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The original price of the guitar was $1250, calculated by taking the sale price of $825 and dividing it by 0.66, which represents the remaining percentage of the price after a 34% discount.

To find the original price of the guitar that has been marked down by 34% and sold for $825, we first need to consider that after the markdown, the guitar's price is equivalent to 100% - markdown percentage of the original price. In this case, it's 100% - 34% = 66% of the original price.

Let's denote the original price by P. Since 66% of P equates to $825, we can set up the following equation:

0.66  imes P = $825

Solving for P:

P = $825 / 0.66

P = $1250

Therefore, the original price of the guitar was $1250.

Answer:

[tex](\, \cos(\frac{\pi}{16}) + i\sin(\frac{\pi}{16}) \,)^{1/2} = \cos(\frac{\pi}{32}) + i\sin(\frac{\pi}{32}) = 0.99 + i0.09[/tex]

Step-by-step explanation:

The complex number given is

[tex]z = (\, \cos(\frac{\pi}{16}) + i\sin(\frac{\pi}{16}) \,)^{1/2}[/tex]

Now, remember that the DeMoivre's theorem states that

[tex]( \cos(x) + i\sin(x) )^n = \cos(nx) + i\sin(nx)[/tex]

Then for this case we have that

[tex](\, \cos(\frac{\pi}{16}) + i\sin(\frac{\pi}{16}) \,)^{1/2} = \cos(\frac{\pi}{32}) + i\sin(\frac{\pi}{32}) = 0.99 + i0.09[/tex]

Answer:14

Step-by-step explanation:

The amount the florist earns for each delivery, with 'x' being the miles away from the flower shop, can be modeled with the equation: Earnings = (7.5 + 0.75x). This represents the fixed net income of $7.5 and $0.75 per mile after paying the driver.

The florist charges ​$12.00 for delivery and an additional ​$1.50 per mile from the flower shop. However, the florist also has costs to cover, namely $0.75 per mile to pay the driver, and ​$4.50 for gas per delivery. The net earning per delivery, with 'x' representing the number of miles a delivery location is from the flower shop, can be modeled by the following algebraic expression: Earnings = (12 + 1.5x) - (0.75x + 4.5).

This actually simplifies to: Earnings = (7.5 + 0.75x). The 7.5 is the fixed net income for each delivery (gross earnings minus the gasoline cost) and 0.75x is the per-mile net income after the driver is paid.

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Answer:

There is no relationship between your year in school and having a job.

Step-by-step explanation:

In this instance, the chi sq test need to be performed.

Chi sq is used to determine if there is a significant relationship between two categorical variables.

The two variables here are year in school and employment status.

The two variables are independent(no relationship exists)

This implies that there is null hypothesis

Therefore, the Null Hypothesis is

There is no relationship between your year in school and having a job.

Equation used to determine the surface area of the box : 2( lb + bh + lh)

Surface area of the box : 94in²

Given, A box is 4 inches wide, 5 inches long and 3 inches tall.

Formula of surface area of cuboid : 2( lb + bh + lh)

Here,

l = 5in

b = 4in

h = 3in

Substitute the values,

Surface area = 2(5×4 + 4×3 + 5×3)

Surface area = 94in²

Know more about surface area,

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Answer:

[tex]Z = -2.865[/tex] means that it would be unusual to see 32% who rarely or never use academic advising services in one of these samples

Step-by-step explanation:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore of a measure X is given by:

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

Z scores below -2 are considered unusually low.

Z scores above 2 are considered unusually high.

For a sample proportion p in a sample of size n, we have that [tex]\mu = p, \sigma = \sqrt{\frac{p(1-p)}{n}}[/tex]

In this problem, we have that:

[tex]\mu = 0.42, \sigma = \sqrt{\frac{0.42*0.58}{200}} = 0.0349[/tex]

Is it unusual to see 32% who rarely or never use academic advising services in one of these samples

What is the z-score for X = 0.32?

