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GenaCL600 [577]
3 years ago
11

PLEASE ANSWER ASAP FOR BRAINLEST WITH WORK!!!!!!!!!!!!!!!

Mathematics
1 answer:
alexira [117]3 years ago
4 0

Answer:

The answer is 43 because 600/14 = 42.8571429 rounded to the nearest whole number is 43.

Step-by-step explanation:

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Answer:

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Step-by-step explanation:

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Plz help I don't understand this question.
ladessa [460]
<h3>Answer:  12</h3>

===========================================================

Explanation:

  • j = number of jerseys ordered
  • c = total cost in dollars

For company X, the equation is c = 10j + 115 where the 10j is the cost of buying all the jerseys (j of them). Then we tack on the flat fee of $115

The equation for company Y is c = 20j for similar reasoning as the previous company, but now each jersey costs $20. We don't add on anything else because there isn't an extra flat fee.

----------------

We have this system

\begin{cases}c = 10j+115\\c = 20j\end{cases}

which are the equations for company X and company Y in that order.

Both equations have 'c' on the left side. So we can equate the right hand sides and solve for j

20j = 10j+115

20j-10j = 115

10j = 115

j = 115/10

j = 11.50

Recall that j is the number of jerseys. It's not possible to order half a jersey, so we have to round to the nearest whole number. The question is: which way? We're right at the midpoint between j = 11 and j = 12. Let's check both values of j to see what the costs are for each company.

  • j = 11 for company X leads to c = 10j+115 = 10*11+115 = 225
  • j = 11 for company Y leads to c = 20j = 20*11 = 220

Company Y is the cheaper option if you bought 11 jerseys. But we want to know when company X is the better deal. Let's try j = 12

  • j = 12 for company X leads to c = 10j+115 = 10*12+115 = 235
  • j = 12 for company Y leads to c = 20j = 20*12 = 240

This time company X is the winner in terms of the best deal (aka cheapest cost). If j gets any bigger, then company X is still the winner. So we can see that the minimum order must be 12 jerseys if we want company X to be the cheapest option.

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