Explain which trig functions and identities produce circles, ellipses and hyperbolas.
1 answer:
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Choice A is the answer which is the point (1,-1)
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How I got this answer:
Plug each point into the inequality. If you get a true statement after simplifying, then that point is in the solution set and therefore a solution. Otherwise, it's not a solution.
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checking choice A
plug in (x,y) = (1,-1)
This is true because -3 is equal to itself. So this is the answer.
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checking choice B
plug in (x,y) = (2,4)
This is false because 0 is not to the left of -3, nor is 0 equal to -3. We can cross this off the list.
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checking choice C
plug in (x,y) = (-2,3)
This is false because 7 is not to the left of -3, nor is 7 equal to -3. We can cross this off the list.
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checking choice D
plug in (x,y) = (3,4)
This is false because -2 is not to the left of -3, nor is -2 equal to -3. We can cross this off the list.
Answer:
Step-by-step explanation:
The total number of permutations of boys and girls on the team are:
¹⁶P₅*¹³P₄
1.
Bob will be one of the fixed boys to be picked. Hence, actually 4 boys are to be picked from 15. The permutations of girls being picked remains the same.
Probability = Permutations with Bob as one of the boys / Total permutations
Probability = (¹⁵P₄*¹³P₄) / (¹⁶P₅*¹³P₄) = ¹⁵P₄ / ¹⁶P₅
Probability = = 15! / 16! = 1/16
2.
Now, Bob is one of the fixed boys and Jane is one of the fixed girls. Hence, actually 4 boys are to be picked from 15 and 3 girls are to be picked from 12.
Probability = Permutations with bob as one of the boys and jane as one of the girls / Total permutations
Probability = (¹⁵P₄*¹²P₃) / (¹⁶P₅*¹³P₄) = (1/16)*(1/13) = 1/208
3.
Now, the probability that at least Jane or Bob will be picked has been asked. This probability is a combination of three probabilities:
Probability = (Probability that only Bob will be picked) + (Probability that only Jane will be picked) + (Probability that both will be picked)
Probability = 1/16 + 1/13 + 1/208 = 0.123
4.
Total teams possible = ¹⁶P₅*¹³P₄ = 8994585600 teams are possible