Which are solutions of the linear equation?
Select all that apply.

3x + y = 10

(1, 6)
(2, 4)
(3, 1)
(4, –1)
(5, –5)

Answer:

Make a Type I error 15% of the time.

Step-by-step explanation:

The question is incomplete:

In all possible samples of size 100 she will:

Option 2 and 4 are both false because we can not estimate this probabilitites a priori.

Option 1.

The power of a test is defined as the conditional probability of rejecting the null hypothesis, given that the alternative hypothesis is true.

Then, the power of the test is complementary of the probability of failing to reject the null hypothesis, gicen that the alternative hypothesis is true. The last is the definition of the probability of a Type II error.

This means that a power of 0.85 implies a probability of (1-0.85)=0.15 of making a Type II error.

[tex]P(Type \,II\, error)=1-Power=1-0.85=0.15[/tex]

The option 1 ("Make a Type II error 5% of the time") is not precise, so it is not correct.

Option 3

The significance level is 0.05. This is also the probability of making a Type I error.

The option 3 ("Make a Type I error 5% of the time") is correct.

Answer:

18.4 kilograms

Step-by-step explanation:

Since we know that both classes recycled the same amount, and together they recycled a total of 36.8 kilograms, this means we can simply divide the total (36.8 kg) by two, to get how much each class recycled.

To find out how much each class recycled, you divide the total amount, 36.8 kilograms, evenly by the number of classes, 2. This results in each class recycling 18.4 kilograms of paper.

In this Mathematics scenario, we are essentially dividing the total amount of waste recycled, 36.8 kilograms, into two equal parts because there are two classes that contributed to this total equally. To find the amount recycled by each class, we could use division.

To calculate, we divide the total amount of waste, 36.8 kilograms, by the number of classes, which is 2. This division looks like this: 36.8 ÷ 2. If you carry out this calculation, the result is 18.4. So, each fifth-grade class recycled 18.4 kilograms of paper this week.

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Answer:

740°

Step-by-step explanation:

2 *360° + 20° = 720° + 20° = 740°

The volume of the cup is 20 cubic inches.

Step-by-step explanation:

Given that the cup is in the hemispherical shape.

Radius of the cup = 2.125 inches

Volume of the cup = (2/3) π(r x r x r)

= (2/3) (3.14) (2.125 x 2.125 x 2.125)

= (2/3) (3.14) ( 9.59)

= (2/3) (30.13)

= 20 cubic inches

The volume of the cup is 20 cubic inches.

Answer: The minimum reliability for the second stage be 0.979.

Step-by-step explanation:

Since we have given that

Probability for the overall rocket reliable for a successful mission = 97%

Probability for the first stage = 99%

We need to find the minimum reliability for the second stage :

So, it becomes:

P(overall reliability) = P(first stage ) × P(second stage)

[tex]0.97=0.99\times x\\\\\dfrac{0.97}{0.99}=x\\\\0.979=x[/tex]

Hence, the minimum reliability for the second stage be 0.979.

Using probability of independent events, it is found that the minimum reliability for the second stage must be of 97.98%.

If two events, A and B, are independent, the probability of both events happening is the multiplication of the probability of each happening, that is:

[tex]P(A \cap B) = P(A)P(B)[/tex]

In this problem:

Then:

[tex]P(A \cap B) = P(A)P(B)[/tex]

[tex]0.97 = 0.99P(B)[/tex]

[tex]P(B) = \frac{0.97}{0.99}[/tex]

[tex]P(B) = 0.9798[/tex]

Hence, the minimum reliability for the second stage must be of 97.98%.

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Answer:

x = 1/2( ln(4) -5)

Step-by-step explanation:

e ^ 2x+5 =4

Take the natural log of each side

ln(e ^ 2x+5) =ln(4)

2x+5 = ln(4)

Subtract 5 from each side

2x = ln(4) - 5

Divide by 2

2x/2 = (ln(4) -5)/2

x = 1/2( ln(4) -5)

Answer:

its c on edge.

