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Daniel [21]
3 years ago
13

of the 5th grade students, 15/20 went to the book fair. Of the students who went to the book fair, 12/16 bought ay least one boo

k. What fraction of the 5th grade students boughtat least one book?
Mathematics
2 answers:
ycow [4]3 years ago
6 0

Answer:

9/16  of the students

Step-by-step explanation:

first you can simplify 15/20 to 3/4

you can simplify 12/16 to 3/4

3/4*3/4=9/16

kolezko [41]3 years ago
4 0

Answer:

9/16

Step-by-step explanation:

That is 12/16 of 15/20

= 3/4 * 3/4

= 9/16

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Health insurers are beginning to offer telemedicine services online that replace the common office visit. A company provides a v
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Answer:

\text {CI} = (60.54, \: 81.46)\\\\

Therefore, we are 95% confident that actual mean savings for a televisit to the doctor is within the interval of ($60.54 to $81.46)

Step-by-step explanation:

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=AVERAGE(number1, number2,....)

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\bar{x} = \$71  

Let us find out the standard deviation of savings for a televisit to the doctor from the given data.

Using Excel,

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The confidence interval is given by

\text {confidence interval} = \bar{x} \pm MoE\\\\

Where the margin of error is given by

$ MoE = t_{\alpha/2}(\frac{s}{\sqrt{n} } ) $ \\\\

Where n is the sample of 20 online doctor visits, s is the sample standard deviation and t_{\alpha/2} is the t-score corresponding to a 95% confidence level.

The t-score is given by is

Significance level = α = 1 - 0.95 = 0.05/2 = 0.025

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MoE = t_{\alpha/2}(\frac{s}{\sqrt{n} } ) \\\\MoE = 2.093\cdot \frac{22.35}{\sqrt{20} } \\\\MoE = 2.093\cdot 4.997\\\\MoE = 10.46\\\\

So the required 95% confidence interval is

\text {CI} = \bar{x} \pm MoE\\\\\text {CI} = 71 \pm 10.46\\\\\text {CI} = 71 - 10.46, \: 71 + 10.46\\\\\text {CI} = (60.54, \: 81.46)\\\\

Therefore, we are 95% confident that actual mean savings for a televisit to the doctor is within the interval of ($60.54 to $81.46)

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Answer:

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