answer:

h>-12

Step-by-step explanation:

Answer:

101.85ft

Step-by-step explanation:

We can use tan here

[tex]tan\theta=\frac{opp}{adj}\\adj = \frac{opp}{tan \theta} \\adj = \frac{74}{tan(36)} = 101.85[/tex]

The height should be 101.85 ft.

Since it casts a 74-ft shadow when the angle of elevation of the sun is 36°

So, the height should be

Here we can use tan

[tex]= 74\div tan\ 36[/tex]

= 101.85 ft

Therefore, we can conclude that The height should be 101.85 ft.

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Answer:

B = 58

Step-by-step explanation:

Complementary angles add to 90 degrees

A+B = 90

If one of the angles is 32

32+ B = 90

Subtract 32 from each side

32-32+B = 90-32

B = 58

Answer:

C

Step-by-step explanation:

The samples are independent because there is not a natural pairing between the two samples.

Since Independent samples are samples that are selected randomly so that its observations do not depend on the values other observations also data set in which each data point in one sample is not paired to a data point in the second sample

To multiply fractions, multiply the numerators and denominators separately. For [tex]\frac{3}{4} \times \frac{16}{19}[/tex], the result is [tex]\frac{12 }{ 19}[/tex] after simplifying.

To multiply fractions, you simply multiply the numerators together and the denominators together. In this case, [tex]\frac{3}{4} \times \frac{16}{19}[/tex] would be calculated as: [tex]\frac{3 \times 16}{4 \times 19} = \frac{48 }{ 76}[/tex] This fraction can be simplified by dividing both the numerator and denominator by their greatest common factor, which in this case is 4: [tex]\frac{48 \div 4 }{ 76 \div 4 }= \frac{12 }{ 19}[/tex].

Answer:

The resulting residual value is e=-40.24.

Step-by-step explanation:

The residual value e for a regression model is defined as the difference between the real value y and the predicted value yp:

[tex]e=y-y_p[/tex]

The predicted value for DISTANCE=0.43 miles is:

[tex]RENT = 1200.4326 - 256.2567\cdot DISTANCE\\\\RENT(0.43) = 1200.4326 - 256.2567\cdot 0.43\\\\RENT(0.43) = 1200.4326 - 110.1904=1090.2422\\\\[/tex]

Then, if the real value is $1,050, the residual value is calculated as:

[tex]e=y-y_p\\\\e=1050-1090.2422=-40.2422[/tex]

Answer:

  7.42 years (7 years 5 months)

Step-by-step explanation:

The future value of Carmen's account can be modeled by

  FV = P(1 +r/12)^(12t)

where P is the principal invested, r is the annual rate, and t is the number of years.

Solving for t, we have ...

  FV/P = (1 +r/12)^(12t)

  log(FV/P) = 12t·log(1 +r/12)

  t = log(FV/P)/(12·log(1 +r/12))

For FV = 3560, P=2000, r = 0.078, the time required is ...

  t = log(3560/2000)/(12·log(1 +.078/12))

  t ≈ 7.42

It will take Carmen about 7 years 5 months to reach her savings goal.

Answer:

[tex](y,z)=(3,5)\\\\ |(y,z)|=\sqrt{34}[/tex]

Step-by-step explanation:

Given y(3, 1), z(0, 4)

-The ordered pair is written by summing their corresponding coordanates as below:

[tex](y,z)=y(3,1)+z(0,4)\\\\=(3+0,1+4)\\\\=(3,5)[/tex]

-We then calculate the magnitude of this ordered pair as following:

[tex]|y,z|=\sqrt{y^2+z^2}\\\\=\sqrt{3^2+5^2}\\\\=\sqrt{34}\\\\\therefore |(y,z)|=\sqrt{34}[/tex]

Hence, the magnitude of the ordered pair is [tex]\sqrt{34}[/tex]

Answer:

0.23 or 3/13.

Step-by-step explanation:

There are 52 cards in a deck. There are four different suits, dividing the decks into 4 sets of 13. There are three face cards for each suit so 3/13. 3 divided by 13 is 0.23. Use fractions if you can because they are easier and more accurate.

To find the probability that at least one of the 4 cards drawn is a face card, calculate the probability of all cards not being face cards and subtract that from 1, resulting in approximately 64.9%.

The problem can be approached by finding the probability that none of the 4 cards drawn is a face card and then subtracting that from 1 to find the probability that at least one is a face card. There are 12 face cards in a standard deck of 52 cards, leaving 40 non-face cards. When the audience members draw and replace the cards, each draw is independent of the previous draw.