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

[tex]Z = \frac{0.32 - 0.42}{0.0349}[/tex]

[tex]Z = -2.865[/tex]

[tex]Z = -2.865[/tex] means that it would be unusual to see 32% who rarely or never use academic advising services in one of these samples

No, it is not unusual to see 32% of students rarely or never using academic advising services in one of these samples.

To determine if it is unusual to see 32% of students rarely or never using academic advising services in one of these samples, we can compare it to the range of values that would be considered usual. In this case, we can use the 95% confidence interval provided, which states that the true proportion of community college students who rarely or never use academic advising services is between 0.113 and 0.439. If the observed proportion falls within this interval, it would be considered usual; otherwise, it would be considered unusual.

Since 32% falls within the range of 0.113 and 0.439, it is considered a usual value. Therefore, it is not unusual to see 32% of students rarely or never use academic advising services in one of these samples.

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Final answer:

To find out how many people will be in Houston, Texas in 20 years, we can use the doubling time of 35 years. Currently, the population is 2.4 million. Using the formula New Population = Current Population x 2^(Number of Doubling Cycles), the new population will be approximately 4.8 million. Option A .

Explanation:

To find out how many people will be in Houston, Texas in 20 years, we can use the doubling time of 35 years. Currently, the population is 2.4 million. Since the city doubles in size every 35 years, in 20 years it will go through 20/35 of a doubling cycle.

To calculate how many people there will be, we use the formula:

New Population = Current Population x 2^(Number of Doubling Cycles)

Plugging in the values, we have:

New Population = 2.4 million x 2^(20/35)

Using a calculator, we find that the new population will be approximately 4.8 million. Therefore, the answer is 4.8 million.

Answer:

The slope is -1

Step-by-step explanation:

Let's find the slope between your two points.

(1,4);(6,−1)

(x1,y1)=(1,4)

(x2,y2)=(6,−1)

Use the slope formula:

m= y2−y1/x2−x1  = −1−4/6−1

= −5/5

= −1

Hope this is a good explanation :)

Answer:

x=4 y=5

Step-by-step explanation:

5x + 2y = 30 and -5x + 4y = 0

Solve by adding the two equations together to eliminate x

5x + 2y = 30

-5x + 4y = 0

----------------------

   6y = 30

Divide each side by 6

6y/6 = 30/6

y =5

Now we solve for x

5x + 2y = 30

5x +2(5) = 30

5x+10 = 30

Subtract 10 from each side

5x+10-10 = 30-10

5x = 20

Divide each side by 5

5x/5 = 20/5

x = 4

Answer:

1.045

Step-by-step explanation:

Answer:

3/2

Step-by-step explanation:

The volume of the rectangular prism with dimensions 10/3 cm, 4/5 cm, and 1/5 cm is 8/15 cm³.

The volume of a rectangular prism can be found using the formula: Volume = length x width x height. In this case, the length, width, and height of the prism are given as 10/3 cm, 4/5 cm, and 1/5 cm respectively. Replace these dimensions in the formula:

Volume = (10/3 cm) x (4/5 cm) x (1/5 cm) = 8/15 cm³.

Therefore, the volume of the rectangular prism is 8/15 cm³.

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Answer:

The critical value of t for 99% confidence interval is 2.898.

Step-by-step explanation:

The complete question is:

A biologist is trying to determine the average age of a local forest. She cuts down 18 randomly  selected trees and counts the tree rings. They find the average number of tree rings to be 83 with a  variance of 320. What is the critical value for the 99% confidence interval?

The population variance is not known and the sample size is too small. So a t-confidence interval will be used to estimate the population mean age of a local forest.

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

[tex]CI=\bar x\pm t_{\alpha/2, (n-1)}\times \frac{s}{\sqrt{n}}[/tex]

The information provided is:

n = 18

(1 - α)% = 99%

The degrees of freedom of the critical value of t is:

n - 1 = 18 - 1 = 17

Compute the critical value of t as follows:

[tex]t_{\alpha/2, (n-1)}=t_{0.01/2, (18-1)}=t_{0.005, 17}=2.898[/tex]

*Use a t-table.

Thus, the critical value of t for 99% confidence interval is 2.898.