Answer:

90% confidence interval: (22.35,23.45)

Step-by-step explanation:

We are given the following in the question:

Sample mean, [tex]\bar{x}[/tex] = 22.9 years

Sample size, n = 20

Alpha, α = 0.10

Population standard deviation, σ = 1.5 years

90% Confidence interval:

[tex]\bar{x} \pm z_{critical}\dfrac{\sigma}{\sqrt{n}}[/tex]

Putting the values, we get,

[tex]z_{critical}\text{ at}~\alpha_{0.05} = \pm 1.64[/tex]

[tex]22.9 \pm 1.64(\dfrac{1.5}{\sqrt{20}} )\\\\ = 22.9 \pm 0.55 \\\\= (22.35,23.45)[/tex]

(22.35,23.45) is the required 90% confidence interval for population mean age.

The 90% confidence interval for the population mean age, from the given data, is approximated to be between 22.46 years and 23.34 years.

To construct a 90% confidence interval of the population mean age, we will use the formula for the confidence interval, which is sample mean ± Z-score * (Standard deviation/ sqrt (sample size)), where the Z-score is based on the confidence level. For 90%, the Z-score is 1.645.

So in this case, the sample mean is 22.9 years, the known standard deviation is 1.5 years, and the sample size (n) is 20. Plugging these values into the formula gives:

22.9 ± 1.645 * (1.5 / sqrt (20))

When you calculate, it will give an interval of about 22.46 years to 23.34 years. Hence the 90% confidence interval for the population mean age is 22.46 years to 23.34 years.

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Wow. That is a nice tuition.

The question is asking, how much money does he need to pay after taking consideration of the grant money and scholarship.

We need to add up the estimated costs for his first year. 1,600+400+1,400=3,400.

Subtract both 1,900 and 1,000 (2,900) because he does not need to pay that amount from the original estimated costs.

3,400-2,900=500

This means that aside from the financial aid help, he will need to pay an additional 500 dollars to pay for his first year of college.

Answer:$500

Step-by-step explanation:

1. 1,600+400+1,400=3,400

2. 1,900+1,000=2,900

3. 3,400-2,900=500

Answer:

The significance level α is 0.02.

Step-by-step explanation:

A hypothesis test for single mean can be performed to determine whether the average starting salary for a 21-25 year old college grad exceeds $52,000 per year.

The hypothesis is defined as follows:

H₀: The average starting salary for a 21-25 year old college grad does not exceeds $52,000 per year, i.e. µ ≤ 52,000.

Hₐ: The average starting salary for a 21-25 year old college grad exceeds $52,000 per year, i.e. µ > 52,000.

The information provided is:

σ = $5,745

n = 65

Also, if [tex]\bar X>\$53,460[/tex] then the null hypothesis will be rejected.

Here, we need to compute the value of significance level α, the type I error probability.

A type I error occurs when we reject a true null hypothesis (H₀).

That is:

α = P (type I error)

α = P (Rejecting H₀| H₀ is true)

   [tex]=P(\bar X>53460|\mu \leq 52000)[/tex]

   [tex]=P[\frac{\bar X-\mu_{0}}{\sigma/\sqrt{n}}>\frac{53460-52000}{5745/\sqrt{65}}][/tex]

   [tex]=P(Z>2.05)\\=1-P(Z<2.05)\\=1-0.97982\\=0.02018\\\approx0.02[/tex]

*Use a z-table for the probability.

Thus, the significance level α is 0.02.

Answer:

A. Negative association, not statistically significant and very weak effect size

B. Negative association, statistically significant and very weak size.

C. Positive association, statistically significant and strong effect size.

Step-by-step explanation:

Cohen's d is an effect size used to indicate the standardised difference between two means. It can be used, for example, to accompany reporting of t-test and ANOVA results. It is also widely used in meta-analysis. Cohen's d is the difference between two group means divided by the pooled standard deviation for the two groups.

A statistically significant result is a result that is always not attributed to chance attributed to chance. The probability value shows the probability of observing the difference if no difference exists.

In statistics , two variables is said to have negative association when the values of one variable seem to decrease as the values of the other variable increase. In statistics, a perfect negative association is represented by the value -1.00, while a 0.00 indicates no association.

A (number of panic attacks in the past month and neuroticism): r = - .03, not sig.

