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.
Step-by-step explanation:
The equation of line A:
y-1=0, 5x-3
y=0, 5x-2. Hence its gradient is 0,5
For line B, its you intercept is -8, so y=mx-8.
When x = -4, y=-4m -8 =0 as shown in the graph. This means that m = -2.
Since |-2| > |0, 5|, line B is steeper than line A.
The are 8 blue bicycles, 6 green bicycles, and 4 yellow bicycles and total they equal 18.
Answer:
x = 24
Step-by-step explanation:
1/6 x + 3 = 7
1/6 x = 4
x = 24
Surprise, there 8 letters.
So, 8*7*6*5*4*3*2*1= 40, 320
