What is the probability that a randomly chosen positive factor of 72 is less than 10?
and
A list consists of all possible three-letter arrangements formed by using the letters A, B, C, D, E, F, G, H such that the first letter is D and either the second or third letter is A. If no letter is used more than once in an arrangement and one three-letter arrangement is randomly selected from the list, what is the probability that the arrangement selected will be DCA?

Problem 1

S = sample space = set of all possible outcomes

S = set of positive factors of 72

S = {1,2,3,4,6,8,9,12,18,24,36,72}

n(S) = number of items in sample space

n(S) = 12

E = event space = set of outcomes we want to happen

E = set of factors of 72 such that they are less than 10

E = {1,2,3,4,6,8,9}

n(S) = number of items in event space

n(S) = 7

P(E) = probability get a value in set E (that is also in set S as well)

P(E) = probability we get a positive factor of 72 that is less than 10

P(E) = n(E)/n(S)

P(E) = 7/12

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Problem 2

n(S) = 12

E = event space

E = {DCA}

n(E) = 1

P(E) = probability we get an item in set S that is in set E also

P(E) = probability we get DCA from set S listed above

P(E) = n(E)/n(S)

P(E) = 1/12

29 is the correct answer

step-by-step explanation:

d

step-by-step explanation:

given

y + 3 = - 4(x - 2) ← distribute the parenthesis

y + 3 = - 4x + 8 ( subtract 3 from both sides )

y = - 4x + 5 → d