Answer:
Step-by-step explanation:
Since no possible correct method is posted, I will suggest a couple.
Method 1: guess and check
Works well for simple problems involving integers like this one.
Victor's age must be zero or greater than one, say one.
Guess v=1, find m=v+6=7, check m=5v-2=5-2=3 no good.
we need to make v bigger
Guess v=2, find m=v+6=2+6=8, check m=5v-2=5*2-2=8 ✔
So v=2, m=8.
Method 2:
Solve the system of two equations.
since the left-hand sides is m in both equations, and since m=m, we just have to equate the right-hand sides to solve for v.
5v-2=v+6
Solve for v
5v-v = 6+2
4v=8
v=2,
so again, v=2, m=v+6=2+6=8.
#1
The uniforms are numbered 0, 1, 2, ..., 99. That's 100 numbers. Half of them are odd and half of them are even. So the probability that any one of the uniforms is odd is 1/2 just like the probability that any one uniform is even is 1/2.
(a) The numbers on the uniforms are independent of one another. That is, the number of her cross-country uniform does not in any way determine the number on her basketball uniform and vice versa. This means that we can find the probability that each is odd and multiply these together using what is called the counting principle. The probability that all are odd is:
(1/2)(1/2)(1/2)=1/8
(b) This is done the same way we did part (a). Since the probability of any one uniform being odd is the same as it being even (1/2), the answer here is the same: (1/2)(1/2)(1/2)=1/8
(c) This problem differs from that in (a) and (b). There is only one way for all three uniforms to be odd numbers: (odd, odd, odd) or all even (even, even, even). However, there are multiple ways for the uniforms to be two odd and one even. If the uniforms are listed in order: cross-country, basketball, softball we can get exactly one even in any of three ways:
even, odd, odd
odd, even, odd
odd, odd, even
The probability for any one of these possibilities is (1/2)(1/2)(1/2)=1/8 but since there are three way the probability that we get even exactly once is equal to (3)(1/8) = 3/8
Answer is D.
Because;
The first equation = the second equation
