Answer:
(w^2 - 4w + 16)
Step-by-step explanation:
Note that w^3 +64 is the sum of two perfect cubes, which are (w)^3 and (4)^3. The corresponding factors are (w + 4)(w^2 - 4w + 16).
Therefore,
(w^3 +64)/(4+ w) reduces as follows:
(w^3 +64)/(4+ w) (4 + w)(w^2 - 4w + 16)
--------------------------- = --------------------------------- = (w^2 - 4w + 16)
4 + w 4 + w
Answer:
C. (1,18)
Explanation
If you don’t know how to find it, just plug in the values into the equation until they are equal to each other
Answer:
3 1/2
Step-by-step explanation:
9514 1404 393
Answer:
- $304
- $91.83
Step-by-step explanation:
1. The finance charge is found from the simple interest formula;
I = Prt
where P is the principal amount, r is the annual rate, and t is the number of years.
24 months is 2 years, so the interest charged is ...
I = $1900×0.08×2 = $304
The finance charge is $304.
__
2. The monthly payment will be the total amount due, divided by the number of months.
payment = ($1900 +304)/24 = $2204/24 ≈ $91.83
The monthly payment is $91.83.
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.
