1 .Fridays and Saturdays, she works 16 hours
16 x $9.70 = $155.20
answer: her gross pay in a week: $155.20
2. 7 hours on Wednesday, 6 hours on Thursday, and 9 hours on Friday. His gross pay for all three days was $196.90
7 + 6 + 9 = 22
196.90 / 22 = $8.95
answer: his hourly rate was $8.95
3.hourly rate of $11.28 per hour worked 46 hours last week. She works overtime for hours exceeding 40 hours in a week. She is paid overtime at a rate of time and a half. What was her gross pay last week?
$11.28 x 40 = $451.20 ---normal 40 hours paid
46 - 40 = 6 (6 hours over time)
11.28 + 11.28/ 2 = $16.92 (overtime pays a rate of time and a half)
6 * $16.92 = $101.52
$451.20(40hours) + $101.52(6hours overtime) = $552.72
answer: her gross pay last week was $552.72
4. Paul received a $15 tip on a meal that cost $120. What percent of the meal cost was the tip?
15 / 120 = .125
.125 * 100 = 12.5%
answer: percent of the meal cost was the tip was 12.5%
5. Carly earns a weekly salary of $720 plus 4% commission. Last week, she sold $3250 worth of products. What was her gross pay?
4% = .04
$3,250 * .04 = $130 (commission on sold products)
$720 + $ 130 = $850
answer: her gross pay was $850
The answers to the questions are:
1. The confidence level is 99 percent.
2. We have to conclude that there is no sufficient evidence available to support this claim because the Confidence interval contains 75 sec.
<h3>How to solve for the confidence level</h3>
1. The confidence level here should be
1- 0.01 = 0.99
= 99 percent
Given that, 99% confidence interval for population mean (μ) is (-34.1 sec u< u < 264.1 ) seconds.
We are to test the claim that the population mean is greater than 75 sec.
2.
The given confidence interval contains the value of 75 sec, so there is not sufficient evidence to support the claim that the mean is greater than 75 sec.
Read more on confidence interval here:
brainly.com/question/20309162
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