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hodyreva [135]
3 years ago
6

The owner of a new restaurant is ordering tables and chairs. he wants to have only tables for two and tables for 4. the total nu

mber of people can be seated in rest is 120, then write an equation to represent the situation. What do the variables represent?
Mathematics
1 answer:
Papessa [141]3 years ago
7 0

Answer:

4 M + 2 N = 120

Step-by-step explanation:

If we name "M" the number of tables for four people, and "N" the number of tables for two people, then the total number of people to be seated using M tables (of four people) would be the product "4 * M".

Similarly, the total number of people to be seated if the owner has N number of two person tables, would be: "2 * N".

Therefore, if the owner has M four people tables and N two person tables, the total number of people seated at the restaurant would be the addition:

4 * M + 2 * N. Since he wants this number to be 120, we complete the equation making this expression equal to 120:

4 * M + 2 * N = 120

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GarryVolchara [31]

Answer:

I did the math and got an answer of 17.5

Step-by-step explanation:

I divided 70 by 4 and got a result of 17.5. This probably means Caleb started his collection about 17 and 1/2 months ago. (I apologize if it's wrong)

3 0
3 years ago
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A resturant sells 10 tacos for 8.49,or 6 of the same kind of taco for 5.40.Which is a better deal
Aliun [14]

Answer:

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3 years ago
ANYONE KNOW THIS NEED ASAP
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B, it’s the point that the lines cross
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2 years ago
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There are 20 machines in a factory. 7 of the machines are defective.
adelina 88 [10]

Answer:

0.1225

Step-by-step explanation:

Given

Number of Machines = 20

Defective Machines = 7

Required

Probability that two selected (with replacement) are defective.

The first step is to define an event that a machine will be defective.

Let M represent the selected machine sis defective.

P(M) = 7/20

Provided that the two selected machines are replaced;

The probability is calculated as thus

P(Both) = P(First Defect) * P(Second Defect)

From tge question, we understand that each selection is replaced before another selection is made.

This means that the probability of first selection and the probability of second selection are independent.

And as such;

P(First Defect) = P (Second Defect) = P(M) = 7/20

So;

P(Both) = P(First Defect) * P(Second Defect)

PBoth) = 7/20 * 7/20

P(Both) = 49/400

P(Both) = 0.1225

Hence, the probability that both choices will be defective machines is 0.1225

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3 years ago
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Evgesh-ka [11]
The correct answer is the fourth one
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3 years ago
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