1/7(1000) = 1/7(7000/7)
1*7000
7*7
7000
49
1000
7
so 1/7(1000) = 1000/7 or 142 6/7
Answer:
C
Step-by-step explanation:
If you have a Ti-84 series calculator, press "stat" then "Edit..." and then fill in the data table values for x and y in two lists. Then press "2nd" and "mode" to quit. Now press "stat" again and right arrow over to "calc" and press down until you find "ExpReg" and set the "Xlist" and "Ylist" that you used and you will get C as the answer. Another way to do this is to manually substitute values into all 4 equations, which is boring.
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: -3x^3 - 7x^2 - x + 39
Step-by-step explanation:
2/(x^2 -9) - 3x/(x^2-5x+6)
get those denominators factored out
2/(x-3)(x+3) - 3x/(x-3)(x-2)
Multiply both by both denominators
2(x-3)(x-2) - 3x(x-3)(x+3)
Multiply it out
(2x^2 - 10x + 12) - (3x^3 + 9x^2 -9x - 27)
Simplify and it becomes
-3x^3 -7x^2 -x + 39
Answer: 72 feet
Step-by-step explanation:
24 yards= 72 feet
