A = 9 (Sorry I can't do the other ones, I haven't learned those ones yet) Hope this helps!! :P
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.
Substitute 4 for y. Then -3x^2 + 16 = 52.
Solve for x. Subtract 16 from both sides, obtaining -3x^2 = 36.
Divide both sides by -3, obtaining x^2 = -12. This last result makes no sense, as no square of a real number could be negative. Probably this is where you're ":getting a negative answer."
If imaginary answers were allowed, then x = i*√12 = i*2√3 or x = -i*2√3.
x =
Well. add all of these up 8+5+5. we know 5+5 is 10. and 10+8 is 18. so 18 is the denominator. There are 8 red markers and 5 blue markers at 8 and 5 up. 8+5 is 13 right? So the numerator will be 13. but the probability of picking a blue then a red is 13 alone or if you are ask for out of the amount of markers there are, its 13/18. so the probability is 13 or 13/18.
x=11
the angles are congruent so 6x-10=56
6x=66
x=11
