Remmeber yo can do antyhing to an equation as long as you do it to both sides
9x>270
divide both sides by 9
x>30
B/4<4
times both sides by 4
B=16
Y/5≤8
times both sides by 5
Y≤40
5c≥125
divide both sides by 5
c≥25
Answer:
Step-by-step explanation: ok so counting my 5’s .. 5, 10, (15) it’s 3 times soo 5 shirts cost 25 you are gonna do 25x3 so it’s 75
There are C(5,2) = 10 ways to choose 2 contraband shipments from the 5. There are C(11, 1) = 11 ways to choose a non-contraband shipment from the 11 that are not contraband. Hence there are 10*11 = 110 ways to choose 3 shipments that have 2 contraband shipments among them.
There are C(16,3) = 560 ways to choose 3 shipments from 16. The probability that 2 of those 3 will contain contraband is
110/560 = 11/56 ≈ 19.6%
_____
C(n,k) = n!/(k!(n-k)!)
Answer:
60 students passed, and 75 appeared in examination.
Step-by-step explanation:
Let's say s is the total number of students and p is the number of students who passed.
80% of the students passed, so:
0.8 s = p
If there were 10 less passers, and 15 less students (5 less failures), then the ratio of passers to failures would be 5/1.
(p − 10) / (s − p − 5) = 5 / 1
Simplify the second equation:
p − 10 = 5 (s − p − 5)
p − 10 = 5s − 5p − 25
6p = 5s − 15
Substitute the first equation.
6 (0.8 s) = 5s − 15
4.8 s = 5s − 15
0.2 s = 15
s = 75
p = 0.8 s
p = 60
60 students passed, and 75 appeared in examination.
-7 -(-7) because it will equal zero and the other one will be -14