Answer: Negative association, not statistically significant and very weak effect size.

B (number of panic attacks in the past month and number of nightmares in the past month): r = - .14 (p = .05).

Answer: Negative association, statistically significant and very weak size.

C (number of nightmares in the past month and neuroticism): r = .48 (p = .003).

Answer: Positive association, statistically significant and strong effect size.

It is important to note that:

There is covariance because there is presence of significant correlation. You cannot say for a fact about temporal precedence because it is not clear if neuroticism or nightmares came first. You also can't rule out the possibility this relationship is due to a third variable, hence Dr. Moynihan cannot say that neuroticism causes nightmares.

Answer:

NUMBER 1

(i) negative direction

(ii) statistically insignificant

(iii) very small effect size

NUMBER 2

(i) negative direction

(ii) the relationship between (A) and (B) is statistically significant

(iii) small effect size

NUMBER 3

(i) positive direction

(ii) the relationship between (B) and (C) is statistically significant

(iii) large effect size

Step-by-step explanation:

Let (A) = the number of panic attacks a person experienced in the past month

(B) = the number of nightmares a person experienced in the past month

(C) = Neuroticism level

1. Association between (A) and (C)

r = -0.03, p = not significant

2. Association between (A) and (B)

r = -0.14, p = 0.05

3. Association between (B) and (C)

r = 0.48, p = 0.003

Let's now see what Cohen's Benchmark is all about.

Cohen's Benchmarks are specified for various Effect Sizes. Effect size is the quantitative measure of the magnitude (how great or small) of a certain phenomenon of scientific or psychological interest.

The terms 'small', 'medium' and 'large' are relative to another and to the particular content and research design or method. For this reason, Jacob Cohen gave conventional scales or benchmarks for effect sizes.

He set small effect size at d=0.2

This corresponds to an r of 0.1

He set medium effect size at d=0.5

This corresponds to an r of 0.3

He set large effect size at d=0.8

This corresponds to an r of 0.5

Based on this, we can now answer the questions.

1. Association between (A) and (C)

(i) Direction of the association is negative. This implies that as one variable increases, the other decreases. If plotted, the curve or graph would be downward sloping from left to right. If (A) comes first - if (A) is on the vertical axis - and (B) is on the horizontal axis, then as (A) increase, (B) will decrease.

(ii) The association is not significant, as already stated in the question. But then this means that the p-value is very high or is higher than 0.05 (same as 5%). This implies that the relationship between both variables is largely caused by chance.

* p-value is the probability that a relationship between or among variables is caused or is explainable by chance.

(iii) * r is the correlation coefficient and it shows the effect size.

According to Cohen's benchmarks,

The ES here is very small.

r = 0.03 is much smaller than r = 0.1

2. Association between (A) and (B)

(i) The direction of the association is negative. As one variable increases, the other decreases and vice versa.

(ii) Statistical significance exists. The results from the data collected (on variables (A) and (B)) are largely explained by statistics, as p=0.05

* A p-value of 0.05 (5%) or below is usually considered to describe the relationship among variables as statistically significant.

(iii) According to Cohen's benchmarks,

The ES here is small.

r = 0.14 is close to r = 0.1

3. Association between (B) and (C)

(i) The direction of the association is positive. There is no negative sign before the r value of 0.48. In this case, both variables increase simultaneously. If a graph were to be plotted, the shape of the curve would be upward sloping from left to right.

(ii) The relationship is statistically significant. The p-value of 0.003 is very small and is less than the p-value benchmark of 0.05. Hence there is very minute probability that chance explains the research results.

(iii) The ES is large, when placed on a Cohen scale. r of 0.48 is approximately r = 0.5 (to 1 decimal place) and this is the r value for which an effect size is considered to be large.

KUDOS!

Answer:

better golf scores

Step-by-step explanation:

i did the quiz just a few secs ago

According to the excerpt, seeing the "hole as bigger" has the benefit of improving gold scores. The right answer is D.

Visualization is the capacity to create mental images of situations and things using one's imagination.

Visualization enhanced results in the excerpt provided.

Golfers typically earned better gold scores when they saw the golf hole as larger circles.

