A supervisor records the repair cost for 17 randomly selected TVs. A sample mean of $76.76 and standard deviation of $24.02 are subsequently computed. Determine the 99% confidence interval for the mean repair cost for the TVs. Assume the population is approximately normal.

Step-by-step explanation:

We want to determine a 99% confidence interval for the mean repair cost for the TVs

Number of sample, n = 17

Mean, u = $76.76

Standard deviation, s = $24.02

For a confidence level of 99%, the corresponding z value is 2.58. This is determined from the normal distribution table.

We will apply the formula

Confidence interval

= mean ± z ×standard deviation/√n

It becomes

76.76 ± 2.58 × 24.02/√17

= 76.76 ± 2.58 × 5.83

= 76.76 ± 15.03

The lower end of the confidence interval is 76.76 - 15.03 = $61.73

The upper end of the confidence interval is 76.76 + 15.03 = $91.79

Therefore, with 99% confidence interval, the mean repair cost for the TVs is between $61.73 and $91.79

b.) 160 miles on a 5 gallons car

step-by-step explanation:

240/10=24 miles/gallon

160/5=32 miles/gallon

30/1=30 miles/gallon

168/6= 28 miles/gallon

32 miles/gallon is the most amount, so this is the best.

so firstly, we need to isolate the y variables to be able to solve these inequalities. firstly, add 0.5x on both sides of the first inequality and subtract 1.5x on both sides of the second inequality:

now since the slope is positive for the first inequality, this means that the line going upwards belongs to the first inequality, and the line going downwards belongs to the second inequality.

next, since y ≥ in the first inequality, this means that the shaded region will be above the first inequality's line, thus shading regions a and b.

next, since y ≤ in the second inequality, this means that the shaded region is going to be below this line, thus shading regions a and c.

since both lines shade region a, this means that region a is the solution.

step-by-step explanation:

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