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Naddik [55]
3 years ago
13

Rita has a number cube and a letter cube. The number cube is numbered 1 through 6. The letter cube is lettered A through F. Rita

will roll each cube once. What is the probability that she rolls an odd number with the number cube and a vowel with the letter cube? I
A 2/3
B 1/6
C 1/3
D 1/4​
Mathematics
1 answer:
netineya [11]3 years ago
3 0
The answer is b I believe
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Answer:

\frac{10q^5w^7}{2w^3}.\ \frac{4\left(q^6\right)^2}{w^{-5}} =20q^{17}w^9

Step-by-step explanation:

Given

\frac{10q^5w^7}{2w^3}.\ \frac{4\left(q^6\right)^2}{w^{-5}}

Required

Determine the equivalent expression

\frac{10q^5w^7}{2w^3}.\ \frac{4\left(q^6\right)^2}{w^{-5}}

Simplify the first fraction

\frac{5q^5w^7}{w^3}.\ \frac{4\left(q^6\right)^2}{w^{-5}}

Apply law of indices on the first fraction;

5q^5w^{7-3}.\ \frac{4\left(q^6\right)^2}{w^{-5}}

5q^5w^4.\ \frac{4\left(q^6\right)^2}{w^{-5}}

\ \frac{5q^5w^4*4\left(q^6\right)^2}{w^{-5}}

Apply law of indices:

5q^5w^4*4\left(q^6\right)^2 * w^{5}

Evaluate the bracket

5q^5w^4*4 * q^{12} * w^{5}

Collect Like Terms

5*4q^5* q^{12}*w^4  * w^{5}

20q^5* q^{12}*w^4  * w^{5}

20q^{5+12}*w^{4+5}

20q^{17}w^9

Hence:

\frac{10q^5w^7}{2w^3}.\ \frac{4\left(q^6\right)^2}{w^{-5}} =20q^{17}w^9

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Answer:

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