these are the answers to all systems
Team A) 45 people
Team B) 55 people
A)There are two ways to solve this problem, finding the number of combinations possible for Team B, or the number of combinations possible for Team A.
Team A
It's a given that 20 mathematicians are on team A, which leavs the other 25 people for team A to be chosen from a pool of 80 (100- 20 mathletes)
80-C-25 = 80! / (25!/(80-25)!) =<span>363,413,731,121,503,794,368
</span>or 3.63 x 10^20
Solving using Team B
Same concept, but choosing 55 from a pool of 80 (mathletes excluded)
80-C-25 = 80! / (55!(80-55!) = 363,413,731,121,503,794,368
or 3.63 x 10^20
As you can, we get the same answer for both.
B)
If none of the mathematicians are on team A, then we exclude the 20 and choose 45:
80-C-45 = 80! / (45!(80-45)!) = <span>5,790,061,984,745,3606,481,440
or 5.79 x 10^22
Note that, if you solve from the perspective of Team B (80-C-35), you get the same answer</span>
Answer:
A. It would be wider
Step-by-step explanation:
because... going up one and over one is much more steep then going up 1 over 3
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Answer:
h(-67) = 3158
General Formulas and Concepts:
Order of Operations: BPEMDAS
- Substitution and Evaluation
Step-by-Step Explanation:
Step 1: Define
h(x) = -49x - 125
h(-67)
Step 2: Solve
1. Substitute: h(-67) = -49(-67) - 125
2. Multiply: h(-67) = 3283 - 125
3. Subtract: h(-67) = 3158
