Question 1
Total budget = 180 + 475 + 15 + 50 + 65 + 25 + 150 + 30 = $990
Real spending = 182 + 475 + 12 + 65 + 68 + 12.50 + 150 + 36 = $1000.5
Mae Green didn't stay within the allocated budget
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Question 2
Eleanor:
Income = 380.48
Spending = 16.50
Peter:
Income = 120 + 13.65 + 100 = 233.65
Total income = 233.65 + 380.48 - 16.50 = 597.63
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Question 3
Total spent = 540 + 48.55 + 34.15 + 12.80 + 18.95 + 38.60 + 2 + 6.50 = 701.55
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Question 4
Marie's new balance = 250.65 - [21.95+48.50+75.60] + 55 = $159.50
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Question 5
[235/825] × 100 = 28.48%
Answer:
37.23% probability that randomly selected homework will require between 8 and 12 minutes to grade
Step-by-step explanation:
Problems of normally distributed samples are solved using the z-score formula.
In a set with mean
and standard deviation
, the zscore of a measure X is given by:

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.
In this problem, we have that:

What is the probability that randomly selected homework will require between 8 and 12 minutes to grade?
This is the pvalue of Z when X = 12 subtracted by the pvalue of Z when X = 8. So
X = 12



has a pvalue of 0.4052
X = 8



has a pvalue of 0.0329
0.4052 - 0.0329 = 0.3723
37.23% probability that randomly selected homework will require between 8 and 12 minutes to grade
Given that mean=56.1 and standard deviation=8.2, P(x>67.5) will be found as follows:
The z-score is given by:
z=(x-μ)/σ
thus the z-score will be given by:
z=(67.5-56.1)/8.2
z=11.4/8.2
z=1.39
thus
P(z=1.39)=0.9177
thus:
P(x>67.5)=1-P(z>0.9177)
=1-0.9177
=0.0823
Answer: A. 0.0823
the hundreds place
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