Here's how you solve this. So, x+y=2, right? Let's isolate x. x+y-y=2-y. x=2-y. NOW, if x=2-y, in 3x+2y=5, we can REPLACE x with 2-y and use it to solve for y! 3x+2y=5. 3(2-y)+2y=5. (3*2)+(3*-y)+2y=5. 6+(-3y)+2y=5. 6+(-y)=5. 6+(-y)-6=5-6. -y=-1. -y/-1=-1/-1. y=1.
So, if y=1, we can substitute that back into either equation--but let's go with the easier one, x+y=2. x+1=2. x+1-1=2-1. x=1. 1+1=2, so that works; let's check the other equation. 3(1)+2(1)=5. 3+2=5. 5=5. That's correct!
Answer: x=1, y=1
There are the combinations that result in a total less than 7 and at least one die showing a 3:
[3, 3] [3,2] [2,1] [1,3] [2,3]
The probability of each of these is 1/6 * 1/6 = 1/36
There is a little ambiguity here about whether or not we should count [3,3] as the problem says "and one die shows a 3." Does this mean that only one die shows a 3 or at least one die shows a 3? Assuming the latter, the total probability is the sum of the individual probabilities:
1/36 + 1/36 + 1/36 + 1/36 + 1/36 = 5/36
Therefore, the required probability is: 5/36
Answer:
80
Step-by-step explanation:
the 80 is the tenth part so you do nothing its a trip up qustion
Answer:
10/20
Step-by-step explanation:
10 times 20 is 200 which would give you 10/20
hope I did this right
