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melisa1 [442]
3 years ago
10

You roll a 6-sided die with faces numbered 1 through 6, and toss a coin. What is the probability of rolling a 2 or getting heads

Mathematics
1 answer:
tino4ka555 [31]3 years ago
5 0

P(2 or H) = P(2) + P(H) - P(2 and H)  

What is the probability of getting a 2 P(2)?  = 1/6

What is the probability of getting heads P(H)?  = 1/2

P(2 and H) is the product of those two events since the events are independent. = 1/6 * 1/2 = 1/12

P(2 or H) = P(2) + P(H) - P(2 and H)  

P(2 or H) = 1/6 + 1/2  - 1/12 = 7/12

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A frog sits at one end of a table which is 2m long. In its first jump the frog goes a distance of 1m along the table, with its s
Phantasy [73]
Ah, this is an infinite sum question, or a sum of geometric sequence
when will the sum reach 1cm of the edge or about 1.99m
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2m=200cm
within 1cm means at least 1.99m
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sum of geometric sequence is
S_n= \frac{a_1(1-r^n)}{1-r}
a1=first term=initial jjump=1
r=common ratio=1/2
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we want it to equal 1.99 so

1.99= \frac{1(1- (\frac{1}{2})^n)}{1-\frac{1}{2}}
1.99= \frac{(1- (\frac{1}{2})^n)}{\frac{1}{2}}
1.99= 2(1- (\frac{1}{2})^n)
divide both sides by 2
\frac{1.99}{2} = 1- (\frac{1}{2})^n
times -1
\frac{-1.99}{2} = (\frac{1}{2})^n-1
add 1 or 2/2 to both sides
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7 0
3 years ago
Williiam lost 25% of his marble,gave 1/6 of the remainder to his.Brother and had 120 marbles left.How many marbles did he have a
bija089 [108]

Answer:

Number of marbles with Williams at the starting = 192

Step-by-step explanation:

Let total number of marbles with Williams = m

He lost marbles = 25% of the total number

                           = \frac{25}{100}\times m

                           = \frac{m}{4}

Remaining marbles with Williams = m - \frac{m}{4}

                                                        = \frac{3m}{4}

He gave marbles to his brother = \frac{1}{6}th of the remaining marbles

                                                    = \frac{1}{6}\times \frac{3m}{4}

                                                    = \frac{m}{8}

Remaining marbles after giving marbles to his brother = 120

Therefore, \frac{3m}{4}-\frac{m}{8}=120

\frac{6m}{8}-\frac{m}{8}=120

\frac{5m}{8}=120

m = \frac{120\times 8}{5}

m = 192

Number of marbles with Williams at the starting = 192

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3 years ago
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tatuchka [14]
The first one is A and the second one is B, hope this helps!!
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