Answer:
11
Step-by-step explanation:
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.
This is an incomplete problem. The missing information is
Each bulleted statement describes how the amount of income tax is determined for
yearly income in different ranges.
1) Yearly incomes of $8,925 or less are taxed at a flat rate of 10%.
2) For yearly incomes from $8,926 to $36,250, the first $8,925 is taxed at 10%
and any income beyond $8,925 is taxed at 15%.
3) For yearly income greater than $36,250, the first $8,925 is taxed at 10%, the
next $27,325 is taxed at 15%, and any income beyond $36,250 is taxed at
25%
The taxable income of Mr. Vance corresponds to number 2.
⇒ 8,925 * 10% = 892.50
⇒ 35,675 - 8,925 = 26,750 * 15% = 4,012.50
Total tax = 892.50 + 4,012.50 = 4,905
Answer:
wgwagwagfwqgtfr3qt3qt432r43
Step-by-step explanation:
Answer:
43.20 for the total cost of the item
