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WITCHER [35]
3 years ago
10

An instructor wants to write a test with 25 questions where each question is worth 3, 4, or 5 points based on difficulty. He wan

ts the number of 3-point questions to be 4 less than the number of 4-point questions, and he wants the quiz to be worth a total of 100 points. How many 3, 4, and 5 point questions could there be?
Mathematics
1 answer:
irakobra [83]3 years ago
3 0

ANSWER

Find out the how many 3, 4, and 5 point questions could there be.

To proof

Let us assume that the 3 points based question be = x

Let us assume that the 4 points based question be = y

Let us assume that the 5 points based question be = z

As given

An instructor wants to write a test with 25 questions

than the equation become in the form

x + y + z = 25

he quiz to be worth a total of 100 points.

than the equation is becomes

3x + 4y + 5z =100

As given

He wants the number of 3-point questions to be 4 less than the number of 4-point questions

x = y -4

Than the three equation are

x + y + z = 25 ,3x + 4y + 5z =100 and x = y -4

put  x = y -4 in the x + y + z = 25 ,3x + 4y + 5z =100

than

2y + z = 29, 7y +5z= 112

multiply 2y + z = 29 by 5 and subtracted 7y +5z= 112

10 y -7y + 5y -5y = 145 -112

3y = 33

y = \frac{33}{3}

y =11

put in the  x = y -4

x = 11-4

x= 7

put the value of x ,y in the x + y + z = 25

7 + 11 +z =25

18 +z = 25

z = 25 -18

z=7

therefore

numbers of the 3 point question be = 7

numbers of the 4 point question be = 11

numbers of the 5 point question be = 7

Hence proved



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lilavasa [31]

Answer:

see explanation

Step-by-step explanation:

Given the 3 equations

3x + 5y + 5z = 1 → (1)

x - 2y = 5 → (2)

2x + 4y = 11 → (3)

Use (2) and (3) to solve for x and y

Multiply (2) by 2

2x - 4y = 10 → (4)

Add (3) and (4) term by term

4x = 21 ( divide both sides by 4 )

x = \frac{21}{4\\}

Substitute this value of x into (3)

2 × \frac{21}{4\\} + 4y = 11

\frac{21}{2\\} + 4y = 11 ( subtract \frac{21}{2\\} from both sides )

4y = \frac{1}{2} ( divide both sides by 4 )

y = \frac{1}{8\\}

Substitute the values of x and y into (1) and solve for z

3 × \frac{21}{4\\} + 5 × \frac{1}{8\\} + 5z = 1

\frac{63}{4} + \frac{5}{8} + 5z = 1

\frac{131}{8} + 5z = 1 ( subtract \frac{131}{8} from both sides )

5z = - \frac{123}{8} ( divide both sides by 5 )

z = - \frac{123}{40}

Solution is

x = \frac{21}{4\\}, y = \frac{1}{8\\}, z = - \frac{123}{40}

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A spherical ball has a radius of 4 inches. What is the volume of this sphere, to the
stealth61 [152]

Answer:

V = 267.9 in^3

Step-by-step explanation:

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V = 4/3 * 3.14 * 4^3

V = 4/3 * 3.14 * 64

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I really, REALLY need help. I will give brainliest to whoever figures it out.
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Answer:

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Sumo wrestler gained 5.5 kg per month

After 11 month, he weighed 140 kg.

Let x be his current weight.

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If Y is the weight of the wrestler after t months, then the linear equation would be:

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5 0
3 years ago
I NEED HELP WITH THIS ASAP! ITS DUE IN 20 MINS
amm1812

Answer:

b

Step-by-step explanation:

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3 years ago
Help! Kinda on a rush with this quiz!
Natasha2012 [34]
B



Step by step explanation This is how I got the answer to your question and I gave you the solution I hope this helps you out
8 0
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