The result of seeing the "hole as greater" is that you will receive better gold scores.

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See the attached picture

Answer:

a) 0.75

b) 0.4375

c) 0.5714

Step-by-step explanation:

Solution:-

- The events are defined as follows:

Event A : The Asian project is successful

Event B : The European project is successful

- The two given events are independent. Their respective probabilities are:

 P ( A ) = 0.25

 P ( B ) = 0.25

- The conditional probability for European project to fail given that asian project also failed.

- The probability can be expressed as:

            P ( B ' / A ' ) = P ( A' & B' ) / P ( A' )

- According to the property of independent events we have:

            P ( A ' & B ' ) = ( 1 - P ( A ) )* ( 1 - P ( B ) )

Therefore,

            P ( B ' / A ' ) = [ ( 1 - P ( A ) )* ( 1 - P ( B ) ) ] /  ( 1 - P ( A ) )

            P ( B ' / A ' ) = 1 - P ( B )

Answer: The probability is simply the failure of event (B) : The european project fails = 0.75.

b) The probability that at-least one of the two projects will successful consists of (either A or B is successful) or  ( Both are successful). We can mathematically express it as:

          P ( At-least 1 project is success ) = P ( A U B ) + P ( A & B )

          P ( At-least 1 project is success ) = P ( A )*(1 - P ( B )) + P ( B )*(1 - P ( A )]+ P ( A ) * P ( B )

                                                                = 2*0.25*0.75 + 0.25^2

                                                                = 0.4375

c ) Given that at least one of the two projects is successful, what is the probability that only the Asian project is successful?

- We can mathematically express the required conditional probability as follows with help of  part b):

           P ( A / At-least 1 project is success ) = P ( A & any at-least 1 is success) / P ( At-least 1 project is success )

- The probability of P ( A & any at-least 1 is success), consists of event A success and event B fails or both are a success:

          P ( A & any at-least 1 is success) = P ( A )*( 1 - P ( B ) ) + P ( A )*P ( B )

                                                                 = 0.25*0.75 + 0.25^2

                                                                 = 0.25

- The conditional probability can now be evaluated:

          = P ( A & any at-least 1 is success) / P ( At-least 1 project is success )

          = 0.25 / 0.4375

          = 0.5714

Polynomials are classified by their degree and number of terms. x³-8 is a 3rd degree binomial, 24 is a 0th degree monomial, 2x⁴-x³+5x²+x-7 is a 4th degree quintic, and 10x is a 1st degree monomial.

Polynomials can be classified by their degree (the highest power of the variable) and by the number of terms they contain. This classification is done as follows:

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Polynomials are classified by degree and number of terms: [tex]x^3[/tex]- 8 is a cubic binomial, 24 is a zero-degree monomial, [tex]2x^4 - x^3 + 5x^2[/tex] + x - 7 is a quartic quintic, and 10x is a linear monomial.

To classify polynomials by degree and number of terms, we check its highest exponent and count its terms:

[tex]x^3[/tex] - 8: This is a binomial (two terms) and its degree is 3 because the highest exponent of x is 3.

24: This is a monomial (one term) and its degree is 0 since it is a constant.

[tex]2x^4 - x^3 + 5x^2[/tex] + x - 7: This is a polynomial of five terms, so it's called a quintic. Its degree is 4 because the highest exponent of x is 4.

10x: This is a monomial (one term) and its degree is 1 because x is to the first power.

The classification is based on the degree of the polynomial, which is determined by the highest power of the variable x present in the equation, and the number of terms present in the polynomial (monomial for one term, binomial for two terms, and so on).

D : 2350

Step-by-step explanation:

Given,

The grocery price: 350

The amount lady gave: 2000

The amount returned to the lady: 1650

Also, given that the shopkeeper of the next shop comes with the Rs.2000 note saying "duplicate" *and takes his money back.

Therefore, the loss would be the sum of two items:

1) The shopkeeper have to return the 2000 rupees to the next shopkeeper. Therefore, he suffers the loss of 2000.

2) He also sold the grocery worth 350 rupees. Therefore, it would be the extra loss of 350 rupees.

