In a certain population, body weights are normally distributed with a mean of 152 pounds 30) and a standard deviation of 26 pounds. how many people must be surveyed if we want to estimate the percentage who weigh more than 180 pounds? assume that we want 96% confidence that the error is no more than 4 percentage points.

We are given the following variables:

μ = the sample mean = 152 pounds

σ = the standard deviation = 26 pounds

x = the sample value we want to test = 180 pounds

n = the sample size = unknown

MOE = margin of error = 4% = 0.04

Confidence level = 96%

The first thing we can do is to find for the value of z using the formula:

z = (x – μ) / σ

z = (180 – 152) / 26

z = 1.0769 = 1.08

Since we are looking for the people who weigh more than 180 pounds, therefore this is a right tailed z test. The p value is:

p = 0.1401

Then we can use the formula below to solve for n:

n = z^2 * p * (1 – p) / (MOE)^2

n = 1.08^2 * 0.1401 * (1 – 0.1401) / (0.04)^2

n = 87.82 = 88

Therefore around 88 people must be surveyed.

he used the algebra tiles correctly.

step-by-step explanation:

given : 3x(2x-1)

solution :

since he used algebra tiles .

armando's solution is the same in the solution shown in figure

as he distributed 2x into x and x and wrote them vertically

and also write   -1 vertically

and 3x into x and x and x and wrote it horizontally

performing the multiplication in the algebra tiles we get

and solving the given expression :

3x(2x-1)

since solving the expression using tiles   and solving the expression algebraically we get the same answer

thus   we can choose the first statement as the answer to this problem which is "he used the algebra tiles correctly".

hello, never fear papaguy is here!

c. histogram

step-by-step explanation: