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AURORKA [14]
2 years ago
6

A data set includes body temperatures of healthy adult humans having a mean of f and a standard deviation of f. construct a ​% c

onfidence interval estimate of the mean body temperature of all healthy humans. what does the sample suggest about the use of 98.6f as the mean body​ temperature?
SAT
1 answer:
MaRussiya [10]2 years ago
5 0

What the sample suggests is that; the mean body temperature couldvery possibly bevery possibly be 98.6 °F.

<h3>What is the Confidence Interval?</h3>

For a 99% Confidence Interval and DF = 103 - 1 = 102, the value of z = 2.33.

The formula for confidence interval is;

CI = x⁻ ± z(σ/√n))

We are given;

Sample mean; x⁻ = 98.9

standard deviation; σ = 0.67

sample size; n = 103

Thus the confidence Interval is;

CI = [98.9 - 2.33(0.67/√103)], [98.9 + 2.33(0.67/√103)]

CI = (98.9 - 0.154, 98.5 + 0.154)

CI = (98.36, 98.64)

Thus, 98.6 is contained within the 99% CI,and so we can conclude that the the mean body temperature could very possibly bevery possibly be 98.6 °F.

Read more about Confidence Interval at; brainly.com/question/17097944

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8 0
3 years ago
A photographic stop bath contains 160ml of pure acetic acid.
NeTakaya

Answer:

Solving for the Percent by Volume

The v/v concentration of acetic acid in the stop bath is 24.6%.

Explanation:

To answer the question, we need to identify certain information stated in the question. Let us write down what’s given:

Volume of solute = 160 mL

Volume of solution = 650 mL

Next, we need to determine what is required, in this case, the percent by volume. Now, let us look at the formula:

v/v concentration = (solute volume / volume of the solution) * 100

Solution:

Percent by Volume = (160 / 650) * 100

Percent by Volume = 0.2461538461538462 * 100

Percent by Volume = 24.61538461538462

Rounded off = 24.6%

Therefore, the final answer is 24.6%

Explanation:

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A. He is shy. I read the book so many times.

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