Since this problem deals with a probability with two potential outcomes, it is a binomial distribution, and so the probability of an event is given as:

Where is the number of events, is the number of "successes" (in this case, a "heads" outcome), and is the probability of success (in this case, fifty percent).

One approach is to calculate the probability of flipping no heads, one head, two heads, etc., all the way to six heads, and adding those probabilities together, but that would be time consuming. Rather, calculate the probability of flipping seven heads. The complement to that would then be the sum of all other flip probabilities, which is what the problem calls for:

Therefore, the probability of six or fewer heads is:

Since this problem deals with a probability with two potential outcomes, it is a binomial distribution, and so the probability of an event is given as:

Where is the number of events, is the number of "successes" (in this case, a "heads" outcome), and is the probability of success (in this case, fifty percent).

Per the question, we're looking for the probability of at least three heads; three head flips or four head flips would satisfy this:

Thus the probability of three or more flips is:

Since this problem deals with a probability with two potential outcomes, it is a binomial distribution, and so the probability of an event is given as:

Where is the number of events, is the number of "successes" (in this case, a "heads" outcome), and is the probability of success (in this case, fifty percent).

One approach is to calculate the probability of flipping no heads, one head, two heads, etc., all the way to six heads, and adding those probabilities together, but that would be time consuming. Rather, calculate the probability of flipping seven heads. The complement to that would then be the sum of all other flip probabilities, which is what the problem calls for:

Therefore, the probability of six or fewer heads is:

Since this problem deals with a probability with two potential outcomes, it is a binomial distribution, and so the probability of an event is given as:

Where is the number of events, is the number of "successes" (in this case, a "heads" outcome), and is the probability of success (in this case, fifty percent).

Per the question, we're looking for the probability of at least three heads; three head flips or four head flips would satisfy this:

Thus the probability of three or more flips is:

A face card constitutes a Jack, Queen, or King, and there are twelve in a deck, so the probability of drawing a face card is .

There are thirteen spades in the deck, so the probability of drawing a spade is .

Keep in mind that there are also three cards that fit into both categories: the Jack, Queen, and King of Spades; the probability of drawing one is

Thus the probability of drawing a face card or a spade is:

A face card constitutes a Jack, Queen, or King, and there are twelve in a deck, so the probability of drawing a face card is .

There are thirteen spades in the deck, so the probability of drawing a spade is .

Keep in mind that there are also three cards that fit into both categories: the Jack, Queen, and King of Spades; the probability of drawing one is

Thus the probability of drawing a face card or a spade is:

To find her average grade for the class, we need to multiply Mary's test scores by their corresponding weights and then add them up.

The five quizzes were each worth 10%, or 0.1, of her grade, and the midterm and final were both worth 25%, or 0.25.

average = (0.1 * 89) + (0.1 * 74) + (0.1 * 84) + (0.1 * 92) + (0.1 * 90) + (0.25 * 92) + (0.25 * 91) = 88.95 = 89.

Looking at the answer choices, they are all spaced 2 percentage points apart, so clearly the closest answer choice to 88.95 is 89.

To find her average grade for the class, we need to multiply Mary's test scores by their corresponding weights and then add them up.

The five quizzes were each worth 10%, or 0.1, of her grade, and the midterm and final were both worth 25%, or 0.25.

average = (0.1 * 89) + (0.1 * 74) + (0.1 * 84) + (0.1 * 92) + (0.1 * 90) + (0.25 * 92) + (0.25 * 91) = 88.95 = 89.

Looking at the answer choices, they are all spaced 2 percentage points apart, so clearly the closest answer choice to 88.95 is 89.

Summing total money and dividing by the number of suspects gives us an average of approximately $2.73. By comparing it to the amounts held by the suspects, we can see that Suspect #2 is guilty.

Summing total money and dividing by the number of suspects gives us an average of approximately $2.73. By comparing it to the amounts held by the suspects, we can see that Suspect #2 is guilty.