Assume that all​ grade-point averages are to be standardized on a scale between 0 and 6. How many​ grade-point averages must be obtained so that the sample mean is within 0.013 of the population​ mean? Assume that a ​98% confidence level is desired. If using the range rule of​ thumb, σ can be estimated as range/4=(6-0)/2=1.5. Does the sample size seem​ practical?

The sample size 'n' = 72,030

Step-by-step explanation:

Step(i):-

Given that the Estimate Error = 0.013

Given that the standard deviation of the Population = 1.5

The estimated error is determined by

Step(ii):-

Given that the Level of significance = 0.98 or 0.02

Z₀.₀₂ =   2.326

The estimated error is determined by

Squaring on both sides, we get

    n =  72,030

Final -

The sample size 'n' = 72,030

answer: the price of one pound cabbage = $ 0.8

the point (10,8) represents that the cost of 10 pounds cabbage is $ 8

step-by-step explanation:

since, here is a proportional relation between the number of pounds of cabbage (x) and the price in dollars (y).

therefore, we can write,

⇒ y = k x

where k is the constant of proportionality.

now, point (5,4) is in the line.

thus, it will satisfies the above equation of line.

⇒ 4 = 5 k

⇒ k = 0.8

thus, the equation of the line is,

y = 0.8 x

a) for x = 1,

y = 0.8

therefore, the cost of one pound of cabbage is 0.8 dollars.

b) in the below graph (10,8) represents that the cost of 10 pounds of cabbage is 8 dollars.

1/3 + 1/5 = 8/15.

step-by-step explanation:

hope it