Answer:
0.0918
Step-by-step explanation:
We know that the average amount of money spent on entertainment is normally distributed with mean=μ=95.25 and standard deviation=σ=27.32.
The mean and standard deviation of average spending of sample size 25 are
μxbar=μ=95.25
σxbar=σ/√n=27.32/√25=27.32/5=5.464.
So, the average spending of a sample of 25 randomly-selected professors is normally distributed with mean=μ=95.25 and standard deviation=σ=27.32.
The z-score associated with average spending $102.5
Z=[Xbar-μxbar]/σxbar
Z=[102.5-95.25]/5.464
Z=7.25/5.464
Z=1.3269=1.33
We have to find P(Xbar>102.5).
P(Xbar>102.5)=P(Z>1.33)
P(Xbar>102.5)=P(0<Z<∞)-P(0<Z<1.33)
P(Xbar>102.5)=0.5-0.4082
P(Xbar>102.5)=0.0918.
Thus, the probability that the average spending of a sample of 25 randomly-selected professors will exceed $102.5 is 0.0918.
Answer:
9.6
Step-by-step explanation:
Answer:
10^20
3^11
3^24
10^12
Step-by-step explanation:
(10^2)^10=10^2*10=10^20
3^3*3^8=3^3+8=3^11
(3^3)^8=3^3*8=3^24
10^2*10^10=10^2+10=10^12
Answer:
270 inches³
Step-by-step explanation:
Volume of carton = wlh
Where w is width, h is height and length is l
V = (9)(5)(6)
V = 270 inches³
