Assume that a random sample is used to estimate a population proportion p. find the margin of error e that corresponds to the given statistics and confidence level. nbsp 90 % confidence; the sample size is 3200 comma of which 15 % are successes 90% confidence; the sample size is 3200, of which 15% are successes the margin of error eequals=nothing. (round to four decimal places as needed.)

The margin of error E = 0.0104

Step-by-step explanation:

The formula to find the margin of error that corresponds to the given statistics and confidence level for population proportion is given by :-

, where

n= Sample size

= Sample proportion

z* = critical value.

As per given , we have

n= 3200

Confidence level : 90%

The critical z-value for 90% confidence is z* = =1.645[By z-table]

Substitute all values in the formula , we get

Hence, the margin of error E = 0.0104

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.

homozygous. one is homozygous, while two is heterozygous.