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6-3 Homework Help MAT 133 | Assignment Solutions 2023
Exam (elaborations) • 10
pages
• 2023
Available in package deal
6-3 Homework Help MAT 133 | Assignment Solutions 2023.
Joan’s finishing time for the Bolder Boulder 10K race was 1.75
standard deviations faster than the women’s average for her age
group. There were 405 women who ran in her age group. Assuming a
normal distribution, how many women ran faster than Joan?
We need to figure out what percent (technically, proportion) of women had
a shorter run time. Then we multiply that by 405 to figure out the number of
women that were faster.
If Joan...
6-3 Homework Help MAT 133 | Assignment Solutions 2023
Last document update:
ago
6-3 Homework Help MAT 133 | Assignment Solutions 2023.
Joan’s finishing time for the Bolder Boulder 10K race was 1.75
standard deviations faster than the women’s average for her age
group. There were 405 women who ran in her age group. Assuming a
normal distribution, how many women ran faster than Joan?
We need to figure out what percent (technically, proportion) of women had
a shorter run time. Then we multiply that by 405 to figure out the number of
women that were faster.
If Joan...
4-3 Homework Help MAT 133-J4254 | Questions with Solutions
Exam (elaborations) • 10
pages
• 2023
Available in package deal
4-3 Homework Help MAT 133-J4254 | Questions with Solutions.
Suppose the probability of an IRS audit is 1.4 percent for U.S. taxpayers who file form 1040
and who earned $100,000 or more.
(a) What are the odds that such a taxpayer will be audited?
P(A) = P(the taxpayer will be audited): 1.4% → .014
P(A’) = P(the taxpayer won’t be audited) = 1 - .014 = .986
b) What are the odds against such a taxpayer being audited?
Here is a video you can watch for Problem 2:
Homework Problem 3
Reca...
4-3 Homework Help MAT 133-J4254 | Questions with Solutions
Last document update:
ago
4-3 Homework Help MAT 133-J4254 | Questions with Solutions.
Suppose the probability of an IRS audit is 1.4 percent for U.S. taxpayers who file form 1040
and who earned $100,000 or more.
(a) What are the odds that such a taxpayer will be audited?
P(A) = P(the taxpayer will be audited): 1.4% → .014
P(A’) = P(the taxpayer won’t be audited) = 1 - .014 = .986
b) What are the odds against such a taxpayer being audited?
Here is a video you can watch for Problem 2:
Homework Problem 3
Reca...
