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.
180-53 is 127 angle 6 and 3 would be the same so your answer is 127
Answer:
Step-by-step explanation:
3x^4(−6x)
=3*x^4*(−6)*x
=−18*x^5
=−18x^5
Hope this helps if I doesn’t helps I will b happy to help u again
Answer: f=-2
Step-by-step explanation:
math
Answer:
Model A is 9
Model B is 5
Step-by-step explanation:
At Lisa's printing Company LLC there are two kinds of printing presses model.
Let x = the number of model A printing press.
Let y = the number of the model B printing press.
model A can print 70 bucks per day and model B can print 55 bucks per day.
The total number of both models prints 905 bucks per day.
This means
70x + 55y = 905 - - - - - - - - -1
The company owns 14 total printing presses. This means
x + y = 14 - - - - - - - - - -2
x = 14-y
Put x = 14-y in equation 1
70(14-y) + 55y = 905
980 -70y + 55y = 905
-15y = 905-980= -75
y = -75 / -15= 5
x = 14-y = 14-5
= 9
