Answer:
Step-by-step explanation:
Paula, a college student, earned one summer = $3,600
We have to show the amount in fraction that Paula spent during the following semester,
she spent for tuition = $2,500 = × 100
=0.6944 × 100
= 69.44% or
She spent for rent = $650 = × 100
= 0.1806 × 100
= 18.06% or
Paula spent for food = $430 = × 100
=0.1194 × 100
= 11.94% or
She Spent in tuition, rent and food = 2,500 + 650 + 430 = $3580
Now she spent in miscellaneous expenses = 3600 - 3580 = $20.00 × 100
= 0.056 × 100
= 0.56% or
Answer: The required equation is,
Step-by-step explanation:
Let x be the size of the larger bottle,
Since, the size of smaller bottle of lotion = 5.4 ounces,
According to the question,
The larger bottle gives 20% more free lotion,
The size of the larger bottle = The size of smaller bottle of lotion + 20 % of the size of smaller bottle of lotion
= 120 % of the size of smaller bottle of lotion
= 120 % of 5.4
Which is the required equation that could be used to find the size of the larger bottle.
Answer:
No, at the 0.05 significance level, the number of units produced on the night shift is not larger.
Step-by-step explanation:
We are given that the mean number of units produced by a sample of 54 day-shift workers was 345. The mean number of units produced by a sample of 60 night-shift workers was 351.
Assume the population standard deviation of the number of units produced on the day shift is 21 and 28 on the night shift.
Let = population mean number of units produced on the day shift
= population mean number of units produced on the night shift
So, <u>Null Hypothesis</u>, :
or
{means that the mean number of units produced on the night shift is same or lesser on the day shift}
<u>Alternate Hypothesis,</u> :
or
{means that the mean number of units produced on the night shift is larger}
The test statistics that will be used here is <u>Two-sample z test statistics</u> as we know about population standard deviations;
T.S. = ~ N(0,1)
where, = sample mean number of units produced by a sample of 54 day-shift workers = 345
= sample mean number of units produced by a sample of 60 night-shift workers = 351
= population standard deviation of the number of units produced on the day shift = 21
= population standard deviation of the number of units produced on the day shift = 28
= sample of day-shift workers = 54
= sample of night-shift workers = 60
So, <em><u>test statistics</u></em> =
= -1.302
Now at 0.05 significance level, the z table gives critical value of -1.6449 for left-tailed test. Since our test statistics is more than the critical value of z as -1.302 > -1.6449 so we have insufficient evidence to reject our null hypothesis as it will not fall in the rejection region.
Therefore, we conclude that the mean number of units produced on the night shift is same or lesser than those produced on the day shift.