Answer:
Step-by-step explanation:
We want to determine a 95% confidence interval for the mean salary of all graduates from the English department.
Number of sample, n = 400
Mean, u = $25,000
Standard deviation, s = $2,500
For a confidence level of 95%, the corresponding z value is 1.96. This is determined from the normal distribution table.
We will apply the formula
Confidence interval
= mean ± z × standard deviation/√n
It becomes
25000 ± 1.96 × 2500/√400
= 25000 ± 1.96 × 125
= 25000 ± 245
The lower end of the confidence interval is 25000 - 245 =24755
The upper end of the confidence interval is 25000 + 245 = 25245
Therefore, with 95% confidence interval, the mean salary of all graduates from the English department is between $24755 and $25245
<span>.99(15) = 1.35x
14.85 = 1.35x
14.85/1.35 = 11
I think it is 15. I am not sure about the proper way to solve this....I just know that if he sold 15 brownies, he could buy 11 packs with exactly no change left over.</span>
Answer: 1.76
1.54
Step-by-step explanation:
we have to divide 800 by 453.592, and the result is 1.763699... rounding it: 1.76
now we do the same for 700 :
700 divided by 453.592 is 1.5432... so rounding it down: 1.54
It would be D………………………….,
