Write an equation that expresses the relationship. Then solve the equation for B. T varies directly as the cube of A and inversely as B.
T= what?
​(Use k as the constant of​ variation.)

Step-by-step explanation:

a. Predicted Amount = $109.46

b. Confidence Interval = (94.84,124.08)

c. Interval = (110.6883,188.8517)

Step-by-step explanation:

Given

ŷ = 17.49 + 1.0334x.

SSE = 1541.4

a.

ŷ = 17.49 + 1.0334(89)

ŷ = 109.4626

ŷ = 109.46 --- Approximated

Predicted Amount = $109.46

b.

ŷ = 17.49 + 1.0334(89)

ŷ = 109.4626

ŷ = 109.46

First we calculate the standard deviation

variance = SSE/(n-2)

v = 1541.4/(9-2)

v = 1541.4/7

v = 220.2

s = √v

s = √220.2

s = 14.839

Then we calculate mean(x) and ∑(x - (mean(x))²

X --- Y -- Mean(x) --- ∑(x - (mean(x))²

148-- 161 -- 43-- 1849

96|| 105|| -9 || 81

91||101 || -14 ||196

110 || 142|| 5 || 25

90 ||100 ||-15|| 225

102 ||||120||-3||9

136|| 167||31||961

90|| 140||-15||225

82|| 98||-23 ||529

Sum945 ||1134||0||4100

Mean (x) = 945/9 = 105

∑(x - (mean(x))² = 4100

α = 1 - 95% = 5%

α/2 = 2.5% = 0.025

tα,df = n − 2 = t0.025,7 =2.365

Confidence interval = 109.46 ± 2.365 * 14.839 √((1/9)+ (89-105)²/4100

Confidence Interval = (109.46 ± 14.62)

Confidence Interval = (94.84,124.08)

c.

ŷ = 17.49 + 1.0334(128)

ŷ = 149.7652

ŷ = 149.77

Interval = 149.77 ± 2.365 * 14.839 √((1/9)+ (128-105)²/4100

Interval = 149.77 ± 39.0817

Interval = (110.6883,188.8517)