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 show the results of John’s experiment after he rolled the die and tossed the coin.

Based on the results of the experiment, how many times would John expect to roll a number on the die that is a factor of 12 and toss a coin that lands heads-up out of the next 144 trials?

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

n = 60

1260=3*6*7*10=3^2*2^2*7*5 perfect cude = 3^3*2^

step-by-step explanation: