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
Answer: 3.71 miles (approximately)
Explanation: 1/7 • 26 miles = 3.71428571429
Answer:
= 4
Step-by-step explanation:
first lets find the value of x in 3x+2=5/9
3x+2=5/9
3x = -2
3x =
3x =
x =
x =
can also be written as
−3x+8
-3 () +8
+8
= 4
you will be left with 12 dollars.
<u>
</u><u>Step-by-step explanation:
</u>
Given:
The actual money you have = 10 dollars.
The bet amount for each trade= 1$
For each win, you get extra +1 dollar.
Step 1:
So, you won 6 trades= (6*1)= 6$
Step 2:
For each lose, you loss -1 dollar.
Step 3:
So, you lose 4 trades= (4)*(-1)= -4$
Step 4:
After 10 trades, the money you earned is= 6$ -4$= 2 dollars.
Finally, the total money left after 10 trades is= actual money + earned money
= $10 + $2= 12 dollars.
53= -19z+5+7z
53= -12z+5 (combine -19z and 7z)
48= -12z. ( subract 5 on both sides)
-4=z. (divide by -12 both sides)