Answer:

2.28% probability that the mean amount of their monthly cell phone bills is more than $60

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal probability distribution

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore of a measure X is given by:

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]s = \frac{\sigma}{\sqrt{n}}[/tex].

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

In this problem, we have that:

[tex]\mu = 58.90, \sigma = 3.64, n = 44, s = \frac{3.64}{\sqrt{44}} = 0.54875[/tex]

What is the probability that the mean amount of their monthly cell phone bills is more than $60?

This is 1 subtracted by the pvalue of Z when X = 60. So

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

By the Central Limit Theorem

[tex]Z = \frac{X - \mu}{s}[/tex]

[tex]Z = \frac{60 - 58.90}{0.54875}[/tex]

[tex]Z = 2[/tex]

[tex]Z = 2[/tex] has a pvalue of 0.9772

1 - 0.9772 = 0.0228

2.28% probability that the mean amount of their monthly cell phone bills is more than $60

Answer:

A) probability that an arriving customer will wait in line is 67%

B)the probability that both windows are idle is 0.33

C) the average length of the waiting line is 1.33

D)it would not be possible to offer a reasonable service with only one window

Step-by-step explanation:

arrival rate: δ = 20 x 0.80 = 16 customers per hour

service rate: μ = 2 × (60/5) = 24 customers/hour

Utilization factor is given as;

Φ = δ/μ

So, Φ = 16/24 ≈ 0.67

A) the probability that an arriving customer will wait in line is;

16/24 x 100% ≈ 67%

B) probability that both windows are idle is;

P(x=0) = 1 - 0.67 = 0.33

C) The average number of customers in the post office will be;

L_s = Φ/(1 - Φ)

L_s = 0.67/(1 - 0.67)

L_s = 0.67/0.33

L_s ≈ 2 customers

Thus, the average length of the waiting line is;

L_w = L_s - Φ

L_w = 2 - 0.67

L_w = 1.33

D) this part demands that we find the utilization factor with only one window.

Thus;

arrival rate: δ = 20 x 0.80 = 16 customers per hour

And

service rate: μ = 1 × (60/5) = 12 customers/hour

Thus, Utilization factor = 16/12 = 1.33

Thus, it would not be possible to offer a reasonable service with only one window

Answer:

[tex] ME = 1.833 * \frac{4.367}{\sqrt{10}}= 2.531[/tex]

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

[tex]\bar X[/tex] represent the sample mean for the sample  

[tex]\mu[/tex] population mean (variable of interest)

s represent the sample standard deviation

n represent the sample size  

Solution to the problem

The confidence interval for the mean is given by the following formula:

[tex]\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}[/tex]   (1)

In order to calculate the mean and the sample deviation we can use the following formulas:  

[tex]\bar X= \sum_{i=1}^n \frac{x_i}{n}[/tex] (2)  

[tex]s=\sqrt{\frac{\sum_{i=1}^n (x_i-\bar X)}{n-1}}[/tex] (3)  

The mean calculated for this case is [tex]\bar X=24.8[/tex]

The sample deviation calculated [tex]s=4.367[/tex]

In order to calculate the critical value [tex]t_{\alpha/2}[/tex] we need to find first the degrees of freedom, given by:

[tex]df=n-1=10-1=9[/tex]

Since the Confidence is 0.90 or 90%, the value of [tex]\alpha=0.1[/tex] and [tex]\alpha/2 =0.005[/tex], and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.05,9)".And we see that [tex]t_{\alpha/2}=1.833[/tex]

And the margin of error would be given by:

[tex] ME = 1.833 * \frac{4.367}{\sqrt{10}}= 2.531[/tex]

The correct question is:

Write out the following sums, one term for each value of k. Simplify each term as much as possible, but do not enter decimals. For example, enter 1 + 4 + 9 instead of 1² + 2² + 3² or 14, or enter 1/2 + 1/2 instead of 0.5 + 0.5 or 1.

The purpose of this problem is for you to show that you know how to interpret summation notation and write all of the terms in a sum, which is why you are being told not to reduce your answers very much.

