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Luda [366]
3 years ago
10

The temperature in an oven changes from 350 degrees to 362 degrees. What is the percent increase in temperature, to the nearest

tenth of a percent?
Group of answer choices

A 12%

B 35%

C 3.5%

D 3.4%


PLEASE HELP FASTTTTTTTTTT
Mathematics
1 answer:
Aleksandr-060686 [28]3 years ago
5 0

Answer:

it's 3.4%

Step-by-step explanation:

You take the difference of 362 and 350 (which is 12) and divide it by the starting number (350) getting you 3.4

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A random sample of 21 observations is used to estimate the population mean. The sample mean and the sample standard deviation ar
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Answer:

a. CI=[128.79,146.41]

b. CI=[122.81,152.39]

c. As the confidence level increases, the interval becomes wider.

Step-by-step explanation:

a. -Given the sample mean is 137.6 and the standard deviation is 20.60.

-The confidence intervals can be constructed using the formula;

\bar X\pm \ z\frac{s}{\sqrt{n}},

where:

  • s is the sample standard deviation
  • z is the s value of the desired confidence interval

we then calculate our confidence interval as:

\bar X\pm z\frac{s}{\sqrt{n}}\\\\=137.60\pm z_{0.05/2}\times\frac{20.60}{\sqrt{21}}\\\\=137.60\pm1.960\times \frac{20.60}{\sqrt{21}}\\\\=137.60\pm8.8108\\\\\\=[128.789,146.411]

Hence, the 95% confidence interval is between 128.79 and 146.41

b. -Given the sample mean is 137.6 and the standard deviation is 20.60.

-The confidence intervals can be constructed using the formula in a above;

\bar X\pm z\frac{s}{\sqrt{n}}\\\\=137.60\pm z_{0.01/2}\times\frac{20.60}{\sqrt{21}}\\\\=137.60\pm3.291\times \frac{20.60}{\sqrt{21}}\\\\=137.60\pm 14.7940\\\\\\=[122.806,152.394]

Hence, the variable's 99% confidence interval is between 122.81 and 152.39

c. -Increasing the confidence has an increasing effect on the margin of error.

-Since, the sample size is particularly small, a wider confidence interval is necessary to increase the margin of error.

-The 99% Confidence interval is the most appropriate to use in such a case.

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we know that    

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