Find a. Round it to the nearest tenth

40. In a statistics class of 30 students, there were 13 men and 17 women. Two of the men and three of the women received an A in

Harrizon [31]

Answer:

a) 56.67% probability that the student is a woman

b) 16.67% probability that the student received an A

c) 63.33% probability that the student is a woman or received an A.

d) 83.33% probability that the student did not receive an A.

Step-by-step explanation:

A probability is the number of desired outcomes divided by the number of total outcomes.

We have that:

30 students

13 men

17 women

2 men that got an A and 11 men that did not get an A.

3 women that got an A and 14 women that did not get an A.

a. Find the probability that the student is a woman.

30 students, of which 17 are women.

$P = \frac{17}{30} = 0.5667$

56.67% probability that the student is a woman

b. Find the probability that the student received an A.

30 students, of which 5 received an A

$P = \frac{5}{30} = 0.1667$

16.67% probability that the student received an A

c. Find the probability that the student is a woman or received an A.

30 students, of which 17 are women and 2 are men who received an A. So

$P = \frac{19}{30} = 0.6333$

63.33% probability that the student is a woman or received an A.

d. Find the probability that the student did not receive an A.

30 students, of which 25 did not receive an A.

$P = \frac{25}{30} = 0.8333$

83.33% probability that the student did not receive an A.

7 0

3 years ago

Help you don’t have to explain

kolezko [41]

Answer:

-4

Step-by-step explanation:

umm you might want to learn this. I cheated in middleschool/ highschool and was an honor student. I am barely surving college though, you'll need math for your general education requirement.

6 0

3 years ago

Sherry has taken 10 quizzes. Her combined total for the 10 quizzes is 83 points. Nine of her quiz scores are shown on the plot.

castortr0y [4]

Answer:

9

Step-by-step explanation:

4 0

3 years ago

Read 2 more answers

In an experiment, college students were given either four quarters or a $1 bill and they could either keep the money or spend it

gavmur [86]

Answer:

a) $P(A|B) = \frac{15/83}{44/83} =\frac{15}{44}=0.341$

b) $P(B|A) = \frac{29/83}{44/83} =\frac{29}{44}=0.659$

c) A. A student given a $1 bill is more likely to have kept the money.

Because the probability 0.659 is atmoslt two times greater than 0.341

Step-by-step explanation:

Assuming the following table:

Purchased Gum Kept the Money Total

Students Given 4 Quarters 25 14 39

Students Given $1 Bill 15 29 44

Total 40 43 83

a. find the probability of randomly selecting a student who spent the money, given that the student was given a $1 bill.

For this case let's define the following events

B= "student was given $1 Bill"

A="The student spent the money"

For this case we want this conditional probability:

$P(A|B) =\frac{P(A and B)}{P(B)}$

We have that $P(A)= \frac{40}{83} , P(B)= \frac{44}{83}, P(A and B)= \frac{15}{83}$

And if we replace we got:

$P(A|B) = \frac{15/83}{44/83} =\frac{15}{44}=0.341$

b. find the probability of randomly selecting a student who kept the money, given that the student was given a $1 bill.

For this case let's define the following events

B= "student was given $1 Bill"

A="The student kept the money"

For this case we want this conditional probability:

$P(A|B) =\frac{P(A and B)}{P(B)}$

We have that $P(A)= \frac{43}{83} , P(B)= \frac{44}{83}, P(A and B)= \frac{29}{83}$

And if we replace we got:

$P(B|A) = \frac{29/83}{44/83} =\frac{29}{44}=0.659$

c. what do the preceding results suggest?

For this case the best solution is:

A. A student given a $1 bill is more likely to have kept the money.

Because the probability 0.659 is atmoslt two times greater than 0.341

3 0

3 years ago

PLEASE HELP SHAWTYYY

Serjik [45]

Answer:

Vertical

x=49

Step-by-step explanation:

5 0

2 years ago