[tex](a) \sum_{k=0}^5 2^k \\ \\(b) \sum_{k=2}^7 \frac{1}{k} \\ \\(c) \sum_{k=1}^5 k^2 \\ \\(d) \sum_{k=1}^6 \frac{1}{6} \\ \\(e) \sum_{k=1}^6 2k[/tex]

Answer:

[tex](a) \sum_{k=0}^5 2^k = $1 + 2 + 4 + 8 + 16 + 32$ \\ \\(b) \sum_{k=2}^7 \frac{1}{k} = \frac{1}{2} + \frac{1}{3} + \frac{1}{4}+ \frac{1}{5}+ \frac{1}{6}+ \frac{1}{7} \\ \\(c) \sum_{k=1}^5 k^2 = 1 + 4 + 9 + 16 + 25 \\ \\(d) \sum_{k=1}^6 \frac{1}{6} = \frac{1}{6} + \frac{1}{6} + \frac{1}{6} + \frac{1}{6} + \frac{1}{6} + \frac{1}{6} \\ \\(e) \sum_{k=1}^6 2k = 2 +4 +6 +8 +10 +12[/tex]

Step-by-step explanation:

[tex](a) \sum_{k=0}^5 2^k\\For k = 0: 2^k = 2^0 = 1\\For k = 1: 2^1 = 2\\For k = 2: 2^2 = 4\\For k = 3: 2^3 = 8\\For k = 4: 2^4 = 16\\For k = 5: 2^5 = 32\\\sum_{k=0}^5 2^k = 1 + 2 + 4 + 8 + 16 + 32[/tex]

[tex](b) \sum_{k=2}^7 \frac{1}{k}\\For k = 2: 1/2\\For k = 3: 1/3\\For k = 4: 1/4\\For k = 5: 1/5\\For k = 6: 1/6\\For k = 7: 1/7\\ \sum_{k=2}^7 \frac{1}{k} = 1/2 + 1/3 + 1/4 + 1/5 + 1/6+ 1/7[/tex]

[tex](c) \sum_{k=1}^5 k^2\\For k = 1: 1^2 = 1\\For k = 2: 2^2 = 4\\For k = 3: 3^2 = 9\\For k = 4: 4^2= 16\\For k = 5: 5^2 = 25\\\sum_{k=1}^5 k^2 = 1 + 4 + 9 + 16 + 25[/tex]

[tex](d) \sum_{k=1}^6 \frac{1}{6}\\For k = 1: 1/6\\For k = 2: 1/6\\For k = 3: 1/6\\For k = 4: 1/6\\For k = 5: 1/6\\For k = 6: 1/6\\ \sum_{k=1}^6 \frac{1}{6} = 1/6 + 1/6 + 1/6 + 1/6 + 1/6 + 1/6[/tex]

[tex](e) \sum_{k=1}^6 2k\\For k = 1: 2\times1 = 2\\For k = 2: 2\times2 = 4\\For k = 3: 2\times3 = 6\\For k = 4: 2\times4 = 8\\For k = 5: 2\times5 = 10\\For k = 6: 2\times6 = 12\\\sum_{k=1}^6 2k = 2 +4 +6 +8 +10 +12[/tex]

Answer:

86.65% probability that more than 70 workers will meet with an accident during the 1-yr period

Step-by-step explanation:

Binomial probability distribution

Probability of exactly x sucesses on n repeated trials, with p probability.

Can be approximated to a normal distribution, using the expected value and the standard deviation.

The expected value of the binomial distribution is:

[tex]E(X) = np[/tex]

The standard deviation of the binomial distribution is:

[tex]\sqrt{V(X)} = \sqrt{np(1-p)}[/tex]

Normal probability distribution

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore of a measure X is given by:

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

When we are approximating a binomial distribution to a normal one, we have that [tex]\mu = E(X)[/tex], [tex]\sigma = \sqrt{V(X)}[/tex].

