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 $25245
D. 16 clusters
Hope it helps
Consider the number 38,288
Now, we have to round this number to the nearest hundreds place.
The place value to the extreme right of the number is ones, then tens, hundreds, thousands, ten thousands and so on.
So, the digit at hundreds place = 2
Consider the digit to the right of the hundred place which is '8', which is greater than 5.
Therefore, we will add '1' to the digit at hundreds place.
Therefore, the number 38,288 rounded to nearest hundreds is 38,300.
If Melanie get's 4.10$, then Jacob get's 20.50$.
We can divide to get the unit rate:
20.50 ÷ 4.10 = 5
This means that if Jacob gets 5$, then Melanie will get 1$.
So in this case, Melanie get's 1$.
