The high temperatures (in degrees Fahrenheit) of a random sample of 5 small towns are: 99.6 99.7 97.9 98.6 97.7 Assume high temperatures are normally distributed. Based on this data, find the 99% confidence interval of the mean high temperature of towns. Enter your answer as an open-interval (i. e., parentheses) accurate to two decimal places (because the sample data are reported accurate to one decimal place).

(97.63, 99.77)

Step-by-step explanation:

Given the data:

99.6 99.7 97.9 98.6 97.7

Using calculator, we can obtain the mean and standard deviation of the sample data:

Mean(m) = 98.7

Standard deviation = 0.93

Sample size (n) = 5

Using the relation to find confidence interval :

Mean ± Zcrit * (s/√n)

Zcrit at 99% = 2.576

98.7 ± 2.576 * (0.93 / √5)

Lower limit : 98.7 - (2.576 * 0.4159086) = 97.6286194464 = 97.63 ( 1 decimal place)

Upper limit : 98.7 + (2.576 * 0.4159086) = 99.7713805536 = 99.77 ( 1 decimal place)

(97.6, 99.8)

the answer is "add 2y and y"

step-by-step explanation:

you need to collect the like terms first.