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lidiya [134]
3 years ago
15

A test is worth 50 points. Multiple-choice are worth 1 point and short answer questions are worth 3 points. If a test has 20 que

stions how many multiple choice questions are there
Mathematics
1 answer:
tankabanditka [31]3 years ago
6 0

Answer:

5 multiple choice questions

Step-by-step explanation:

Write equations to represent the situation.

let "x" be the number of multiple choice questions

let "y" be the number of short answer questions

Equation for the total number of questions:

x + y = 20        # of multiple choice questions + # of short answer questions

Equation for the total points: (Multiply the type of question with how many points you get.)

x + 3y = 50     You don't need to write '1' beside 'x' for multiplying by 1.

The system of equations we need to solve is:

x + y = 20 and x + 3y = 50

We can solve using the substitution method.

Rearrange x + y = 20 to isolate one variable.

Isolate 'y'.

x + y = 20

x - x + y = 20 - x     Subtract 'x' from both sides

y = 20 - x

Take the other equation x + 3y = 50. You can replace 'y' with the equation that equals 'y' that we got when rearranging.

Substitute y for 20 - x

x + 3y = 50

x + 3(20 - x) = 50    Use distributive property. Multiply the 3 outside the bracket by each number inside the bracket.

x + 60 - 3x = 50     Combine like terms. 'x' and '-3x' are alike because they both have 'x'.

60 - 2x = 50          Start isolating 'x'.

60 - 60 - 2x = 50 - 60     Subtract 60 from both sides.

-2x = 50 - 60     60-60 cancels out on the left side.

-2x = -10

-2x/-2 = -10/-2       Divide both sides by -2 to isolate 'x'.

x = -10/-2           -2x/-2 becomes 'x'. -2/-2 cancels out.

x = 5          Number of multiple choice questions

Therefore there are 5 multiple choice questions.

If you needed the number of short answer questions too, you would substitute x for 5 in any of the other equations.

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Then,

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Mean is represented by the following expression:

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Population variance formula looks like this:

\begin{gathered} \sigma^2=\frac{\sum^{}_{}(x-\mu)^2}{N} \\ \text{where,} \\ \sigma^2=\text{population variance} \\ \sum ^{}_{}=addition\text{ of} \\ x=\text{each value} \\ \mu=population\text{ mean} \\ N=\text{ number of values in the population} \end{gathered}

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Which Answer is a solution to the inequality?
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Answer:

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