Answer:
143 because if you subtract 49 and 28 you get 21 the you divide 21 and 1623
4(m + 2) expanded is 4 x m and 4 x 2
simplified: 4m + 8
Answer:
5
Step-by-step explanation:
In order to solve this problem, we need to use Pythagorem Theorem (a^2+b^2=c^2)
When using the pythagorem theorem the hypoteneuse would be the c^2 and the two other sides would be a and b (the order doesn't matter).
So, our formula would be:
4^2+3^2=c^2
Simplifies into:
25=c^2
In order to remove the exponent 2, we would have to square root both sides by 2. After doing so, you get the answer 5.
Hope this helps! :)
Answer:
a = 1/2
Step-by-step explanation:
a - 3/10 = 1/5
Add 3/10 to each side
a - 3/10 +3/10 = 1/5+3/10
a = 1/5+3/10
Get a common denominator
a= 1/5*2/2 +3/10
a = 2/10 +3/10
a = 5/10
Divide the top and bottom by 5
a = 1/2
Complete question :
Suppose that of the 300 seniors who graduated from Schwarzchild High School last spring, some have jobs, some are attending college, and some are doing both. The following Venn diagram shows the number of graduates in each category. What is the probability that a randomly selected graduate has a job if he or she is attending college? Give your answer as a decimal precise to two decimal places.
What is the probability that a randomly selected graduate attends college if he or she has a job? Give your answer as a decimal precise to two decimal places.
Answer:
0.56 ; 0.60
Step-by-step explanation:
From The attached Venn diagram :
C = attend college ; J = has a job
P(C) = (35+45)/300 = 80/300 = 8/30
P(J) = (30+45)/300 = 75/300 = 0.25
P(C n J) = 45 /300 = 0.15
1.)
P(J | C) = P(C n J) / P(C)
P(J | C) = 0.15 / (8/30)
P(J | C) = 0.5625 = 0.56
2.)
P(C | J) = P(C n J) / P(J)
P(C | J) = 0.15 / (0.25)
P(C | J) = 0.6 = 0.60