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Matthew and Ashley went a candy store, but for some reason wanted to purchase nuts. They could not decide among peanuts, cashews, or almonds. They then decided to create a mix of all three! They bought 2.5 pounds of peanuts for $1.30 per pound, 4 pounds of cashews for $4.50 per pound, and 2 pounds of almonds for $3.75 per pound. Determine the price per pound of the mix. (Hint: use the weighted mean formula to solve correctly here.)

Matthew and Ashley went a candy store, but for some reason wanted to purchase nuts. They could not decide among peanuts, cashews, or almonds. They then decided to create a mix of all three! They bought 2.5 pounds of peanuts for $1.30 per pound, 4 pounds of cashews for $4.50 per pound, and 2 pounds of almonds for $3.75 per pound. Determine the price per pound of the mix. (Hint: use the weighted mean formula to solve correctly here.)

MATLAB: An Introduction with Applications
MATLAB: An Introduction with Applications
6th Edition
ISBN:9781119256830
Author:Amos Gilat
Publisher:Amos Gilat
Chapter1: Starting With Matlab
Section: Chapter Questions
Problem 1P
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Matthew and Ashley went a candy store, but for some reason wanted to purchase nuts. They could not decide among peanuts, cashews, or almonds. They then decided to create a mix of all three! They bought 2.5 pounds of peanuts for $1.30 per pound, 4 pounds of cashews for $4.50 per pound, and 2 pounds of almonds for $3.75 per pound. Determine the price per pound of the mix. (Hint: use the weighted mean formula to solve correctly here.)
Transcribed Image Text:Matthew and Ashley went a candy store, but for some reason wanted to purchase nuts. They could not decide among peanuts, cashews, or almonds. They then decided to create a mix of all three! They bought 2.5 pounds of peanuts for $1.30 per pound, 4 pounds of cashews for $4.50 per pound, and 2 pounds of almonds for $3.75 per pound. Determine the price per pound of the mix. (Hint: use the weighted mean formula to solve correctly here.)
For a normal distribution, approximately deviation of the mean, and approximately percent of the data lie within one standard percent of the data lie within two standard deviations of the mean.
Transcribed Image Text:For a normal distribution, approximately deviation of the mean, and approximately percent of the data lie within one standard percent of the data lie within two standard deviations of the mean.
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