Use the t-distribution to find a confidence interval for a mean mu given the relevant sample results. Give the best point estimate for mu, the margin of error, and the confidence interval. Assume the results come from a random sample from a population that is approximately normally distributed. A 95% confidence interval for mu using the sample results x-bar equals 76.4, s = 8.6, and n = 42. Point estimate = ?
Margin of error = ?

Point estimate = 76.4

Margin of Error = 2.680

Step-by-step explanation:

Given that distribution is approximately normal;

The point estimate = sample mean, xbar = 76.4

The margin of error = Zcritical * s/√n

Tcritical at 95%, df = 42 - 1 = 41

Tcritical(0.05, 41) = 2.0195

Margin of Error = 2.0195 * (8.6/√42)

Margin of Error = 2.0195 * 1.327

Margin of Error = 2.67989

Margin of Error = 2.680

d, x=8

step-by-step explanation:

simplifying

4x + -5 = 2x + 11

reorder the terms:

-5 + 4x = 2x + 11

reorder the terms:

-5 + 4x = 11 + 2x

solving

-5 + 4x = 11 + 2x

solving for variable 'x'.

move all terms containing x to the left, all other terms to the right.

add '-2x' to each side of the equation.

-5 + 4x + -2x = 11 + 2x + -2x

combine like terms: 4x + -2x = 2x

-5 + 2x = 11 + 2x + -2x

combine like terms: 2x + -2x = 0

-5 + 2x = 11 + 0

-5 + 2x = 11

add '5' to each side of the equation.

-5 + 5 + 2x = 11 + 5

combine like terms: -5 + 5 = 0

0 + 2x = 11 + 5

2x = 11 + 5

combine like terms: 11 + 5 = 16

2x = 16

divide each side by '2'.