7 x 500,000 = 3,500,000.
The towns are 3,500,000 units apart (I don't know what the cm were scale too, sorry.)
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.
Answer:
I think 1/4
Step-by-step explanation:
So, you first should find a common denominator so you can add the fractions together.
So, 1/4 + 1/3 + 1/6 becomes 3/12+4/12+2/12.
Add that up, you get 9/12.
Simplify.
9/12 = 3/4
3/4 is how much is dug out. 4/4 - 3/4 is 1/4. That's how much for digging is left, I believe.
Answer:
E(X ¯)=210,000.
Step-by-step explanation:
A sampling distribution for samples of size n=49 from a population with means μ=210,000 and standard deviation σ=35,000, has the following means anda standard deviation:

If X ¯ is the mean sales price of the sample, it will have a mean value of E(X ¯)=210,000.