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)

answer is 8x+24

area of vanessa's patio = length × width

given, length = (3x² + 5x + 10)

width = (x² - 3x - 1)

area of vanessa's patio = (3x² + 5x + 10)(x² - 3x - 1)

= 3x²(x² - 3x - 1) + 5x(x² - 3x - 1) + 10(x² - 3x - 1)

= 3x⁴ - 9x³ - 3x² + 5x³ - 15x² - 5x + 10x² - 30x - 10

= 3x⁴ - 4x³ - 8x² - 35x - 10

hence, expression of vanessa's patio = (3x⁴ - 4x³ - 8x² - 35x - 10)

i’m sorry but i don’t know sorry