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:
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).
About 13%,
0.20 is the additional price added,
0.20/1.5, however it could also be 11% depending on which rate you start from.
Answer:
PQ = 14 MQ = 7 PM = 7
Step-by-step explanation:
PQ= PM + MQ = PMQ
If M is the MIDpoint it dissects PQ in half so that means PM = MQ
PQ= 5x-3+11-2x
PQ = 3x + 8
5x - 3 = 11 -2x
+2x +2x
7x - 3 = 11
+3 +3
7x = 14
7x / 7 = 14/7
x = 2
Plug in 2 for X
PQ = 3x + 8 = 3 x 2 +8 = 6 + 8 = 14
PM = 5x -3 + 5 x 2 - 3 = 10 - 3 = 7
MQ = 11 - 2x = 11 - 2 x 2 = 11 - 4 = 7
The answer is C.
Hope this helps :)