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ch4aika [34]
3 years ago
7

the scores in physics test of 10 students are as follows:10,11,8,10,12,8,10,11,12,10. find its median,mean,mode

Mathematics
1 answer:
Olin [163]3 years ago
7 0

Answer:

mean= 10+11+8+10+12+8+10+11+12+10/10

=92/10=9.2

mode=10

hope it helps

plz mark as brainliest

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Eliminating by multiplying <br> X+3y=1<br> -5x+4y=-24
gavmur [86]

Answer:

y = -1         x = 4

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(X+3y=1) 5            Multiply this equation by 5 to cancel out the x

-5x+4y=-24

-5x + 4y= -24

+ 5x + 15y = 5       Add both equations together

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y = -1

Plug -1 for y into one of the equations

x + 3(-1) = 1             Multiply 3(-1)

x - 3 = 1

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3 years ago
Which of the following are true statements.
Mariulka [41]

Answer:

Second statement is true.

The lengths 7, 40 and 41 can not be sides of a right triangle. The lengths 12, 16, and 20 can be sides of a right triangle.

Step-by-step explanation:

for first part of statement

The lengths 7, 40 and 41 can not be sides of a right triangle.

If the square of long side is equal to the sum of square of other two sides

then the given length can be sides of a right triangle.

Check the given length by Pythagoras Theorem.

c^{2} =a^{2} +b^{2}----------(1)

Let c=41 and a = 7 and b=40

Put all the value in equation 1.

41^{2} =7^{2} +40^{2}

1681=49+1600

1681=1649

Therefore, the square of long side is not equal to the sum of square of other two sides, So given lengths 7, 40 and 41 can not be sides of a right triangle.

for second part of statement.

The lengths 12, 16, and 20 can be sides of a right triangle.

Check the given length by Pythagoras Theorem.

Let c=20 and a = 12 and b=16

20^{2} =12^{2} +16^{2}

400=144+256

400=400

Therefore, the square of long side is equal to the sum of square of other two sides, So given the lengths 12, 16, and 20 can be sides of a right triangle.

Therefore, The lengths 7, 40 and 41 can not be sides of a right triangle. The lengths 12, 16, and 20 can be sides of a right triangle.

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3 years ago
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