Hence, the shopkeeper faced the loss of 2000 + 350 = 2350 rupees.  

Answer:

What does this mean 31>28

Step-by-step explanation:

Final answer:

The expression 31>28 means that 31 is greater than 28, which is a basic mathematical comparison used widely across various contexts, from simple arithmetic to data analysis and the mnemonic for remembering the number of days in each month.

Explanation:

The expression 31>28 is a mathematical comparison indicating that the number 31 is greater than the number 28. This can be applied in various contexts, such as comparing quantities, ages, temperatures, or even the number of days in different months. For example, using the knuckle mnemonic, months with 31 days are more than those with 28 days (February, typically), illustrating a practical application of this comparison.

In statistics and data analysis, understanding comparative values like this is crucial. It helps in analyzing data intervals, as in the provided reference, where different age intervals are being compared based on the number of data values they contain. This fundamental concept of comparison underpins much of mathematical reasoning and data analysis.

Answer:

2/15 cups of cheese

Step-by-step explanation:

Because you are eating 2/5 of the recipe you have to multiply the values

2/5*1/3 = 2/15

Answer:

[tex]t=\frac{6.5-6.7}{\sqrt{\frac{0.5^2}{100}+\frac{0.7^2}{110}}}}=-2.398[/tex]  

[tex] df = n_1 +n_2 -2= 100+110-2= 208[/tex]

Since is a bilateral test the p value would be:

[tex]p_v =2*P(t_{208}<-2.398)=0.01110[/tex]

Comparing the p value with the significance level given [tex]\alpha=0.05[/tex] we see that [tex]p_v<\alpha[/tex] so we can conclude that we can reject the null hypothesis, and we have significant differences between the two groups at 5% of significance.

Step-by-step explanation:

Data given and notation

[tex]\bar X_{1}=6.5[/tex] represent the sample mean for Atlanta

[tex]\bar X_{2}=6.7[/tex] represent the sample mean for Chicago

[tex]s_{1}=0.5[/tex] represent the sample deviation for Atlanta

[tex]s_{2}=0.7[/tex] represent the sample standard deviation for Chicago

[tex]n_{1}=100[/tex] sample size for the group Atlanta

[tex]n_{2}=110[/tex] sample size for the group Chicago

t would represent the statistic (variable of interest)

[tex]\alpha=0.01[/tex] significance level provided

Develop the null and alternative hypotheses for this study?

We need to conduct a hypothesis in order to check if the meanfor atlanta is different from the mean of Chicago, the system of hypothesis would be:

Null hypothesis:[tex]\mu_{1}=\mu_{2}[/tex]

Alternative hypothesis:[tex]\mu_{1} \neq \mu_{2}[/tex]

Since we don't know the population deviations for each group, for this case is better apply a t test to compare means, and the statistic is given by:

[tex]t=\frac{\bar X_{1}-\bar X_{2}}{\sqrt{\frac{s^2_{1}}{n_{1}}+\frac{s^2_{2}}{n_{2}}}}[/tex] (1)

t-test: Is used to compare group means. Is one of the most common tests and is used to determine whether the means of two groups are equal to each other.

Calculate the value of the test statistic for this hypothesis testing.

Since we have all the values we can replace in formula (1) like this:

[tex]t=\frac{6.5-6.7}{\sqrt{\frac{0.5^2}{100}+\frac{0.7^2}{110}}}}=-2.398[/tex]  

What is the p-value for this hypothesis test?

The degrees of freedom are given by:

[tex] df = n_1 +n_2 -2= 100+110-2= 208[/tex]

Since is a bilateral test the p value would be:

[tex]p_v =2*P(t_{208}<-2.398)=0.01110[/tex]

Based on the p-value, what is your conclusion?

Comparing the p value with the significance level given [tex]\alpha=0.05[/tex] we see that [tex]p_v<\alpha[/tex] so we can conclude that we can reject the null hypothesis, and we have significant differences between the two groups at 5% of significance.

Answer:

No, I can't say precisely that. Because there are common values within both working hours intervals.

Step-by-step explanation:

1)Let's do it by parts. For a Confidence Interval of 95%, i.e. covering 95% of the area of the Graph of this distribution, with known mean and Standard Deviation we have to plug it in the formula below.

