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Alina [70]
3 years ago
6

Mitchell is a big coffee fan, so he always takes care of coffee brewing at the office. Normally he uses 100100 grams of Robusta

coffee to prepare 1010 cups of coffee. His friend brings him a packet of Arabica coffee and tells him that he should use 20\%20% more than usual when brewing Arabica coffee. How many grams of coffee should he use to make a 1515-cup pot of Arabica coffee?
Mathematics
2 answers:
Nata [24]3 years ago
8 0
That is always the third option
Vaselesa [24]3 years ago
7 0

Answer:

180 grams

Step-by-step explanation:

Mitchell has to use more coffee than usual for two reasons: First, he's making a larger pot than usual (15(15left parenthesis, 15 cups rather than 10)10)10, right parenthesis. And second, he's using the Arabica coffee, which requires 20\%20%20, percent more coffee for the same strength.

2

Instead of brewing \blue{10}10start color blue, 10, end color blue cups, Mitchell wants to brew \red{15}15start color red, 15, end color red cups, which means he has to use \dfrac{\red{15}}{\blue{10}}=\purple{1.5}  

​  =1.5start fraction, start color red, 15, end color red, divided by, start color blue, 10, end color blue, end fraction, equals, start color purple, 1, point, 5, end color purple times more coffee.

3

As his friend said, Mitchell also has to use \pink{20\%}20%start color pink, 20, percent, end color pink more coffee than usual, so he'll have to multiply the amount of coffee by \green{1.2}1.2start color green, 1, point, 2, end color green, or \green{120\%}120%start color green, 120, percent, end color green.

4

The total amount of coffee he has to use is 100100100 grams \times×times \purple{1.5} \times \green{1.2} = 1801.5×1.2=180start color purple, 1, point, 5, end color purple, times, start color green, 1, point, 2, end color green, equals, 180 grams.

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irga5000 [103]

The testing of the four vertices of the objective function gives;

C(5, 0, 55) = 430

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<h3>How to test objective functions?</h3>

The objective function is seen as attached.

Now, it is noticed that just two of the vertices are listed. The other two vertices are intersections of x + y + z = 60 with x = 5 and the constraints on y.

We will assume that; 0 ≤ y ≤ 30

Thus, the missing vertices are;

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The two given vertices are;

(x, y, z) = (15, 30, 15)  and  (45, 0, 15)

Therefore, the objective function values are;

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C(5, 30, 25) = 9·5 +6·30 +7·25 = 45 +180 +175 = 400

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C(45, 0, 15) = 9·45 +6·0 +7·15 = 405 +0 +105 = 510

Read more about Objective Functions at; brainly.com/question/16826001

#SPJ1

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