The sum can be rewritten as y=4x, where y=f(x).
Now, we can rewrite the equation a x=y/4
Therefore, inv(f(x))=x/4
Answer:
No, the on-time rate of 74% is not correct.
Solution:
As per the question:
Sample size, n = 60
The proportion of the population, P' = 74% = 0.74
q' = 1 - 0.74 = 0.26
We need to find the probability that out of 60 trains, 38 or lesser trains arrive on time.
Now,
The proportion of the given sample, p = 
Therefore, the probability is given by:
![P(p\leq 0.634) = [\frac{p - P'}{\sqrt{\frac{P'q'}{n}}}]\leq [\frac{0.634 - 0.74}{\sqrt{\frac{0.74\times 0.26}{60}}}]](https://tex.z-dn.net/?f=P%28p%5Cleq%200.634%29%20%3D%20%5B%5Cfrac%7Bp%20-%20P%27%7D%7B%5Csqrt%7B%5Cfrac%7BP%27q%27%7D%7Bn%7D%7D%7D%5D%5Cleq%20%5B%5Cfrac%7B0.634%20-%200.74%7D%7B%5Csqrt%7B%5Cfrac%7B0.74%5Ctimes%200.26%7D%7B60%7D%7D%7D%5D)
P![(p\leq 0.634) = P[z\leq -1.87188]](https://tex.z-dn.net/?f=%28p%5Cleq%200.634%29%20%3D%20P%5Bz%5Cleq%20-1.87188%5D)
P![(p\leq 0.634) = P[z\leq -1.87] = 0.0298](https://tex.z-dn.net/?f=%28p%5Cleq%200.634%29%20%3D%20P%5Bz%5Cleq%20-1.87%5D%20%3D%200.0298)
Therefore, Probability of the 38 or lesser trains out of 60 trains to be on time is 0.0298 or 2.98 %
Thus the on-time rate of 74% is incorrect.
A has fixed one time fee of $12 and if you go to it say "m" months you pay $28 for each month, so your total cost at A is really 12 + 28m.
B has a fixed one time fee of $20 and if you go to it "m" months you pay $26 for each month, so you total cost at B is 20 + 26m.
how many months for the cost to be the same?

well, since the cost for both is the same, we can just get A's, knowing that B is the same

Convert to equivalent fractions to see
Both bottom s can be 6
So times the half by 3
And times the third by 2
1/2=3/6
1/3=2/6
So she has a total of 5/6 of the movie accounted for...
So the left over is 1/6
1/6 at nighttime
Answer:
I don't know how to do this. I'm so sorry that I couldn't help!!
Step-by-step explanation: