Anew york restaurant would like to estimate the average amount customers spend for dinner. a sample of size 49 customers resulted an average of $32.60. assume the population standard deviation is $5.what is the point estimation for the mean of the population? at 90% level of confidence, what is the margin of error? what is the 90% confidence level for the mean of the population?

Point of estimation = 32.60

Margin of error = 1.175.

95% confidence interval would be (81.95,84.45).

Step-by-step explanation:

Since we have given that

n = 49

Average = $32.60

Standard deviation = $5

We need to find:

1) Point of estimation for the mean of the population:

2) At 90% level of confidence, z = 1.645

Margin of error would be

So, Margin of error = 1.175.

3) At 90% level of confidence, interval would be

Hence, 90% confidence interval would be (31.43,33.76).

the figures below are made out of circles, semicircles, quarter … find the area and the perimeter of each figure and give your answers as a completely simplified exact value in terms of π (no approximations). remember that we'll also have to cut whatever areas we get in half, since we're only dealing

step-by-step explanation:

let x be the money she spent so that she can earn the free t-shirt.

since, she already spent $112.49

⇒

also, as per the statement: jennifer must spend $250 or more in order to qualify for a free t-shirt at the store.

then,

we have an inequality i.,e

subtract 112.49 from both sides we get;

therefore, $137. 51 more money does jennifer need to spend so that she can earn the free t-shirt