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Sloan [31]
3 years ago
13

1)In a sample of 100 households, the mean number of hours spent on social networking sites during the month of January was 50 ho

urs. In a much larger study,
the standard deviation was determined to be 6 hours.
Assume the population standard deviation is the same. What is the 98% confidence interval for the mean hours devoted to social networking in January?

2)Two studies were completed in Florida. One study in northern Florida involved 2,000 patients; 64% of them experienced flu-like symptoms during the month of December.
The other study, in southern Florida, involved 3,000 patients;
54% of them experienced flu-like symptoms during the same month. Which study has the smallest margin of error for a 95% confidence interval?
Mathematics
1 answer:
ra1l [238]3 years ago
8 0
So the following is the answer of the following question in your problem and please do not post to many question in just one posting. 
#1 The 98% confidence interval for the mean hours is between 6 and 50.
#2 The smallest margin of error between the two is southern florida with 1.8% margin of error.
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Answer:

Option A. 5

Step-by-step explanation:

From the question given above, the following data were obtained:

First term (a) = –3

Common ratio (r) = 6

Sum of series (Sₙ) = –4665

Number of term (n) =?

The number of terms in the series can be obtained as follow:

Sₙ = a[rⁿ – 1] / r – 1

–4665 = –3[6ⁿ – 1] / 6 – 1

–4665 = –3[6ⁿ – 1] / 5

Cross multiply

–4665 × 5 = –3[6ⁿ – 1]

–23325 = –3[6ⁿ – 1]

Divide both side by –3

–23325 / –3 = 6ⁿ – 1

7775 = 6ⁿ – 1

Collect like terms

7775 + 1 = 6ⁿ

7776 = 6ⁿ

Express 7776 in index form with 6 as the base

6⁵ = 6ⁿ

n = 5

Thus, the number of terms in the geometric series is 5.

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Identify the leading coefficient in the following polynomial:<br> – 5x² + 3x - 7
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