In this problem, we have that:

[tex]p = 0.08, n = 1000[/tex]

So

[tex]\mu = E(X) = np = 1000*0.08 = 80[/tex]

[tex]\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{1000*0.08*0.92} = 8.58[/tex]

What is the probability that more than 70 workers will meet with an accident during the 1-yr period

Using continuity correction, this is [tex]P(X \geq 70 + 0.5) = P(X \geq 70.5)[/tex], which is 1 subtracted by the pvalue of Z when X = 70.5. So

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

[tex]Z = \frac{70.5 - 80}{8.58}[/tex]

[tex]Z = -1.11[/tex]

[tex]Z = -1.11[/tex] has a pvalue of 0.1335

1 - 0.1335 = 0.8665

86.65% probability that more than 70 workers will meet with an accident during the 1-yr period

The probability that more than 70 workers will be involved in an accident is 0.8665

The given parameters are:

[tex]\mathbf{n = 1000}[/tex] --- population

[tex]\mathbf{p = 0.08}[/tex] --- the probability that a worker meets an accident

[tex]\mathbf{x = 70}[/tex] -- the number of workers

Start by calculating the mean and the standard deviation

[tex]\mathbf{\mu = np}[/tex] --- mean

So, we have:

[tex]\mathbf{\mu = 1000 \times 0.08}[/tex]

[tex]\mathbf{\mu = 80}[/tex]

[tex]\mathbf{\sigma = \sqrt{\mu(1 - p)}}[/tex]

So, we have:

[tex]\mathbf{\sigma = \sqrt{80 \times (1 - 0.08)}}[/tex]

[tex]\mathbf{\sigma = \sqrt{73.6}}[/tex]

[tex]\mathbf{\sigma = 8.58}[/tex]

The probability is then represented as

[tex]\mathbf{P(x > 70) = P(x > 70.5)}[/tex] ---- By continuity correction

Calculate the z-score for x = 70.5

[tex]\mathbf{z = \frac{x - \mu}{\sigma}}[/tex]

So, we have:

[tex]\mathbf{z = \frac{70.5 - 80}{8.58}}[/tex]

[tex]\mathbf{z = -1.11}[/tex]

So, we have:

[tex]\mathbf{P(x > 70) = P(z > -1.11)}[/tex]

Using z-scores of probabilities, we have:  

[tex]\mathbf{P(x > 70) = 0.8665}[/tex]

Hence, the probability that more than 70 workers will be involved in an accident is 0.8665

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Answer: No, these results are not statistically significant because

p > 0.05

Step-by-step explanation:

The null hypothesis is

H0 : μ = 17

The alternative hypothesis is

H 1 : μ ≠ 17

where μ is the mean amount of cereal in each box.

The p value that he got is 0.1499. This is greater than alpha = 0.05 which is the given level of significance.

If the level of significance is lesser than the p value, we would accept the null hypothesis.

Therefore, the correct option is

No, these results are not statistically significant because p>0.05

Answer:

The calculated z- value = 1.479 <  1.645 at 0.10 or 90% level of significance.

The null hypothesis is accepted at 90% level of significance.

There is no significant difference in the proportion of E. Coli in organic vs. conventionally grown produce.

Step-by-step explanation:

Step:-(i)

Given first sample size n₁ = 200

The first sample proportion     [tex]p_{1} = \frac{5}{200} = 0.025[/tex]

Given first sample size n₂= 500

The second sample proportion     [tex]p_{2} = \frac{25}{500} = 0.05[/tex]

Step:-(ii)

Null hypothesis :H₀:There is no significant difference in the proportion of E. Coli in organic vs. conventionally grown produce

Alternative hypothesis:-H₁

There is  significant difference in the proportion of E. Coli in organic vs. conventionally grown produce

level of significance ∝=0.10

Step:-(iii)

          The test statistic    

                                        [tex]Z =\frac{p_{1} - p_{2} }{\sqrt{pq(\frac{1}{n_{1} }+\frac{1}{n_{2} } } }[/tex]

     where         p =   [tex]\frac{n_{1} p_{1} + n_{2}p_{2} }{n_{1}+n_{2} }= \frac{200X0.025+500X0.05 }{500+200}[/tex]

                     p = 0.0428

                     q = 1-p =1-0.0428 = 0.9572

                                      [tex]Z =\frac{0.025- 0.05}{\sqrt{0.0428X0.9571(\frac{1}{200 }+\frac{1}{500 } } }[/tex]

                                     Z = -1.479

                                    |z| = |-1.479|

                                    z = 1.479

The tabulated value z= 1.645 at 0.10 or 90% level of significance.

