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sukhopar [10]
2 years ago
12

Mike will roll two number cubes labeled 1 through 6. After each roll, he adds the two numbers showing and records the result. He

will roll the cubes 500 times. About how many times can he expect a sum greater than 8?
Mathematics
1 answer:
xeze [42]2 years ago
8 0

Answer:

In 500 rolls he can expect to get this outcome about  27.77...%* 500 =  139 times.

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Dan has 3 ⅔ cups of milk. If he used ¼ of the milk for his breakfast, how much milk does Dan have left?
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There would be 3 2/12 left.

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How can i estimate 75.78 - 49.381 by first rounding each number to the nearest whole number
Andre45 [30]
First you look at the tenths place of each number.
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Trigonometry help!! - double angle formulae
ivolga24 [154]

Answer:

The two rules we need to use are:

Sin(a + b) = sin(a)*cos(b) + sin(b)*cos(a)

cos(a + b) = cos(a)*cos(b) - sin(a)*sin(b)

And we also know that:

sin^2(a) + cos^2(a) = 1

To solve the relations, we start with the left side and try to construct the right side.

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sin(A + A)*cos(A) + sin(A)*cos(A + A)

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3*sin(A)*cos^2(A) - sin(A)*sin^2(A)

sin(A)*(3*cos^2(A) - sin^2(A))

Now we can add and subtract 4*sin^3(A)

sin(A)*(3*cos^2(A) - sin^2(A)) + 4*sin^3(A) -  4*sin^3(A)

sin(A)*(3*cos^2(A) + 3*sin^2(A)) - 4*sin^3(A)

sin(A)*3*(cos^2(A) + sin^2(A)) - 4*sin^3(A)

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b) Here we do the same as before:

cos(3*A) = 4*cos^3(A) - 3*cos(A)

We start with:

Cos(2*A + A) =  cos(2*A)*cos(A) - sin(2*A)*sin(A)

= cos(A + A)*cos(A) - sin(A + A)*sin(A)

= (cos(A)*cos(A) - sin(A)*sin(A))*cos(A) - ( sin(A)*cos(A) + sin(A)*cos(A))*sin(A)

= (cos^2(A) - sin^2(A))*cos(A) - sin^2(A)*cos(A) - sin^2(A)*cos(A)

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=  cos(A)*(cos^2(A) - 3*sin^2(A))

now we subtract and add 4*cos^3(A)

= cos(A)*(cos^2(A) - 3*sin^2(A)) + 4*cos^3(A) - 4*cos^3(A)

= cos(A)*(-3*cos^2(A) - 3*sin^2(A)) + 4*cos^3(A)

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Answer:

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8 0
3 years ago
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