Answer:
sorry I don't know about this
Step-by-step explanation:
sorry I don't know about this
Answer:
a=10.1
k=3
Step-by-step explanation:
Lets solve the first equation first.
94=10a-7
Lets add 7 to both sides.
101=10a
Divide both sides by 10 to isolate a
101/10=a
a=10.1
Lets do the second equation.
7k+10=31
Subtract 10 from both sides
7k=21
Divide 7 by both sides to isolate k
k=3
He would have to multiply 2x12x2.54 (so that would be the third answer)
Answer:
The 95% confidence interval for the mean savings is ($60.54, $81.46).
Step-by-step explanation:
As there is no information about the population standard deviation of savings and the sample is not large, i.e. <em>n</em> = 20 < 30, we will use a <em>t</em>-confidence interval.
The (1 - <em>α</em>)% confidence interval for population mean (<em>μ</em>) is:

For the data provided compute the sample mean and standard deviation as follows:
![\bar x=\frac{1}{n}\sum\limits^{n}_{i=1}{ X_{i}}=\frac{1}{20}\times [92+34+40+...+53+82]=71](https://tex.z-dn.net/?f=%5Cbar%20x%3D%5Cfrac%7B1%7D%7Bn%7D%5Csum%5Climits%5E%7Bn%7D_%7Bi%3D1%7D%7B%20X_%7Bi%7D%7D%3D%5Cfrac%7B1%7D%7B20%7D%5Ctimes%20%5B92%2B34%2B40%2B...%2B53%2B82%5D%3D71)

The critical value of <em>t</em> for <em>α</em> = 0.05 and (n - 1) = 19 degrees of freedom is:

*Use a <em>t</em>-table for the value.
Compute the 95% confidence interval for the mean savings as follows:

Thus, the 95% confidence interval for the mean savings is ($60.54, $81.46).