No, this is not valid. In the picture below,

is a vertical angle with

. But

is not
(a) Average time to get to school
Average time (minutes) = Summation of the two means = mean time to walk to bus stop + mean time for the bust to get to school = 8+20 = 28 minutes
(b) Standard deviation of the whole trip to school
Standard deviation for the whole trip = Sqrt (Summation of variances)
Variance = Standard deviation ^2
Therefore,
Standard deviation for the whole trip = Sqrt (2^2+4^2) = Sqrt (20) = 4.47 minutes
(c) Probability that it will take more than 30 minutes to get to school
P(x>30) = 1-P(x=30)
Z(x=30) = (mean-30)/SD = (28-30)/4.47 ≈ -0.45
Now, P(x=30) = P(Z=-0.45) = 0.3264
Therefore,
P(X>30) = 1-P(X=30) = 1-0.3264 = 0.6736 = 67.36%
With actual average time to walk to the bus stop being 10 minutes;
(d) Average time to get to school
Actual average time to get to school = 10+20 = 30 minutes
(e) Standard deviation to get to school
Actual standard deviation = Previous standard deviation = 4.47 minutes. This is due to the fact that there are no changes with individual standard deviations.
(f) Probability that it will take more than 30 minutes to get to school
Z(x=30) = (mean - 30)/Sd = (30-30)/4.47 = 0/4.47 = 0
From Z table, P(x=30) = 0.5
And therefore, P(x>30) = 1- P(X=30) = 1- P(Z=0.0) = 1-0.5 = 0.5 = 50%
We can find the acceleration via

We have


Then by definition of average acceleration,

so that


We alternatively could have found the time without knowing the acceleration. Since acceleration is constant, the average velocity is

Then


Answer:
B.
Step-by-step explanation:
Well we know that

so we can get the 2 outside of the radical

and we can get the x^2 outside too.

and we also can get y outside.
so we have:
![2x^{2}y\sqrt[5]{7xy^3}](https://tex.z-dn.net/?f=2x%5E%7B2%7Dy%5Csqrt%5B5%5D%7B7xy%5E3%7D)
Answer:
yea its in the book
Step-by-step explanation: