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:
x = 24
Step-by-step explanation:
x + 
x = 36
Make the denominator the same by multiplying the first fraction by 2 to make both denominators 6.
x + x = 36
Now we can calculate the numerators and keep the denominators the same.
x = 36
Find x.
x = 
× 36
x = 24
Answer:
I have left $88
Step-by-step explanation:
Step two:
Given data
Initial amount= $56
deposited amount= $42
The balance after deposit will be= 56+42
=$98
Step two:
If I withdraw 10 dollars
The new balance will be
=98-10
=$88
Hence I have left $88
123.5%
= 123.5/100
= 1235/1000
= (1235/5) / (1000/5)
= 247/200
= 1 47/200
The final answer is 247/200 or 1 47/200~