Notice that in the formula the part:

[tex]Z\frac{s}{\sqrt{n}}[/tex]

2)We can find the margin of error.

Atlanta:

Average Daily Work Time

100 workers

[tex]\bar{x}=6.5[/tex]

[tex]s=0.5[/tex]

[tex]\bar{x}\pm Z\frac{s}{\sqrt{n}}\\6.5\pm 1.96\frac{0.5}{\sqrt{100}}\\6.5 \pm 0.098\\6.40\: to\: 6.59[/tex]

Atlanta workers may have a an average of 6.4 to 6.59 daily working hours

Chicago

110 workers (observations)

[tex]\bar{x}:6.7\\s=0.7[/tex]

[tex]\bar{x}\pm Z\frac{s}{\sqrt{n}}\\6.7\pm 1.96\frac{0.7}{\sqrt{110}}\\6.7\pm 0.13\\6.57\: to\: 6.83[/tex]

Chicago workers may have worked an average of 6.57 to 6.83 daily working hours

Answer:

True

Step-by-step explanation:

Perishable is a term for a product that has limited time to sell. A product like meat, fruit, and most fresh food will have a shelf life and you have to sell the product before the shelf life expires. Some preservation techniques can help to extend food products like freezing or curing them.

Room night of a hotel or any rent product is also considered as perishable since you can't store the rent time and sell it another day. So, every unoccupied room is treated as expired food. Unlike food products, room nights for the hotel can't be preserved. There are other strategies to maximize the revenue of rent products such as overselling or selling promise/reservation.

Answer:

d. No, because the two populations from which the samples are selected do not appear to be normally distributed.

Step-by-step explanation:

First, the assumptions of an independent-measures t test are as follows:

1. The data is continuous

2. Only two groups are compared

3. The two groups should be independent

4. The groups should have equal variance

5. The data should be normally distributed

In this case, the 5th assumption has been violated because scores in the two samples are distributed in different ranges in two samples. So the outliers in the scores may be exist. Therefore, it would not be valid for the professor to use the independent-measures t test because the two populations from which the samples are selected do not appear to be normally distributed.

Answer:

The t-test for one mean can be safely used for this situation.

Step-by-step explanation:

A statistical experiment is conducted to determine if a particular training program would improve physical fitness.

The objective of the experiment was to determine if there is some evidence that the maximum oxygen uptake of the sample of students differed from the general population of untrained subjects, whose maximum oxygen uptake is known to be 45 ml/kg/min on average.

The hypothesis is defined as:

H₀: The population mean is 45 ml/kg/min, i.e. μ = 45.

Hₐ: The population mean is different from 45 ml/kg/min, i.e. μ ≠ 45.

The test for single mean can be either using the z-distribution or the t-distribution.

The z-distribution is used if it provided that:

The t-distribution is used if :

To use a t-distribution for a single mean test the following assumptions are to made:

In this case we will use a t-distribution since there is no information about the population standard deviation.

The sample of 35 students are randomly selected for the training program.

The maximum oxygen uptake of every student is independent of the others.

Thus, the t-test for one mean can be safely used for this situation.

Using Chebyshev's theorem, we conclude that at least 88.9% of recent graduates have salaries between $22,800 and $30,000, given a mean salary of $26,400 and a standard deviation of $1200.

The question is asking for the minimum percentage of recent graduates who have salaries within a specific range using Chebyshev's Theorem. By definition, Chebyshev's theorem states that at least 1 - 1/k^2 of data from a sample will fall within k standard deviations from the mean, where k is any number greater than 1. The range in this question can be represented as being within 3 standard deviations from the mean (because ($30,000 - $26,400)/$1200 = 3 and ($26,400 - $22,800)/$1200 = 3). Thus, the minimum percentage of recent graduates having salaries within this range is at least 1 - (1/3^2) = 1 - 1/9 = 8/9 = 88.9%. So, at least 88.9% of the recent graduates fall within this salary range according to Chebyshev's theorem.

