SOLUTION: I started at x(2.6) + 4y (4*4.4) =4444
That is not going to work to get to how many standard version downloads there were-Help- i am totally stuck!
A Web music store offers two
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Question 533905: I started at x(2.6) + 4y (4*4.4) =4444
That is not going to work to get to how many standard version downloads there were-Help- i am totally stuck!
A Web music store offers two versions of a popular song. The size of the standard version is 2.6 megabytes (MB). The size of the high-quality version is 4.4 MB. Yesterday, the high-quality version was downloaded four times as often as the standard version. The total size downloaded for the two versions was 4444 MB. How many downloads of the standard version were there?
Answer by solver91311(24713) (Show Source): You can put this solution on YOUR website!
That is because you are guessing and you are trying to make this too difficult. Let 
represent the number of standard version downloads, just like you did. Then the number of megabytes downloaded for standard version downloads was 
, just like you have it. But then you chose to add the variable 
into the mix. That's ok, in and of itself, but why did you multiply by 4?
You were given that the high quality downloads outnumbered the standard downloads by a factor of 4. So if you insist on having a variable to represent the high-quality download quantity, then you can always say that 
What you have going on in parentheses after the 
makes no sense. You are putting in still another factor of 4!?
Let's recap: is the number of standard downloads, and is the total size of the standard downloads. is the number of high quality downloads, and 
is the total size of the high quality downloads. We know that the sum of these two sizes is 4444 megabytes, so:
But we know that , so:
\ =\ 4444)
Solve for
John
My calculator said it, I believe it, that settles it
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