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1 year ago

If the average of x and y is 11 and the average of x, and y z is 5 what is the value of z

Mathematics

1 answer:

qwelly [4]1 year ago

6 0

If the average of x and y is 11 and the average of x, y and z is 5, then the value of z is -7

Given,

The average of x and y = 11

We know average = Sum of the terms / Number of terms

Then, the equation will become

$\frac{x+y}{2}=11$
x+y =11×2

x+y =22

Average of x, y and z = 5

$\frac{x+y+z}{3}=5$

x+y+z = 5×3

x+y+z =15

Substitute the value of x+y

22+z=15

z =15-22

x = -7

Hence, If the average of x and y is 11 and the average of x, y and z is 5, then the value of z is -7

Learn more about average here

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The Internal Revenue Service (IRS) provides a toll-free help line for taxpayers to call in and get answers to questions as they

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Answer:

From the result of the one-tailed z-test, we can conclude that the mean waiting time is not significantly less than the the 15 minute claim by the taxpayer advocate.

Step-by-step explanation:

Sample size = 50

Mean waiting time = 13 minutes

Standard deviation = 11 minutes

We need to perform one hypothesis test that the actual mean waiting time turned out to be significantly less than the 15-minute claim made by the taxpayer advocate.

The null hypothesis would be that there is no significant difference.

H₀: μ₀ = 15 minutes

The alternative hypothesis would be that the mean waiting time is indeed significantly less than the 15-minute claim by the taxpayer advocate.

Hₐ: μ₀ < 15 minutes.

This is evidently a one tail hypothesis test (we're investigating only in one direction; less than the claim). Hence, we can use the z-test

z = (x - μ)/σₓ

σₓ = (σ/√n) = (11/√50) = 1.556

z = (13 - 15)/1.556 = - 1.29

Using the z-table at significance level of 0.05.

p-value = 1 - 0.9115 (from the z-tables)

p-value = 0.0885

Since the p-value is more than the significance level (0.0885 > 0.05), we do not reject the null hypothesis.

Hence, we accept the null hypothesis and can conclude that the mean waiting time is not significantly less than the the 15 minute claim by the taxpayer advocate.

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