I need help with the area and perimeter

A recent Gallup poll found that 36% of U.S. teens aged from 13 to 17 years old have a computer with [(6)] Internet access in the

In-s [12.5K]

Answer:

Step-by-step explanation:

a) The population parameter is the population proportion.

b) Confidence interval is written as

Sample proportion ± margin of error

Margin of error = z × √pq/n

Where

z represents the z score corresponding to the confidence level

p = sample proportion. It also means probability of success

q = probability of failure

q = 1 - p

p = x/n

Where

n represents the number of samples

x represents the number of success

From the information given,

n = 1028

p = 36/100 = 0.36

q = 1 - 0.36 = 0.64

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

α = 1 - 0.99 = 0.1

α/2 = 0.01/2 = 0.005

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

1 - 0.05 = 0.995

The z score corresponding to the area on the z table is 2.58. Thus, confidence level of 99% is 2.58

Therefore, the 99% confidence interval is

0.36 ± 2.58 × √(0.36)(0.64)/1028

The lower limit of the confidence interval is

0.36 - 0.039 = 0.321

The upper limit of the confidence interval is

0.36 + 0.039 = 0.399

Therefore, with 99% confidence interval, the proportion of U.S. teens aged from 13 to 17 years old that have a computer with Internet access in their rooms is between 0.321 and 0.399

c) for a margin of error of 1%, that is 1/100 = 0.01, then

0.01 = 2.58 × √(0.36)(0.64)/n

0.01/2.58 = √0.2304/n

0.00387596899 = √0.2304/n

Square both sides

0.00001502314 = 0.2304/n

n = 0.2304/0.00001502314

n = 15336

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The graph is stretch/shrunk by a factor of a. The Domain is h=

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Pete's Taco Stand finds that the cost to produce 400 tacos is $650, while the cost to

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Answer:

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Step-by-step explanation:

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sorry i want points

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Please help me.What is an equation that represents this situation?

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The weight of National Football League (NFL) players has increased steadily, gaining up to 1.5 lb. per year since 1942. Accordin

GuDViN [60]

Answer:

The probability that the sample mean weight will be more than 262 lb is 0.0047.

Step-by-step explanation:

The random variable X can be defined as the weight of National Football League (NFL) players now.

The mean weight is, μ = 252.8 lb.

The standard deviation of the weights is, σ = 25 lb.

A random sample of n = 50 NFL players are selected.

According to the Central Limit Theorem if we have an unknown 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 means will be approximately normally distributed.

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

$\mu_{\bar x}=\mu$

And the standard deviation of the sample means is given by,

$\sigma_{\bar x}=\frac{\sigma}{\sqrt{n}}$

The sample of players selected is quite large, i.e. n = 50 > 30, so the central limit theorem can be used to approximate the distribution of sample means.

$\bar X\sim N(\mu_{\bar x}=252.8,\ \sigma_{\bar x}=3.536)$

Compute the probability that the sample mean weight will be more than 262 lb as follows:

$P(\bar X>262)=P(\frac{\bar X-\mu_{\bar x}}{\sigma_{\bar x}}>\frac{262-252.8}{3.536})\\\\=P(Z>2.60)\\\\=1-P(Z$

*Use a z-table for the probability.

Thus, the probability that the sample mean weight will be more than 262 lb is 0.0047.

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3 years ago