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EleoNora [17]
3 years ago
15

Suppose you roll a 6-faced die 90 times. About how many times would you expect to get a 5?

Mathematics
1 answer:
irinina [24]3 years ago
7 0
15 over 90
3over18
yes
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7 0
4 years ago
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Brut [27]

Answer:

Step-by-step explanation:                

\displaystyle\Large\boldsymbol{} x^4+2x^3-2x^2+2x-3= \\\\\\x^4+2x^2(x-1)+2x-\underbrace{2-1}_{-3}=  \\\\\\x^4-1+2x^2(x-1)+2x-2 = \\\\\\(x^2-1)(x^2+1)+2x^2(x-1)+2(x-1) = \\\\\\\underline{(x-1)}(x+1)(x^2+1)+2x^2\underline{(x-1)} +2\underline{(x-1)} =\\\\\\(x-1)((x+1)(x^2+1)+2x^2+2)=\\\\\\(x-1)(x^3+x^2+2x^2+x+2+1)=\\\\\\(x-1)(x^3+3x^2+x+3) =\\\\\\(x-1)( \underline{(x+3)}x^2+\underline{(x+3)})=(x^2+1)\underbrace{(x-1)(x+3)}_{x^2+2x-3}= \\\\\\(x^2+1)(x^2+2x-3) : (x^2+2x-3)=x^2+1

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3 years ago
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Savatey [412]

Answer:

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Step-by-step explanation:

this is how i did it though:

12(a+b)

12a+60

5 0
3 years ago
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