Supposethat z is a normally distributed variable with variance 4. You collect a sample of size n for which you get a sample average value of 3. You are asked to test the null hypothesis that the mean is smaller than 2. Find the sample size n above which you can reject the null hypothesiswith a 95% of confidence.

The sample size is 4

Step-by-step explanation:

Null hypothesis: The mean is 2

Alternate hypothesis: The mean is less than 2

Mean = 3

sd = sqrt(variance) = sqrt(4) = 2

At 95% confidence level, t-value is 1.960

Assuming the lower bound of the mean is 1.04

Lower bound = mean - (t×sd/√n)

1.04 = 3 - (1.96×2/√n)

3.92/√n = 3 - 1.04

3.92/√n = 1.96

√n = 3.92/1.96

√n = 2

n = 2^2

n = 4

maximum area = xy = (16)(32) = 512 ft2

to maximize the area of the garden, the side of the fence parallel to the side of the house should be 32 ft long, and the other two sides should both have length 16 ft.'

source

b: |12 – x| ≤ 0.25

step-by-step explanation:      

let x be the possible lengths in inches, of the ribbons jaime cut.

we have been given that jaime is cutting strips of ribbon for a project. he tries to cut each piece to be 12 inches long, but sometimes he cuts it too long or too short by as much as 0.25 inch.  

we can represent this information in an inequality as:

by the definition of absolute value function we will get,

upon combining our inequalities we will get,

therefore, the inequality   describes all the possible lengths (x), in inches, of the ribbons jaime cut.

step-by-step explanation: