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Nezavi [6.7K]
3 years ago
11

HELPPPP!!! What is the measure of JKL?

Mathematics
1 answer:
cricket20 [7]3 years ago
3 0

the answer is b 220


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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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4 years ago
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Len [333]
The rule of law I think
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