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Answer:

21/10

Step-by-step explanation:

Answer:

21/10

Step-by-step explanation:

Answer: $350 because you multiply 140 and 100 then divide 40

Answer:

There is 95% confidence that the population proportion of people from Vermont who exercised for at least 30 minutes a day 3 days a week is between 55.9% and 74.7%.

Step-by-step explanation:

We have to answer the population proportion for Vermont.

We can only do it by a confidence interval, as we only have information from a sample.

This sample, of size n=100, has a proportion p=0.653.

The degrees of freedom are:

[tex]df=n-1=100-1=99[/tex]

We will calculate a 95% confidence interval, which for df=99 has a critical value of t of t=1.984.

The margin of error can be calculated as:

[tex]E=t*\sigma_p=t\sqrt{\dfrac{p(1-p)}{n}}=1.984\sqrt{\dfrac{0.653*0.347}{100}}\\\\\\E=1.984*\sqrt{0.00226}=1.984*0.0476=0.094[/tex]

Then, the upper and lower bounds of the confidence interval are:

[tex]LL=p-E=0.653-0.094=0.559\\\\UL=p+E=0.653+0.094=0.747[/tex]

Then, we can say that there is 95% confidence that the population proportion of people from Vermont who exercised for at least 30 minutes a day 3 days a week is between 55.9% and 74.7%.

Final answer:

The estimated population proportion of Vermont residents who exercised for at least 30 minutes a day 3 days a week is 65.3%, which is based on the sample proportion from the Gallup survey.

Explanation:

The student is asking about the population proportion for people from Vermont who exercised for at least 30 minutes a day 3 days a week based on the Gallup survey results. The survey indicated that 65.3% of the Vermont respondents exercised at the mentioned rate.

To find the value of the population proportion (population proportion), we typically use the sample proportion as an estimate. From the survey, we have that the sample proportion (p-hat) for Vermont is 65.3%, which we express as a decimal, 0.653. Assuming the sample is representative, we would estimate the population proportion to also be 0.653 or 65.3%.

It's important to note that this is an estimate based on the sample and that to infer more confidently about the entire population of Vermont, a larger sample size or additional statistical methods such as confidence intervals or hypothesis testing may be applied. Nevertheless, with the information provided, the best estimate for the population proportion of Vermont residents who exercised according to the guidelines is the sample proportion of 65.3%.

Answer: the sample size should be 39

Step-by-step explanation:

The sample mean is the point estimate for the population mean. Confidence interval is written as

Sample mean ± margin of error

Margin of error = z × σ/√n

Where

σ = population standard Deviation

n = number of samples

z represents the z score corresponding to the confidence level

From the information given,

σ = 4 minutes

Margin of error = 75 seconds. Converting to minutes, it becomes 75/60 = 1.25 minutes

To determine the z score, we subtract the confidence level from 100% to get α

α = 1 - 0.95 = 0.05

α/2 = 0.05/2 = 0.025

This is the area in each tail. Since we want the area in the middle, it becomes

1 - 0.025 = 0.975

The z score corresponding to the area on the z table is 1.96. Thus, confidence level of 95% is 1.96

Therefore,

1.25 = 1.96 × 4/√n

1.25/1.96 = 4/√n

0.6378 = 4/√n

√n = 4/0.6378 = 6.27

n = 6.27² = 39

Answer:

Step-by-step explanation:

The confidence interval for the mean swimming speed of all emperor penguins in the sampled population is (9.4248, 10.1172).

Given the following data:

In Statistics, a confidence interval can be defined as the degree of uncertainty that is associated with a given statistical population.

First of all, we would subtract one from the given sample size for penguins.

[tex]n=31-1=30[/tex]

Next, we would subtract the confidence interval from 1 and then divide by 2:

[tex]\frac{(1-0.95)}{2} =\frac{0.05}{2} =0.025[/tex]

From the t-distribution table, a 30 degree of freedom (df) with 0.025 is equal to 2.042.

Divide the standard deviation by the square root of penguin's population size and then multiply by the t-interval.

[tex]\frac{0.944}{\sqrt{30} } \times 2.042=0.3462[/tex]

For the lower end:

[tex]9.771-0.3462=9.4248[/tex]

For the upper end:

[tex]9.771+0.3462=10.1172[/tex]

Therefore, the confidence interval for the mean swimming speed of all emperor penguins in the sampled population is (9.4248, 10.1172).

