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ipn [44]
2 years ago
7

I don’t know the answer

Mathematics
2 answers:
liberstina [14]2 years ago
5 0

Answer: B

Step-by-step explanation:

taurus [48]2 years ago
3 0

Answer:

b

Step-by-step explanation:

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John has a die and a coin. The die is labeled 1 through 6 and the two sides of the coin are heads and tails. The tables below sh
Alenkasestr [34]

Answer:

n = 60

Step-by-step explanation:

Total outcomes, while tossing a coin are 2, given as:

{Heads, Tails}

The probability of getting a heads up, while tossing the coin is:

P(Heads) = Favorable Outcome/Total Outcomes

P(Heads) = 1/2 = 0.5

Now, the total outcomes, while rolling a die are 6, given as:

{1, 2, 3, 4, 5, 6}

5 of these numbers are factors of 12, which are:

{1, 2, 3, 4, 6}

Thus, the probability of getting a factor of 12, while rolling a die is:

P(Factor of 12) = Favorable Outcome/Total Outcomes

P(Factor of 12) = 5/6 = 0.83

So, the probability of getting both heads and factor of 12 will be:

P(Heads and Factor of 12) = P(Heads ∪ Factor of 12)

P(Heads and Factor of 12) = P(Heads) * P(Factor of 12) = 0.5 * 0.83

P(Heads and Factor of 12) = 0.417 = 41.7%

So, for 144 trials, the number of trials in which we get heads and factor of 12, are given by:

n = P(Heads and Factor of 12) * 144

where,

n = no. of trials John would expect to roll a number on the die that is a factor of 12 and toss a coin that lands heads-up out of next 144 trials

n = 0.417 * 144

<u>n = 60</u>

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What is the lcm of 12 and 18
tekilochka [14]
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12x3 and 18x2 = 36
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