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:
store one
Step-by-step explanation:
Store One:
$720(0.50)=$360
$720-$360=$360
Store Two:
$720(0.30)=$216
$720-$216=$504
$504(0.20)=$100.8
$504-$100.8=$403.2
Your volume would be 27 inches cubed.
This is because Volume = length x width x height. If you multiply 3 x 3 x 3 you get 27. This resulting in your answer being 27.
Two parallel lines never intersect, therefore they will never have a solution.
One line could be x=6 and the other could be x=2
Hope this helps :)
