Answer:
a) P(X=2)= 0.29
b) P(X<2)= 0.59
c) P(X≤2)= 0.88
d) P(X>2)= 0.12
e) P(X=1 or X=4)= 0.24
f) P(1≤X≤4)= 0.59
Step-by-step explanation:
a) P(X=2)= 1 - P(X=0) - P(X=1) - P(X=3) - P(X=4)= 1-0.41-0.18-0.06-0.06= 0.29
b) P(X<2)= P(X=0) + P(X=1)= 0.41 + 0.18 = 0.59
c) P(X≤2)= P(X=0) + P(X=1) + P(X=2)=0.41+0.18+0.29= 0.88
d) P(X>2)=P(X=3) + P(X=4)=0.06+0.06= 0.12
e) P(X=1 or X=4)=P(X=1 ∪ X=4) = P(X=1) + P(X=4)=0.18+0.06= 0.24
f) P(1≤X≤4)=P(X=1) + P(X=2) + P(X=3) + P(X=4)=0.18+0.29+0.06+0.06= 0.59
-5 because it is closest to zero and -17 is farther than zero on a number line :) <span />
Answer:
Option C. 
Step-by-step explanation:
we know that
![A=\frac{P[(1+r)^{n} -1]}{r(1+r)^{n}}](https://tex.z-dn.net/?f=A%3D%5Cfrac%7BP%5B%281%2Br%29%5E%7Bn%7D%20-1%5D%7D%7Br%281%2Br%29%5E%7Bn%7D%7D)
we have



substitute in the formula
![A=\frac{400[(1+0.00625)^{72} -1]}{0.00625(1+0.00625)^{72}}\\ \\A=\frac{226.446972}{0.009788}\\ \\A=\$23,134.61](https://tex.z-dn.net/?f=A%3D%5Cfrac%7B400%5B%281%2B0.00625%29%5E%7B72%7D%20-1%5D%7D%7B0.00625%281%2B0.00625%29%5E%7B72%7D%7D%5C%5C%20%5C%5CA%3D%5Cfrac%7B226.446972%7D%7B0.009788%7D%5C%5C%20%5C%5CA%3D%5C%2423%2C134.61)