The calculated z- value = 1.479 <  1.645 at 0.10 or 90% level of significance.

The null hypothesis is accepted at 90% level of significance.

Conclusion:-

There is no significant difference in the proportion of E. Coli in organic vs. conventionally grown produce

Answer:

25.5 cm²

Step-by-step explanation:

5.5 cm × 20 cm = 25.5 cm²

Answer:i think the answer is 110 but it would be more helpful if i knew what kind of shape it is

Step-by-step explanation:

to find the area of a square or rectangle(assuming this is a square or rectangle) you multiply the base by the height

Answer:

You will need 10 boxes

Step-by-step explanation:

Nine boxes will only hold 54 trophies so you need one more than nine.

Answer:

You will need 348 boxes for the trophies

Step-by-step explanation:

just multiply 58 and 6

Answer:

0.625

Step-by-step explanation:

Answer:

12.75

Step-by-step explanation:

Answer: The 95% confidence interval will be wider.

Step-by-step explanation:

Confidence interval for population proportion is written as

Sample proportion ± margin of error

margin of error = z score × √pq/n

The z score is determined by the confidence level. The z score for a confidence level of 95% is higher than the z score for a confidence level of 90%

This means that with all other things being equal, a 95% confidence level will give a higher margin of error compared to a 90% confidence level.

The higher the margin of error, the wider the confidence interval. Therefore,

The 95% confidence interval will be wider.

The correct statement is provided by Trevor: 'The 95% confidence interval will be wider.'

Given that,

If construct a 90% confidence interval for the population proportion,

And a 95% confidence interval for the population proportion,

Tim: 'The 90% confidence interval will be wider.'

Trevor: 'The 95% confidence interval will be wider.

Tracy: 'There is not enough information to tell.

Either interval could be wider.'

We have to determine,

Which of these intervals will be wider.

According to the question,

Three students provide their answers as follows:

Tim: 'The 90% confidence interval will be wider.

'Trevor: 'The 95% confidence interval will be wider.

'Tracy: 'There is not enough information to tell. Either interval could be wider.'

Therefore, Confidence interval for population proportion is written as

Sample proportion ± margin of error

Margin of error  [tex]= z score \times \frac{ \sqrt{pq}}{n}[/tex]

The z score for a confidence level of 95% is higher than the z score for a confidence level of 90%.

Other things being equal, a 95% confidence level will give a higher margin of error compared to a 90% confidence level.

The higher the margin of error, the wider the confidence interval.

Therefore, The 95% confidence interval will be wider.

Hence, The correct statement is provided by Trevor: 'The 95% confidence interval will be wider.'

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Answer:

The average speed was 48.7 miles per hour.

Step-by-step explanation:

The average speed v is given by the following formula:

[tex]v = \frac{d}{t}[/tex]

In which d is the distance, and t is the time.

The Harrisons drove 304.2 miles in 6.25 hours

This means that [tex]d = 304.2, t = 6.25[/tex]

We have the distance is miles and the time in hours, so the distance is in miles per hour.

So

[tex]v = \frac{304.2}{6.25} = 48.7[/tex]

The average speed was 48.7 miles per hour.

Answer:

(A) Type I error is convicting an innocent person.

Step-by-step explanation:

When someone is on trial for suspicion of committing a crime, the hypotheses are:

[tex]H_0[/tex] : innocent

[tex]H_A[/tex] : guilty

From the given options, A Type I error is convicting an innocent person. If the null hypothesis holds (i.e. a person is innocent) nut we still go ahead to convict the person, we have rejected a true null hypothesis.

Answer:

c

Step-by-step explanation:

Answer:

43/91 or 47%

Step-by-step explanation: