Records of 40 used passenger cars and 40 used pickup trucks (none used commercially) were randomly selected to investigate whether there was any difference in the mean time in years that they were kept by the original owner before being sold. For cars the mean was 5.3 years with standard deviation 2.2 years. For pickup trucks the mean was 7.1 years with standard deviation 3.0 years. Construct the 95% confidence interval for the difference in the means based on these data

The 95% confidence interval for the difference in means = (-2.953, -0.647)

Step-by-step explanation:

The confidence interval for difference in means formula =

μ1 - μ2 ± z × √(σ²1/n1 + σ²2/n2)

Where

σ = standard deviation

n = number of samples

μ = mean

Passenger cars = 1

Pick up trucks = 2

μ1 =5.3 , σ1 =2.2 , n2 = 40

μ1 =7.1, σ1 =3.0 , n2 = 40

z = 95 % confidence interval = 1.96

μ1 - μ2 ± z × √(σ²1/n1 + σ²2/n2)

= 5.3 - 7.1 + 1.96 × √2.2²/40 + 3.0²/40

= -1.8 ± 1.96× √0.121 + 0.225

= -1.8 ± 1.96 ×√0.346

= -1.8 ± 1.96 × 0.5882176468

Confidence interval = -1.8 ± 1.1529065877

= -1.8 - 1.1529065877

= -2.9529065877

≈ -2.953

= -1.8 + 1.1529065877

= -0.6470934123

≈ -0.647

Therefore, the 95% confidence interval for the difference in means = (-2.953, -0.647)

g(x) = 3*(8(3^(x−2))+2)

g(x) = 24 (3^(x-2) )+6

the factor is 2

step-by-step explanation:

let f(x)=8(3^(x−2)+2)

to stretch a graph by 3, it would be 3f(x)

g(x) =3f(x)

g(x) = 3*(8(3^(x−2))+2)

g(x) = 24 (3^(x-2) )+6

x     −2     0     2     4     6

f(x) 1/64 1/16   1/4     1     4

1/64 * z * z = 1/16

i multiplies by z twice because it its 2 steps from -2 to 0

1/64 z^2 = 1/16

multiply by 64 to clear the fraction on the left side

1/4 * 64 z^2 =1/16 *64

z^2 = 4

take the square root of each side

sqrt(z^2) = ±sqrt(4)

z = ±2  

we are multiplying by 2 each time.