Horizontal translation: 2 units right.
Vertical translation: 5 units up
Stretch/compression: stretched by a factor of 3.
Reflection: not reflected.
Answer:
71.123 mph ≤ μ ≤ 77.277 mph
Step-by-step explanation:
Taking into account that the speed of all cars traveling on this highway have a normal distribution and we can only know the mean and the standard deviation of the sample, the confidence interval for the mean is calculated as:
≤ μ ≤ 
Where m is the mean of the sample, s is the standard deviation of the sample, n is the size of the sample, μ is the mean speed of all cars, and
is the number for t-student distribution where a/2 is the amount of area in one tail and n-1 are the degrees of freedom.
the mean and the standard deviation of the sample are equal to 74.2 and 5.3083 respectively, the size of the sample is 10, the distribution t- student has 9 degrees of freedom and the value of a is 10%.
So, if we replace m by 74.2, s by 5.3083, n by 10 and
by 1.8331, we get that the 90% confidence interval for the mean speed is:
≤ μ ≤ 
74.2 - 3.077 ≤ μ ≤ 74.2 + 3.077
71.123 ≤ μ ≤ 77.277
Add (-2X) to both sides:
2X + B + (-2X) = C + (-2X)
If we (rearrange) it into B + 2X + (-2X) = C + (-2X)
B + 2X - 2X = C + (-2X)
Since +2X - 2X = 0
B + 0 = C + (-2X)
B = C + (-2X)
B = C - 2X
41, because the ratio is the same