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:
Step-by-step explanation:
Hypotenuse^2=3^2+8^2
Hypotenuse^2=9+64=73
Hypotenuse =square root of 73
Hypptenuse =8.54
Y=(kx)/z
15=(k10)/4
15=(5k/2)
15=k(5/2)
ties both sides by 2/5 to clear fraction
6=k
y=(6x)/z
x=20
z=6
y=(6*20)/6
y=20
Answer:
8
Step-by-step explanation:
Y) -8 x -4 = 32
X) -2 x -4 = 8
