121 is big enough to assume normality and not worry about the t distribution. By the 68-95-99.7 rule a 95% confidence interval includes plus or minus two standard deviations. So 95% of the cars will be in the mph range
![(73 - 2 \cdot 11, 73 + 2 \cdot 11) = (51,95)](https://tex.z-dn.net/?f=%2873%20-%202%20%5Ccdot%2011%2C%2073%20%2B%202%20%5Ccdot%2011%29%20%3D%20%2851%2C95%29)
The question is a bit vague, but it seems we're being asked for the 95% confidence interval on the average of 121 cars. The 121 is a hint of course.
The standard deviation of the average is in general the standard deviation of the individual samples divided by the square root of n:
![\sigma = \dfrac{ 11}{\sqrt{121}} = 1](https://tex.z-dn.net/?f=%5Csigma%20%3D%20%5Cdfrac%7B%2011%7D%7B%5Csqrt%7B121%7D%7D%20%3D%201)
So repeating our experiment of taking the average 121 cars over and over, we expect 95% of the averages to be in the mph range
![(73 - 2 \cdot 1, 73 + 2 \cdot 1) = (71,75)](https://tex.z-dn.net/?f=%2873%20-%202%20%5Ccdot%201%2C%2073%20%2B%202%20%5Ccdot%201%29%20%3D%20%2871%2C75%29%20)
That's probably the answer they're looking for.
5(2y-4) - 3y = 1. 10y-20-3y=1. 7y-20=1. 7y=21. y =3. x = 2(3)-4. x = 2. x*y =6.
Answer:
-5x + 56
Step-by-step explanation:
Use distributive property to refine.
20 - 2x + 36 - 3x
Combine Like Terms
-5x + 56