The weekly earnings of students in one age group are normally distributed with a standard deviation of $110. A researcher wishes to estimate the mean weekly earnings of students in this age group. Find the sample size needed to assure with 99 percent confidence that the sample mean will not differ from the population mean by more than $10.

806

Step-by-step explanation:

Given that:

Standard deviation (s) = $110

Error = $10

Zcritical ; Zα/2 at 99% confidence interval = 2.58

Using the relation :

[(Zα/2 * s) / sqrt(n)] < 10

[(2.58 * 110) /sqrt(n)] = 10

283.8 / sqrt(n) = 10

283.8 = 10 * sqrt(n)

Sqrt(n) = 283.8/ 10

Sqrt(n) = 28.38

Square both sides

n = 28.38^2

n = 805.4244

n = 806

Hence, sample size = 806

yes because a quadrilateral is equal to 4 sides and a rhombus has 4 sides so technacly it is a quadrilateral.

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