Of all individual tax returns, 37 percent include errors made by the taxpayer. If IRS examiners are assigned randomly selected r

Step One

Deduct 32 from both sides.

F - 32 = \frac{9}{5}(k - 273.15)

Step Two

Multiply each side by \frac{5}{9}.

\frac{5}{9}(F - 32) = \frac{5}{9} \times \frac{9}{5}(k - 273.15)

\frac{5}{9}(F - 32) = k - 273.15

Step Three

Add 273.15 to both sides.

\frac{5}{9}(F - 32) + 273.15 = k

Problem B

F = 180

Solve for k

k = \frac{5}{9}(F - 32) + 273.15

k = \frac{5}{9}(180 - 32) + 273.15

k = \frac{5}{9} \times 148 + 273.15

k = 82.2222 + 273.15

k = 355.3722

k = 355.4  <<< Answer

Step-by-step explanation:

The prices he received quotes for are as follows: $663, $273, $410, $622, $174, $374

To begin, we will find the average.

Average = total of data points/ number of data points.

Total of data points =

663 + 273 + 410 + 622 + 174 + 374

= 2516

Total count = 6

Average = 2516/6 = 419.33

Standard deviation = √summation(x - m)^2/n

summation(x - m)^2/n = (663 - 419.33)^2 + (273 - 419.33)^2 + (410 - 419.33)^2 + (622 - 419.33)^2 + (174 - 419.33)^2 + (374 - 419.33)^2

= 179417.9334/6 = 29902.9889

Standard deviation = √29902.9889

= 172.9

Answer:

a) Null hypothesis:

Alternative hypothesis:

b)

Given that the p-value is lower than the significance threshold in this situation, we have sufficient grounds to reject the null hypothesis.

c)

In this case, since the p-value exceeds the significance threshold, we have adequate evidence to FAIL to reject the null hypothesis.

d)

Here again, with the p-value being less than the significance level, we can reject the null hypothesis.

Step-by-step explanation:

1) Provided data and references

represents the average of the samples

denotes the standard deviation of the samples

indicates the number of samples

is the value we are examining

defines the significance level for the test.

t represents the specific statistic of interest

indicates the p-value relevant to the test (the variable of concern)

Define the null and alternative hypotheses.

To assess if the true mean is at least 10 hours, we must set up a hypothesis:

Part a

Null hypothesis:

Alternative hypothesis:

If we consider the sample size being less than 30 and the population deviation unknown, it’s more appropriate to use a t-test to compare the actual mean with the reference value, calculated as:

(1)

Part b

In this scenario t=-2.3,

Initially, we need to calculate the degrees of freedom

Since this is a left-tailed test, the p-value is determined by:

In this instance, with the p-value being less than the significance level, we have sufficient evidence to reject the null hypothesis.

Part c

For this situation t=-1.8,

We need to find the degrees of freedom

For the left-tailed test, the p-value is given by:

In this case, since the p-value is above the significance level, we have enough grounds to FAIL to reject the null hypothesis.

Part d

For this case t=-3.6,

Firstly, we find the degrees of freedom

Since we are conducting a left-tailed test, the p-value is calculated as:

Here, with the p-value being lower than the significance threshold, we can reject the null hypothesis.

Solution:

x=8

Detailed explanation:

This isosceles triangle consists of two right triangles with sides equal to 8,, and y

Considering we possess two sides of a right triangle, determining the third side can be achieved using the Pythagorean Theorem

Because it is an isosceles triangle, these two right triangles are the same, thus x=2y

Hence, x=2(4)=8