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allsm [11]
2 years ago
15

A science test, which is worth 100 points, consists of 24 questions. Each question is worth either 3 points or 5 points. If x is

the number of 3-point questions and y is the number of 5-point questions, the system shown represents this situation. x + y = 24 3x + 5y = 100 What does the solution of this system indicate about the questions on the test? The test contains 4 three-point questions and 20 five-point questions. The test contains 10 three-point questions and 14 five-point questions. The test contains 14 three-point questions and 10 five-point questions. The test contains 20 three-point questions and 8 five-point questions.
Mathematics
2 answers:
KiRa [710]2 years ago
3 0
You can solve the system of equations.

Multiply the first equation by -3 and add to the second equation.

                    -3x - 3y = -72
                    3x + 5y = 100
(add)     --------------------------
                            2y = 28

y = 14

x + y = 24
x + 14 = 24
x = 10

Answer:
There are 10 3-points questions and 14 5-point questions.

Check: 10 + 14 = 24   There are 24 questions
10 * 3 + 14 * 5 = 30 + 70 = 100   The total number of points is 100.
Our answer is correct.
Sloan [31]2 years ago
3 0
From looking at the second equation 3x + 5y = 100 you can see that x is the number of 3 point questions, y is the number of 5 points questions.

Use the first equation y + x = 24 to substitute into the second equation. You both equations combined into one equation in one variable.
y = 24 - x
sub 24- x in for y

3x + 5(24-x) = 100
3x + 120 - 5x = 100
-2x = 100 - 120
-2x = -20
x = -20/-2
x = 10

then plug this into ether equation to solve for y.

y + 10 = 24
y = 14

10 three pts questions
14 five pts questions

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2. Jamie is about 56 meters away from Cameron.

3. Arthur is closer to Cameron.

Step-by-step explanation:

We have been given the locations of Cameron (70,10), Arthur (20,30) and Jamie (45,60) on the the grid in meters.

1. To find the distances of Arthur and Jamie from Cameron we will use distance formula.

\text{Distance}=\sqrt{(x_2-x_1)^{2}+(y_2-y_1)^{2}}

\text{Distance between Cameron and Arthur}=\sqrt{(70-20)^{2}+(10-30)^{2}}

\text{Distance between Cameron and Arthur}=\sqrt{(50)^{2}+(-20)^{2}}

\text{Distance between Cameron and Arthur}=\sqrt{2500+400}

\text{Distance between Cameron and Arthur}=\sqrt{2900}

\text{Distance between Cameron and Arthur}=53.8516480713450403\approx 54

Therefore, the distance Arthur is about 54 meters away from​ Cameron.

2. Let us find the distance between Cameron and Jamie.

\text{Distance between Cameron and Jamie}=\sqrt{(70-45)^{2}+(10-60)^{2}}

\text{Distance between Cameron and Jamie}=\sqrt{(25)^{2}+(-50)^{2}}

\text{Distance between Cameron and Jamie}=\sqrt{625+2500}

\text{Distance between Cameron and Jamie}=\sqrt{3125}

\text{Distance between Cameron and Jamie}=55.9016994374947424\approx 56

Therefore, Jamie is about 56 meters away from​ Cameron.

3. We can see that 56 is greater than 54, therefore, Arthur is closer to Cameron.


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We are given that a sample of 2,000 licensed drivers revealed the following number of speeding violations;

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<u />

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