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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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BlackZzzverrR [31]

Answer:

(1) The value of P (A) is 0.4286.

(2) The value of P (B) is 0.50.

(3) The value of P (A ∩ B) is 0.2143.

(4) The the value of P (B|A) is 0.50.

(5) The events <em>A</em> and <em>B</em> are independent.

Step-by-step explanation:

The events are defined as follows:

<em>A</em> = a student in the class has a sister

<em>B</em> = a student has a brother

The information provided is:

<em>N</em> = 210

n (A) = 90

n (B) = 105

n (A ∩ B) = 45

The probability of an event <em>E</em> is the ratio of the favorable number of outcomes to the total number of outcomes.

P(E)=\frac{n(E)}{N}

The conditional probability of an event <em>X</em> provided that another event <em>Y</em> has already occurred is:

P(X|Y)=\frac{P(A\cap Y)}{P(Y)}

If the events <em>X</em> and <em>Y</em> are independent then,

P(X|Y)=P(X)

(1)

Compute the probability of event <em>A</em> as follows:

P(A)=\frac{n(A)}{N}\\\\=\frac{90}{210}\\\\=0.4286

The value of P (A) is 0.4286.

(2)

Compute the probability of event <em>B</em> as follows:

P(B)=\frac{n(B)}{N}\\\\=\frac{105}{210}\\\\=0.50

The value of P (B) is 0.50.

(3)

Compute the probability of event <em>A</em> and <em>B</em> as follows:

P(A\cap B)=\frac{n(A\cap B)}{N}\\\\=\frac{45}{210}\\\\=0.2143

The value of P (A ∩ B) is 0.2143.

(4)

Compute the probability of <em>B</em> given <em>A</em> as follows:

P(B|A)=\frac{P(A\cap B)}{P(A)}\\\\=\frac{0.2143}{0.4286}\\\\=0.50

The the value of P (B|A) is 0.50.

(5)

The value of P (B|A) = 0.50 = P (B).

Thus, the events <em>A</em> and <em>B</em> are independent.

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