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Nutka1998 [239]
3 years ago
8

You roll a 6-sided die with faces numbered 1 through 6, and toss a coin. What is the probability of rolling a 3 or getting heads

? A. 7/12 B. 1/12 C. 1/2 D. 1/3
Mathematics
2 answers:
ratelena [41]3 years ago
5 0

You roll a 6 sided die, and am trying to get a number 3. You have a 1/6 chance of getting the number

The probability of getting a heads tossing a coin is half, because there is only two sides.

Multiply the two fractions together:

1/6 x 1/2 = 1/12

B) 1/12 is your answer

hope this helps

In-s [12.5K]3 years ago
5 0

|\Omega|=6\cdot2=12\\ |A|=1\\\\ P(A)=\dfrac{1}{12}\implies\text{B}

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Answer:

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Given

See attachment for model

Required

Determine \frac{4}{7}\times\frac{13}{9} from the model

The model is represented by:

\frac{4}{7}\times\frac{13}{9} = \frac{4}{7}\times\frac{9}{9} + \frac{4}{7}\times\frac{4}{9}

To get: \frac{4}{7}\times\frac{9}{9}, we consider the first partition

The number of shaded box is 63 ---- this represents the denominator

The total boxes shaded at the bottom is 36 ---- this represents the numerator

So, we have:

\frac{4}{7}\times\frac{9}{9} = \frac{36}{63}

To get: \frac{4}{7}\times\frac{9}{9}, we consider the first partition

The number of shaded box is 63 ---- this represents the denominator

The total boxes shaded at the bottom is 16 (do not count the gray boxes) ---- this represents the numerator

So, we have:

\frac{4}{7}\times\frac{4}{9} =\frac{16}{63}

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\frac{4}{7}\times\frac{13}{9} = \frac{4}{7}\times\frac{9}{9} + \frac{4}{7}\times\frac{4}{9}

\frac{4}{7}\times\frac{13}{9} = \frac{36}{63} + \frac{16}{63}

\frac{4}{7}\times\frac{13}{9} = \frac{36+16}{63}

\frac{4}{7}\times\frac{13}{9} = \frac{52}{63}

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