Read more on confidence interval here: https://brainly.com/question/15712887

Answer:

g(25) = -336

Step-by-step explanation:

g(x) = 24(x - 39)

g(25) = 24(25 - 39)

g(25) = 600 - 936

g(25) = -336

Answer:

(-2,1)

Step-by-step explanation:

To find the coordinates of the intersection of the diagonals of parallelogram QRST, we first need to find the midpoints of the diagonals. The midpoint of diagonal QR is (-3, 1) and the midpoint of diagonal ST is (-1, -3). We can find the equation of the line passing through these midpoints, and then find the x and y coordinates of the intersection point.

To find the coordinates of the intersection of the diagonals of parallelogram QRST, we first need to find the midpoints of the diagonals. The midpoint of diagonal QR can be found by averaging the x-coordinates of Q and R, and the y-coordinates of Q and R. Therefore, the midpoint of QR is ((-8 + 2)/2, (1 + 1)/2) = (-3, 1). Similarly, the midpoint of diagonal ST is ((4 - 6)/2, (-3 - 3)/2) = (-1, -3).

Now that we have the midpoints, we can find the equation of the line passing through these two points. Using the slope-intercept form, we have y - 1 = (1 - (-3))/(-3 - (-1))(x - (-3)). Simplifying, we get y - 1 = 2(x + 3). Rearranging the equation gives us y = 2x + 5.

Substituting this equation into the equation of diagonal QR, we can find the x-coordinate of the intersection point. Plugging in y = 2x + 5 into the equation of QR, we get 1 = -x + 5. Solving for x, we get x = 4. Substituting the value of x into the equation of ST, we find the y-coordinate to be y = 2(4) + 5 = 13. Therefore, the coordinates of the intersection point are (4, 13).

Answer:

There is no enough evidence to to support the claim that Portland has more average yearly rainfall than Seattle.

Being μ1: average rainfall in Portland, μ2: average rainfall in Seattle, the null and alternative hypothesis are:

[tex]H_0: \mu_1-\mu_2=0\\\\H_a:\mu_1-\mu_2 > 0[/tex]

P-value = 0.1290

As the P-value is bigger than the significance level, the effect is not significant and the null hypothesis failed to be rejected.

Step-by-step explanation:

We have to test the hypothesis of the difference between means.

The claim is that Portland has more average yearly rainfall than Seattle.

Being μ1: average rainfall in Portland, μ2: average rainfall in Seattle, the null and alternative hypothesis are:

[tex]H_0: \mu_1-\mu_2=0\\\\H_a:\mu_1-\mu_2 > 0[/tex]

The significance level is 0.10.

The sample for Portland, of size n1=45, has a mean of M1=37.50 and standard deviation of s1=1.82.

The sample for Seattle, of size n1=35, has a mean of M1=37.07 and standard deviation of s1=1.68.

The difference between means is:

[tex]M_d= M_1-M_2=37.50-37.07=0.43[/tex]

The standard error for the difference between means is:

[tex]s_{M_d}=\sqrt{\dfrac{\sigma_1^2}{n_1}+\dfrac{\sigma_2^2}{n_2}}=\sqrt{\dfrac{1.82^2}{45}+\dfrac{1.68^2}{35}}=\sqrt{ 0.0736+0.0688 }=\sqrt{0.1424}\\\\\\s_{M_d}=0.3774[/tex]

We can calculate the t-statistic as:

[tex]t=\dfrac{M_d-(\mu_1-\mu_2)}{s_{M_d}}=\dfrac{0.43-0}{0.3774}=1.1393[/tex]

The degrees of freedom are:

[tex]df=n1+n2-2=45+35-2=78[/tex]

Then, the p-value for this one-tailed test with 78 degrees of freedom is:

[tex]P-value=P(t>1.1393)=0.1290[/tex]

As the P-value is bigger than the significance level, the effect is not significant and the null hypothesis failed to be rejected.

There is no enough evidence to to support the claim that Portland has more average yearly rainfall than Seattle.