margin

A companys cost function is C(x) = 16x^2 + 900 dollars, where x is the number of units. Find the marginal cost function. Marginal Cost is the derivative of the Cost function. [B]C'(x) = 32x[/B]

a confidence interval for a population mean has a margin of error of 0.081

Margin of error = Interval Size/2 0.081 = Interval Size/2 Cross Multiply: Interval Size = 0.081 * 2 Interval Size = [B]0.162[/B]

A random sample of 149 students has a test score average of 61 with a standard deviation of 10. Find the margin of error if the confidence level is 0.99. (Round answer to two decimal places) Using our [URL='https://www.mathcelebrity.com/normconf.php?n=149&xbar=61&stdev=10&conf=99&rdig=4&pl=Large+Sample']confidence interval of the mean calculator[/URL], we get [B]58.89 < u < 63.11[/B]

A sample of 71 college students yields a mean annual income of $3595. Assuming that ? = $898, find the margin of error in estimating µ at the 99% level of confidence

A sample of 71 college students yields a mean annual income of $3595. Assuming that ? = $898, find the margin of error in estimating µ at the 99% level of confidence

based on a sample of size 41, a 95% confidence interval for the mean score of all students,µ, on an aptitude test is from 60 to 66. Find the margin of error

This bid ask calculator takes a bid amount and ask amount and calculates the spread and the margin percent

Given N, n, and a confidence percentage, this will calculate the estimation of confidence interval for the population proportion π including the margin of error. confidence interval of the population proportion

Given inputs of sales, fixed costs, variable costs, depreciation, and taxes, this will determine EBIT and Net Income and Profit Margin

The monthly earnings of a group of business students are are normally distributed with a standard deviation of 545 dollars. A researcher wants to estimate the mean monthly earnings of all business students. Find the sample size needed to have a confidence level of 95% and a margin of error of 128 dollars.

fixed cost 500 marginal cost 8 item sells for 30. Set up Cost Function C(x) where x is the number of items sold: C(x) = Marginal Cost * x + Fixed Cost C(x) = 8x + 500 Set up Revenue Function R(x) where x is the number of items sold: R(x) = Revenue per item * items sold R(x) = 30x Set up break even function (Cost Equals Revenue) C(x) = R(x) 8x + 500 = 30x Subtract 8x from each side: 22x = 500 Divide each side by 22: x = 22.727272 ~ 23 units for breakeven

Foster is centering a photo that is 9/1/2 inches wide on a scrapbook pages that is 10 inches wide. How far from each side of the pages should he put the picture? Enter your answer as a mixed number. First, determine your margins, which is the difference between the width and photo width, divided by 2. 10 - 9 & 1/2 = 1/2 1/2 / 2 = [B]1/4[/B]

Given a confidence interval, this determines the margin of error and sample mean.

Marginal propensity to save is 0.3. Calculate MPC. MPC is Marginal Propensity to Consume. And MPS is Marginan Propensity to Save. The relational equation between the two is: MPC + MPS = 1 To get MPC, we have: MPC = 1 - MPS The problem gives us MPS = 0.3, so plug it into this modified MPC equation: MPC = 1 - 0.3 MPC = [B]0.7[/B]

Given an upper bound and a lower bound and a sample size, this calculate the point estimate, margin of error.

Given a cost and a gross margin percentage, this calculator calculates price, gross profit, markup percentage

Using the Profit Equation with inputs (Revenue-Cost-Profit-Tax), this determines the relevant output including gross proft, gross profit margin, net profit, and net profit margin.

based on a sample of size 41, a 95% confidence interval for the mean score of all students,µ, on an aptitude test is from 60 to 66. Find the margin of error

A sample of 71 college students yields a mean annual income of $3595. Assuming that ? = $898, find the margin of error in estimating µ at the 99% level of confidence

[URL]http://www.mathcelebrity.com/marginoferror.php?num=60%2C66&pl=Calculate+Margin+of+Error+and+Sample+Mean[/URL]

a confidence interval for a population mean has a margin of error of 0.081. Determine the length of the confidence interval

standard deviation of 545 dollars. Find the sample size needed to have a confidence level of 95% and margin of error 128

Standard Error (margin of Error) = Standard Deviation / sqrt(n) 128 = 545/sqrt(n) Cross multiply: 128sqrt(n) = 545 Divide by 128 sqrt(n) = 4.2578125 Square both sides: [B]n = 18.1289672852 But we need an integer, so the answer is 19[/B]

The brand manager for a brand of toothpaste must plan a campaign designed to increase brand recognition. He wants to first determine the percentage of adults who have heard of the bran. How many adults must he survey in order to be 90% confident that his estimate is within seven percentage points of the true population percentage? [IMG]https://ci5.googleusercontent.com/proxy/kc6cjrLvUq64guMaArhfiSR0mOnTrBwB9iFM9u9VaZ5YYn86CSDWXr1FNyqxylwytHdbQ3iYsUDnavt-zvt-OK0=s0-d-e1-ft#http://latex.codecogs.com/gif.latex?%5Chat%20p[/IMG] = 0.5 1 - [IMG]https://ci5.googleusercontent.com/proxy/kc6cjrLvUq64guMaArhfiSR0mOnTrBwB9iFM9u9VaZ5YYn86CSDWXr1FNyqxylwytHdbQ3iYsUDnavt-zvt-OK0=s0-d-e1-ft#http://latex.codecogs.com/gif.latex?%5Chat%20p[/IMG] = 0.5 margin of error (E) = 0.07 At 90% confidence level the t is, alpha = 1 - 90% alpha = 1 - 0.90 alpha = 0.10 alpha / 2 = 0.10 / 2 = 0.05 Zalpha/2 = Z0.05 = 1.645 sample size = n = (Z[IMG]https://ci4.googleusercontent.com/proxy/mwhpkw3aM19oMNA4tbO_0OdMXEHt9juW214BnNpz4kjXubiVJgwolO7CLbmWXXoSVjDPE_T0CGeUxNungBjN=s0-d-e1-ft#http://latex.codecogs.com/gif.latex?%5Calpha[/IMG] / 2 / E )2 * [IMG]https://ci5.googleusercontent.com/proxy/kc6cjrLvUq64guMaArhfiSR0mOnTrBwB9iFM9u9VaZ5YYn86CSDWXr1FNyqxylwytHdbQ3iYsUDnavt-zvt-OK0=s0-d-e1-ft#http://latex.codecogs.com/gif.latex?%5Chat%20p[/IMG] * (1 - [IMG]https://ci5.googleusercontent.com/proxy/kc6cjrLvUq64guMaArhfiSR0mOnTrBwB9iFM9u9VaZ5YYn86CSDWXr1FNyqxylwytHdbQ3iYsUDnavt-zvt-OK0=s0-d-e1-ft#http://latex.codecogs.com/gif.latex?%5Chat%20p[/IMG] ) = (1.645 / 0.07)^2 *0.5*0.5 23.5^2 * 0.5 * 0.5 552.25 * 0.5 * 0.5 = 138.06 [B]sample size = 138[/B] [I]He must survey 138 adults in order to be 90% confident that his estimate is within seven percentage points of the true population percentage.[/I]

The monthly earnings of a group of business students are are normally distributed with a standard deviation of 545 dollars. A researcher wants to estimate the mean monthly earnings of all business students. Find the sample size needed to have a confidence level of 95% and a margin of error of 128 dollars.

The monthly earnings of a group of business students are are normally distributed with a standard deviation of 545 dollars. A researcher wants to estimate the mean monthly earnings of all business students. Find the sample size needed to have a confidence level of 95% and a margin of error of 128 dollars.