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Fed [463]
3 years ago
12

For the past 20 years, the high temperature on April 15th has averaged = 92 degrees with a standard deviation of = 4. But on the

recent April 15 date, the 21st year, the high temperature was 102 degrees. Based on this information, the recent year's temperature on April 15th was ________________.
Mathematics
1 answer:
tia_tia [17]3 years ago
5 0

Answer:

96.8degrees.

Step-by-step explanation:

Average is 92 degrees with 20 years of temperature data that means sum of total temperatures (taken on each 15 April) divided by 20 = 92 degree.

Now we want to know the new average temperature on 21 year with new temperature taken on 15 April being equal to 102 degrees.

We can add new temperatures to the previous sum and this time divide by 21 because it is 21st year this time.

\frac{Sum-of-total-temperatures}{20-years} is equal to 92 to get total temperature multiply both sides by 20 which gives sum of total temperature = 1932.

Let's add new temperature in it, 1932+102=2034 and divide it by 21.

2034/21=96.86 this gives us recent year's temperature on April 15.

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A data set includes 103 body temperatures of healthy adult humans having a mean of 98.9degreesf and a standard deviation of 0.67
trasher [3.6K]

Answer:

Step-by-step explanation:

Confidence interval is written in the form,

(Sample mean - margin of error, sample mean + margin of error)

The sample mean, x is the point estimate for the population mean.

Margin of error = z × s/√n

Where

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n = number of samples = 103

From the information given, the population standard deviation is unknown hence, we would use the t distribution to find the z score

In order to use the t distribution, we would determine the degree of freedom, df for the sample.

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Since confidence level = 99% = 0.99, α = 1 - CL = 1 – 0.99 = 0.01

α/2 = 0.01/2 = 0.005

the area to the right of z0.005 is 0.005 and the area to the left of z0.005 is 1 - 0.005 = 0.995

Looking at the t distribution table,

z = 2.6249

Margin of error = 2.6249 × 0.67/√103

= 0.173

Confidence interval = 98.6 ± 0.173

This suggests that the mean body temperature could very possibly be

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3 0
3 years ago
The diagonal length is 16 and the length is 11 what is the width of this rectangle
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