Suppose that a fast food restaurant decides to survey its customers to gauge interest in a breakfast menu. After surveying multiple people, the restaurant created a 95% confidence interval for the proportion of customers interested in a breakfast menu. The confidence interval is ( 0.349 , 0.433 ) . Use the confidence interval to find the point estimate and margin of error for the proportion. Give your answer precise to three decimal places.

The point estimate is 0.391 and the margin of error is 0.042.

Step-by-step explanation:

A confidence interval has two bounds, a lower bound and an upper bound.

A confidence interval is symmetric, which means that the point estimate used is the mid point between these two bounds, that is, the mean of the two bounds.

The margin of error is the absolute value of the difference between the point estimate and any of the two bounds.

In this problem, we have that:

Lower bound: 0.349

Upper bound: 0.433

Point estimate:

(0.349+0.433)/2 = 0.391

Margin of error:

|0.391 - 0.433| = |0.391 - 0.349| = 0.042

The point estimate is 0.391 and the margin of error is 0.042.

.5 lb

step-by-step explanation:

12 in / 6 in =2 in

.25 x 2 = .5 lb

i am currently in the exact same situation rn. algebra2/adv algebra is killing

step-by-step explanation: