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OLEGan [10]
4 years ago
6

A board games game has a spinner divided into sections of equal size. Each section is labeled with a number between 1 and 5. Whi

ch number is a reasonable estimate of the number of times the spinner will land on a section labeled 5 over the course of 150 spins?
Mathematics
1 answer:
Bess [88]4 years ago
7 0
30 would be a reasonable estimate.

Since there are 5 sections, the probability of spinning a 5 is 1/5.

Multiply this probability by the number of spins (150):

1/5(150) = 150/5 = 30
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Hey! Can someone check my answers?
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The greatest common factor is the biggest number taken from the values.

Q1. The answer is <span>A. 5y^6
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20 y^{9} +5 y^{6}= 4*5 y^{9}+5 y^{6}
Since x^{a}* x^{b}  =x^{a+b}, then x^{9}= x^{3}* x^{6}

Back to our expression:
4*5 y^{9}+5 y^{6}=4*5 y^{3}*y^{6}+5 y^{6}=4 y^{3}*5y ^{6}+  5y ^{6}*1=5 y^{6} (4y ^{3} +1)
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Q2. The answer is <span>D. 12xy^2
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12x y^{5}+60 x^{4} y^{2} -24 x^{3} y^{3}=12x y^{5}+5*12 x^{4} y^{2} -2*12 x^{3} y^{3}
We will use the rule  x^{a}* x^{b} =x^{a+b} to factorise the exponents:
12x y^{5}+5*12 x^{4} y^{2} -2*12 x^{3} y^{3}= \\ =12x*y^{2}*y^{3}+5*12*x* x^{3} *y^{2}-2*12x* x^{2} *y*y^{2}= \\ =12xy^{2}*y^{3}+12xy^{2}*5 x^{3} -12xy^{2}*2 x^{2} y= \\ =12xy^{2}(y^{3}+5 x^{3}-2 x^{2} y)
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4 years ago
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yuradex [85]
The answer is x=4.5 because 9÷2=4.5 and 4.5×2=9
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3 years ago
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