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:
y = -2x -4
Step-by-step explanation:
The equation of a line is y = mx + b, where m is the slope of the line, and b is the y-intercept. We are told that the slope of the line is -2, so we can start with
y = mx + b
y = -2x + b
We can figure out the value of b by plugging in the coordinates of the point given (-2, 0)
0 = -2(-2) + b
0 = 4 + b
Subtract 4 from both sides of the equation
-4 = b
Then plug that value of b back into the equation above:
y = -2x + (-4) or y = -2x -4
30 ( x + 10 = 55 ) hope this
