A media downloaded store sells songs for $1 each and movies for $15 each. You have $60 to spend. How many movies and songs can you purchase for $60?

termegolz6
2022-08-01
Answered

You can still ask an expert for help

dasse9

Answered 2022-08-02
Author has **12** answers

Let us assume we can purchase x number of songs and y number of movies for $60

Hence,

x+15y- 60

Possible combinations can be:

60 number of Songs and 0 number of movies

45 number of Songs and 1 number of movies

30 number of Songs and 2 nurber of movies

15 number of Songs and 3 number of movies

0 number of Songs and 4 number of movies

Hence,

x+15y- 60

Possible combinations can be:

60 number of Songs and 0 number of movies

45 number of Songs and 1 number of movies

30 number of Songs and 2 nurber of movies

15 number of Songs and 3 number of movies

0 number of Songs and 4 number of movies

asked 2022-07-01

I need to know how precise a measurement system is, expressed as a standard deviation. I don't really have much of a background in math so please excuse me if I'm a bit unclear or inaccurate about something.

The simplest way I know of doing this is by taking e.g. 100 measurements of the same object and then using the usual sample standard deviation formula.

Unfortunately, I can't use this method. I must measure different objects multiple times. Let's say 10 objects, 10 times each.

The thing that comes to mind is to calculate the mean for each group, subtract the relevant means from each measurement (sort of a normalisation) and then calculate the standard deviation from the resulting 100 numbers. Does this make sense? If not, how do I approach the problem?

The simplest way I know of doing this is by taking e.g. 100 measurements of the same object and then using the usual sample standard deviation formula.

Unfortunately, I can't use this method. I must measure different objects multiple times. Let's say 10 objects, 10 times each.

The thing that comes to mind is to calculate the mean for each group, subtract the relevant means from each measurement (sort of a normalisation) and then calculate the standard deviation from the resulting 100 numbers. Does this make sense? If not, how do I approach the problem?

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Evaluate $$(3{m}^{3})\ufeff(3+m-3{m}^{2}-2{m}^{3})=$$?

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Although Hausdorff measure provides a reasonable way to approach general metric space including fractal geometry, I do not really realize how they actually get meanings in the physical world.

May I ask for some examples where the connection between Hausdorff measure and physical phenomenon has been observed?

I meant some concrete properties like mass, conductivity, or probability distribution associated with Hausdorff measure. Any measures having such realization other than Lebesgue and probability measures will be also helpful. Thank you.

May I ask for some examples where the connection between Hausdorff measure and physical phenomenon has been observed?

I meant some concrete properties like mass, conductivity, or probability distribution associated with Hausdorff measure. Any measures having such realization other than Lebesgue and probability measures will be also helpful. Thank you.

asked 2022-07-02

Is the answer key wrong? or it's me?

ok, so im reviewing for a math test and the following question is from the practice final exam. Rationalize the denominator in the example:

$\frac{\sqrt{2}}{\sqrt{x-3}}$

after multiplying both the numeration and denominator by the conjugate of the denominator I got

$\frac{\sqrt{2x+6}}{x-3}$

But, in the answer key the answer is

$\frac{\sqrt{2x-6}}{x-3}$

The problem looks quite simple, but I'm not sure what is the answer.

ok, so im reviewing for a math test and the following question is from the practice final exam. Rationalize the denominator in the example:

$\frac{\sqrt{2}}{\sqrt{x-3}}$

after multiplying both the numeration and denominator by the conjugate of the denominator I got

$\frac{\sqrt{2x+6}}{x-3}$

But, in the answer key the answer is

$\frac{\sqrt{2x-6}}{x-3}$

The problem looks quite simple, but I'm not sure what is the answer.

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