You are in charge of managing the advertising for a restaurant. you are able to buy up to 62 advertisements a month, and your data from the previous month shows that you need to buy at least 5 newspaper ads for every 3 television ads to draw in the most customers. what is the greatest number of newspaper and television ads you could buy?

Greatest number of newspaper ads = 59

Greatest number of television ads = 23

Step-by-step explanation:

It is given that we should buy at least 5 newspapers for every 3 television ads.

So, after purchasing 3 television ads, we can buy any number of television ads but the number should be greater than 5.

Since, the maximum is 62, we can buy 59 newspaper ads. (Note that we have already bought 3 television ads).

For every 3 television ads, we should buy 5 newspaper ads. So,

For the purchase of 3 television ads, we should buy the minimum of 5 newspaper ads.

For the purchase of 2(3) television ads, we should buy the minimum of 2(5) newspaper ads.

For the purchase of 3(3) television ads, we should buy the minimum of 3(5) newspaper ads.

...

...

For the purchase of 7(3) = 21 television ads, we should buy the minimum of 7(5) = 35 newspaper ads.

So, totally 21+35 = 56 ads.

For the remaining 6 ads, if we buy 3 television ads, we should buy 5 newspaper ads.

So, we can buy 2 television ads and 4 newspaper ads.

Hence, we can buy the maximum of 21+2 = 23 television ads.

$650 per month plus utilities

step-by-step explanation:

just based on the info you gave, it would seem more cost effective to rent at $650 per month plus utilities.   if utilities are, in fact, $150 per month, your monthly living costs would be $800 as opposed to the other option of $825.

a

step-by-step explanation:

the denominator cannot be the same as one of the factors.

8 denominator times 2 denominator = 16

and 7 numerator time 11 = 77

77/16= 77/16 ( they cannot be further simplified)