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.
I think is OB, I’m not too sure tho
Answer:
d. 50°
Step-by-step explanation:
Since you know one angle and its opposite side, use the law of sines.
sin 36°/11.9 = sin D/15.6
sin D = 15.6 × sin 36° / 11.9
sin D = =0.7705
D = sin^-1 0.7705
D = 50.4°
Answer:
147,000N
Step-by-step explanation:
A1= 2m^2
A2= 0.2m^2
F2= 14,700N
Required
F1, the applied force
Applying the formula
F1/A1= F2/A2
substute
F1/2=14700/0.2
2*14700= F1*0.2
29400= F1*0.2
F1= 29400/0.2
F1=147,000N
Hence, the applied force is 147,000N
Answer:
y = 6(x + 6)(x - 4).
Step-by-step explanation:
y = 6x^2 + 12x - 144
y = 6(x^2 + 2x - 24) Factoring the trinomial in the parentheses ,we get:
y = 6(x + 6)(x - 4)
