Stay in school: In a recent school year in the state of Washington, there were 315000 high school students. Of these, were girls and were boys. Among the girls, dropped out of school, and among the boys, dropped out. A student is chosen at random. Round the answers to four decimal places. (a) What is the probability that the student is female? (b) What is the probability that the student dropped out? (c) What is the probability that the student is female and dropped out? (d) Given that the student is female, what is the probability that she dropped out? (e) Given that the student dropped out, what is the probability that the student is female? Part: 0 / 50 of 5 Parts Complete

Step-by-step explanation:

Hello!

Since the data on the text is missing, I've found a similar exercise with the same questions so I'm going to use the data of that one to explain how to calculate the probabilities.

We have a sample of 323000 high school students

154000 are girls, of those 47700 dropped out

16900 are boys, of those 10300 dropped out

a) To calculate this probability you have to divide the total number of girls by the total number of students:

P(G)= 154000/323000= 0.476 ≅ 0.48

b) In this item you have to calculate the probability of dropouts, the total of students that dropped out are the girls + boys that dropped out:

Dropouts: 47700+10300= 58000

Now you divide it by the total of students to reach the probability:

P(D)= 58000/323000= 0.179≅ 0.18

c) Now you have to calculate the probability of the intersection between the events "female" and "dropped out", symbolically:

P(G∩D)= P(G)*P(D)= 0.48*0.18= 0.0864

d) The probability of the student dropping out given that it is female is a conditional probability and you can calculate using the following formula:

P(D/G)= P(G∩D) = 0.0864 = 0.18

                 P(G)        0.48

e) Now you have to calculate the probability of the student being a femal, given that she dropped out of school, symbolically:

P(G/D)= P(G∩D) = 0.0864 = 0.48

                 P(D)        0.88

Note:

As you can see in d) P(D/G)=P(D)=0.18 and in e) P(G/D)=P(G)=0.48, this means that both events "the student is a girl" and "the student dropped out" are independent.

Remember two events are not independent when the occurrence of one modifies the probability of occurrence of the other.

I hope it helps!

{x = -1/2 ,   y = -1

step-by-step explanation:

solve the following system:

{2 x + y = -2

{10 x - 3 y = -2

in the first equation, look to solve for y:

{2 x + y = -2

{10 x - 3 y = -2

subtract 2 x from both sides:

{y = -2 x - 2

{10 x - 3 y = -2

substitute y = -2 x - 2 into the second equation:

{y = -2 x - 2

{10 x - 3 (-2 x - 2) = -2

10 x - 3 (-2 x - 2) = (6 x + 6) + 10 x = 16 x + 6:

{y = -2 x - 2

{16 x + 6 = -2

in the second equation, look to solve for x:

{y = -2 x - 2

{16 x + 6 = -2

subtract 6 from both sides:

{y = -2 x - 2

{16 x = -8

divide both sides by 16:

{y = -2 x - 2

{x = -1/2

substitute x = -1/2 into the first equation:

{y = -1

{x = -1/2

collect results in alphabetical order:

answer:   {x = -1/2 ,   y = -1

$655.61 which is option c

step-by-step explanation:

price of house =   $146,000

initial rate = 3.5%

mortgaged paid in this = 3.5 % of $146,000

                                        =

                                        =$5110

the total years would be = 7

total mortgaged paid = 7 * $5110

                                      =$35770

remaining money = $146000-35770

                                =$110230

new mortgaged payment = 5 % of 110230

                                            =  

                                            =$655.61