You are doing marketing research about yogurt consumption. Based on your previous research, you feel sure that the population standard deviation of the number of yogurts consumed per year is 6. In other words, σ is known and σ = 6. You take a sample of 50 consumers and the sample mean is 32 yogurts consumed per year. In other words, = 32. You want to develop a confidence interval for μ푥(the population mean) such that you are 90% confident that the true value of μ lies within the interval.1. What is the value of ?훼

Step-by-step explanation:

Confidence interval is written in the form,

(Sample mean - margin of error, sample mean + margin of error)

The sample mean, x is the point estimate for the population mean.

Since the sample size is large and the population standard deviation is known, we would use the following formula and determine the z score from the normal distribution table.

Margin of error = z × σ/√n

Where

σ = population standard Deviation

n = number of samples

From the information given

1) x = 32

σ = 6

n = 50

To determine the z score, we subtract the confidence level from 100% to get α

α = 1 - 0.90 = 0.1

α/2 = 0.1/2 = 0.05

This is the area in each tail. Since we want the area in the middle, it becomes

1 - 0.05 = 0.95

The z score corresponding to the area on the z table is 1.645. Thus, confidence level of 90% is 1.645

Margin of error = 1.645 × 6/√50 = 1.4

Confidence interval = 32 ± 1.4

The lower end of the confidence interval is

32 - 1.4 = 30.6

The upper end of the confidence interval is

32 + 1.4 = 33.4

there are many solutions.

step-by-step explanation:

step 1: simplify both sides of the equation.

4x+6=4x^2+10x−4

step 2: subtract 4x^2+10x-4 from both sides.

4x+6−(4x^2+10x−4)=4x^2+10x−4−(4x^2+10x−4)

−4x^2−6x+10=0

step 3: factor left side of equation.

−2(2x+5)(x−1)=0

step 4: set factors equal to 0.

x=-5/2 or x=1