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2 years ago

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d) Jerry, Barry, and Harry went fishing and they each caught a giant fish! Jerry's fish is 62 inches long. Barry's fish is 7 fee

t long. Harry's fish is 2 yards long. Who caught the longest fish?

1 answer:

Sauron [17]2 years ago

7 0

No it’s berry 2yards=6feet so they are all smaller than berrys 7foot fish

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Nady [450]

Answer:

answer is c Hope this helps

Step-by-step explanation:

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2 years ago

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PLEASE HELP NEED DONE FAST!!!

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2 years ago

g A researcher wants to construct a 95% confidence interval for the proportion of children aged 8-10 living in Atlanta who own a

Helga [31]

Answer:

The sample size needed is 2401.

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of $\pi$ , and a confidence level of $1-\alpha$ , we have the following confidence interval of proportions.

$\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}$

In which

z is the zscore that has a pvalue of $1 - \frac{\alpha}{2}$ .

The margin of error of the interval is:

$M = z\sqrt{\frac{\pi(1-\pi)}{n}}$

95% confidence level

So $\alpha = 0.05$ , z is the value of Z that has a pvalue of $1 - \frac{0.05}{2} = 0.975$ , so $Z = 1.96$ .

Estimate the sample size needed if no estimate of p is available so that the confidence interval will have a margin of error of 0.02.

We have to find n, for which M = 0.02.

There is no estimate of p available, so we use $\pi = 0.5$

$M = z\sqrt{\frac{\pi(1-\pi)}{n}}$

$0.02 = 1.96\sqrt{\frac{0.5*0.5}{n}}$

$0.02\sqrt{n} = 1.96*0.5$

$\sqrt{n} = \frac{1.96*0.5}{0.02}$

$(\sqrt{n})^{2} = (\frac{1.96*0.5}{0.02})^{2}$

$n = 2401$

The sample size needed is 2401.

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3 years ago

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I need help ASAP! Can anyone please check my work?

STALIN [3.7K]

A = event the person got the class they wanted

B = event the person is on the honor roll

P(A) = (number who got the class they wanted)/(number total)

P(A) = 379/500

P(A) = 0.758

There's a 75.8% chance someone will get the class they want

Let's see if being on the honor roll changes the probability we just found

So we want to compute P(A | B). If it is equal to P(A), then being on the honor roll does not change P(A).

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

A and B = someone got the class they want and they're on the honor roll

P(A and B) = 64/500

P(A and B) = 0.128

P(B) = 144/500

P(B) = 0.288

P(A | B) = P(A and B)/P(B)

P(A | B) = 0.128/0.288

P(A | B) = 0.44 approximately

This is what you have shown in your steps. This means if we know the person is on the honor roll, then they have a 44% chance of getting the class they want.

Those on the honor roll are at a disadvantage to getting their requested class. Perhaps the thinking is that the honor roll students can handle harder or less popular teachers.

Regardless of motivations, being on the honor roll changes the probability of getting the class you want. So Alex is correct in thinking the honor roll students have a disadvantage. Everything would be fair if P(A | B) = P(A) showing that events A and B are independent. That is not the case here so the events are linked somehow.

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3 years ago

An Article a Florida newspaper reported on the topics that teenagers most want to discuss with their parents. The findings, the

Sergio039 [100]

Answer:

The estimate is $P__{hat}} \pm E = 0.37 \pm 0.0348$

Step-by-step explanation:

From the question we are told that

The sample size is n = 522

The sample proportion of students would like to talk about school is $\r p__{hat}} = 0.37$

Given that the confidence level is 90 % then the level of significance can be mathematically evaluated as

$\alpha = 100 - 90$

$\alpha = 10\%$

$\alpha = 0.10$

Next we obtain the critical value of $\frac{\alpha }{2}$ from the normal distribution table, the value is

$Z_{\frac{\alpha }{2} } =Z_{\frac{0.10}{2} } = 1.645$

Generally the margin of error can be mathematically represented as

$E = Z_{\frac{\alpha }{2} } * \sqrt{\frac{\r P_{hat}(1- \r P_{hat} )}{n } }$

=> $E = 1.645 * \sqrt{\frac{0.37 (1- 0.37 )}{522 } }$

=> $E = 0.0348$

Generally the estimate the proportion of all teenagers who want more family discussions about school at 90% confidence level is

$P__{hat}} \pm E$

substituting values

$0.37 \pm 0.0348$

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3 years ago