Answer:
1.236 × 10^(-3)
Step-by-step explanation:
Let A be the event that the person is a future terrorist
Let B the event that the person is identified as a terrorist
We are told that there are 1,000 future terrorists in a population of 400 million. Thus, the Probability that the person is a terrorist is;
P(A) = 1000/400000000
P(A) = 0.0000025
P(A') = 1 - P(A)
P(A') = 1 - 0.0000025
P(A') = 0.9999975
We are told that the system has a 99% chance of correctly identifying a future terrorist. Thus; P(B|A) = 0.99
Thus, P(B'|A) = 1 - P(B|A)
P(B'|A) = 1 - 0.99
P(B'|A) = 0.01
We are told that there is a 99.8% chance of correctly identifying someone who is not a future terrorist. Thus; P(B'|A') = 0.998
Hence: P(B|A') = 1 - P(B'|A')
P(B|A') = 1 - 0.998
P(B|A') = 0.002
We want to find the probability that someone who is identified as a terrorist, is actually a future terrorist. This is represented by: P(A|B)
We can find it from bayes theorem as follows;
P(A|B) = [P(B|A) × P(A)]/[(P(B|A) × P(A)) + (P(B|A') × P(A')]
Plugging in the relevant values;
P(A|B) = [0.99 × 0.0000025]/[(0.99 × 0.0000025) + (0.002 × 0.9999975)]
P(A|B) = 0.00123597357 = 1.236 × 10^(-3)
We have
so this is indeed a difference of squares. Factorizing would give
making the answer D.
We can further factor the first term here and write
but that's clearly out of the scope of this question.
A combination is an unordered arrangement of r distinct objects in a set of n objects. To find the number of permutations, we use the following equation:
n!/((n-r)!r!)
In this case, there could be 0, 1, 2, 3, 4, or all 5 cards discarded. There is only one possible combination each for 0 or 5 cards being discarded (either none of them or all of them). We will be the above equation to find the number of combination s for 1, 2, 3, and 4 discarded cards.
5!/((5-1)!1!) = 5!/(4!*1!) = (5*4*3*2*1)/(4*3*2*1*1) = 5
5!/((5-2)!2!) = 5!/(3!2!) = (5*4*3*2*1)/(3*2*1*2*1) = 10
5!/((5-3)!3!) = 5!/(2!3!) = (5*4*3*2*1)/(2*1*3*2*1) = 10
5!/((5-4)!4!) = 5!/(1!4!) = (5*4*3*2*1)/(1*4*3*2*1) = 5
Notice that discarding 1 or discarding 4 have the same number of combinations, as do discarding 2 or 3. This is being they are inverses of each other. That is, if we discard 2 cards there will be 3 left, or if we discard 3 there will be 2 left.
Now we add together the combinations
1 + 5 + 10 + 10 + 5 + 1 = 32 choices combinations to discard.
The answer is 32.
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Note: There is also an equation for permutations which is:
n!/(n-r)!
Notice it is very similar to combinations. The only difference is that a permutation is an ORDERED arrangement while a combination is UNORDERED.
We used combinations rather than permutations because the order of the cards does not matter in this case. For example, we could discard the ace of spades followed by the jack of diamonds, or we could discard the jack or diamonds followed by the ace of spades. These two instances are the same combination of cards but a different permutation. We do not care about the order.
I hope this helps! If you have any questions, let me know :)
If a pentagon is regular, then all of the sides are the same length. This means that
So, if 
, the lengths of the sides evaluate to
So, all the sides of the pentagon are 21 units long, and AB is 21 units long as well, because all sides are the same length.
Answer:
=x^2 +4x+11
Step-by-step explanation:
f(x) = x^2 + 7,
Replace x with x+2
f(x+2) = (x+2)^2 + 7
= (x+2)(x+2) +7
FOIL
= x^2 +2x+2x+4 +7
Combine like terms
=x^2 +4x+11
