From the given mean and margin of error, the 99% confidence interval for the mean amount of money spent on lunch per week for all students is:
[$19.5, $22.5].
<h3>How to calculate a confidence interval given the sample mean and the margin of error?</h3>
The confidence interval is given by the sample mean plus/minus the margin of error, hence:
- The lower bound is the sample mean subtracted by the margin of error.
- The upper bound is the sample mean added to the margin of error.
For this problem, we have that:
- The sample mean is of $21.
- The margin of error is of $1.50.
Hence the bounds are given as follows:
- Lower bound: 21 - 1.50 = $19.50.
- Upper bound: 21 + 1.50 = $22.50.
Hence the interval is [$19.50, $22.50].
More can be learned about confidence intervals at brainly.com/question/25890103
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(5x³-7)(2x²+1)
OPTION D is the correct answer
Answer: 0.25
Step-by-step explanation:
The relative frequency of the customers that buy computers is equal to the number of customers that bought a computer divided the total number of customers that entered the shop.
p = 25/100 = 0.25
If we take this as the probability, then the probability that the next customer that enters the shop buys a computer is 0.25 or 25%
Here are some words to make at least 20. Answer attached