First, calculate the probability of drawing a non-face card (P(NF)):

P(NF) = number of non-face cards / total number of cards = 40/52

Since the card is replaced each time, the probability remains the same for each of the four draws. Thus, the probability that all 4 cards are non-face cards is:

P(all four are NF) = [tex]P(NF)^4 = (40/52)^4[/tex]

Then subtract this probability from 1 to get the probability of at least one face card:

P(at least one face card) = 1 - P(all four are NF)

Calculation:

P(at least one face card) = [tex]1 - (40/52)^4 = 1 - (0.7692)^4[/tex]

P(at least one face card) ≈ 1 - 0.3515 ≈ 0.6485

Therefore, the probability that at least one of the 4 cards drawn is a face card is approximately 64.9% (rounded to one decimal place).

Answer:

x=4

Step-by-step explanation:

4x-6 + 2x = 18

Combine like terms

6x -6 = 18

Add 6 to each side

6x-6+6 =18+6

6x = 24

Divide each side by 6

6x/6 = 24/6

x =4

Answer:

Option D is correct.

Failure to reject the null hypothesis means that there is not sufficient evidence to support the claim that the mean attendance is greater than 694.

Step-by-step explanation:

For hypothesis testing, the first thing to define is the null and alternative hypothesis.

The null hypothesis plays the devil's advocate and usually takes the form of the opposite of the theory to be tested. It usually contains the signs =, ≤ and ≥ depending on the directions of the test.

It usually maintains that random chance is responsible for the outcome or results of any experimental study/hypothesis testing.

While, the alternative hypothesis usually confirms the the theory being tested by the experimental setup. It usually contains the signs ≠, < and > depending on the directions of the test.

It usually maintains that other than random chance, there are significant factors affecting the outcome or results of the experimental study/hypothesis testing.

For this question, the owner of a football team claims that the average attendance at games is over 694, and he is therefore justified in moving the team to a city with a larger stadium.

So, the null hypothesis would be that there isn't enough evidence to support the claim that the mean attendance is greater than 694.

That is, the mean isn't greater than 694; the mean is less than or equal to 694.

While the alternative hypothesis would be that there is sufficient evidence to support the claim that the mean attendance is greater than 694.

Mathematically,

The null hypothesis is represented as

H₀: μ₀ ≤ 694

The alternative hypothesis is represented as

Hₐ: μ₀ > 694

Failure to reject the null hypothesis means that the null hypothesis is true. Hence, the answer choice picked is the obvious correct answer.

Hope this Helps!!!

Answer: 1.59 hours elapsed before the body was found

Step-by-step explanation: Please see the attachments below

Answer:

We can therefore conclude that the geographical distribution of hotline callers could be the same as the U.S population distribution.

Step-by-step explanation:

The null Hypothesis: Geographical distribution of hotline callers could be the same as the U.S. population distribution

Alternative hypothesis: Geographical distribution of hotline callers could not be the same as the U.S. population distribution

The populations considered are the Midwest, South, Northeast, and west.

The number of categories, k = 4

Number of recent calls = 200

Let the number of estimated parameters that must be estimated, m = 0

The degree of freedom is given by the formula:

df = k - 1-m

df = 4 -1 - 0 = 3

Let the significance level be, α = 5% = 0.05

For  α = 0.05, and df = 3,

from the chi square distribution table, the critical value = 7.815

Observed and expected frequencies of calls for each of the region:

Northeast

Observed frequency = 39

It contains 18.1% of the US Population

The probability = 0.181

Expected frequency of call = 0.181 * 200 = 36.2

Midwest

Observed frequency = 55

It contains 21.9% of the US Population

The probability = 0.219

Expected frequency of call = 0.219 * 200 =43.8

South

Observed frequency = 60

It contains 36.7% of the US Population

The probability = 0.367

Expected frequency of call = 0.367 * 200 = 73.4

West

Observed frequency = 46

It contains 23.3% of the US Population

The probability = 0.233

Expected frequency of call = 0.233 * 200 = 46

[tex]x^{2} = \sum \frac{(O_{i} - E_{i}) ^{2} }{E_{i} } , i = 1, 2,.........k[/tex]

Where [tex]O_{i} =[/tex] observed frequency

[tex]E_{i} =[/tex] Expected frequency

Calculate the test statistic value, x²

[tex]x^{2} = \frac{(39 - 36.2)^{2} }{36.2} + \frac{(55 - 43.8)^{2} }{43.8} + \frac{(60 - 73.4)^{2} }{73.4} + \frac{(46 - 46.6)^{2} }{46.6}[/tex]

[tex]x^{2} = 5.535[/tex]

Since the test statistic value, x²= 5.535 is less than the critical value = 7.815, the null hypothesis will not be rejected, i.e. it will be accepted. We can therefore conclude that the geographical distribution of hotline callers could be the same as the U.S population distribution.  

Step-by-step explanation:

Let the two-digit number be 10t+u where t is the tens digit and u the units digit.

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EQUATION:

t+u = 11

10u+t = 10t+u + 45

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Rearrange the equations:

t+u = 11

9t-9u = -45

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Simplify:

t+u = 11

t-u = -5

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Add the equations to solve for "t":

2t = 6

t = 3

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Substitute to solve for "u":

3+u = 11

u = 8