Spring break can be a very expensive holiday. A sample of 80 students is surveyed, and the average amount spent by students on travel and beverages is $593.84. The sample standard deviation is approximately $369.34. Is $521.58 ≤μ≤ $666.10 correct?

$521.58 < μ < $666.1

Step-by-step explanation:

Spring break can be a very expensive holiday. A sample of 80 students is surveyed, and the average amount spent by students on travel and beverages is $593.84 at 92% confidence level. The sample standard deviation is approximately $369.34. Is $521.58 ≤μ≤ $666.10 correct?

Given that:

number of samples (n) = 80 students, mean (μ) = $593.84, standard deviation (σ) = $369.34, confidence level (c) = 92% = 0.92.

α = 1 - c = 1 - 0.92 = 0.08

the z score of 0.46 (0.5 - 0.04) is the same as the z score of 0.04. This is gotten from the  Normal Distribution Table.

Therefore,

The margin of error (e) is given as:

The confidence interval = (μ - e, μ + e) = ($593.84 - $72.26, $593.84 + $72.26) = ($521.58, $666.1)

The confidence interval is $521.58 < μ < $666.1

answer: first option i.e. 'a' is correct.

step-by-step explanation:

since we have given that

number of minutes to wrap a small gift box = 4 minutes

we need to find the number of gifts to be wrapped in 5 minutes

since there is a direct variation, as the number of minutes will increase with   number of gifts to be wrapped.

so, let the number of gifts to be wrapped in 5 minutes be x.

so, it becomes,

hence, first option 'a'   is correct.

2+2 obliviously is 4.