Answer:
Step-by-step explanation:
corresponds with 
Answer:he accompanying table shows wind speed and the corresponding wind chill factor
when the air temperature is 29°F. Write a logarithmic regression equation for this set
of data, rounding all coefficients to the nearest thousandth. Using this equation, find
he accompanying table shows wind speed and the corresponding wind chill factor
when the air temperature is 29°F. Write a logarithmic regression equation for this set
of data, rounding all coefficients to the nearest thousandth. Using this equation, find
Step-by-step explanation:
Answer:
between 222 and 223 days
Step-by-step explanation:
This relation includes the point (0, 2).
A proportional relation must include the point (0, 0).
This is not proportional, but it may be linear.
Points: (0, 2), (2, 20)
y = mx + b
20 = (20 - 2)/(2 - 0) × 2 + b
20 = 18/2 × 2 + b
b = 2
y = 18x + 2
When 4000 people know, y = 4000.
y = 18x + 2
4000 = 18x + 2
3998 = 18x
x = 3998/18
x = 222.111...
Answer: between 222 and 223 days
H(X) = (x - 2)²(x + 4)(x - 1) has 3 zeroes, they are -4,2 and 1
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.