Answer: B. The United States might not have been able to stop the attacks, but it could have made a better effort to do so.
Step-by-step explanation:
I think that this is a combination problem. From the given, the 8 students are taken 3 at a time. This can be solved through using the formula of combination which is C(n,r) = n!/(n-r)!r!. In this case, n is 8 while r is 3. Hence, upon substitution of the values, we have
C(8,3) = 8!/(8-3)!3!
C(8,3) = 56
There are 56 3-person teams that can be formed from the 8 students.
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Answer:
The radius of a sphere is 2 millimeters
The surface area of a sphere is 50.24 square millimeters.
The circumference of the great circle of a sphere is 12.56 millimeters.
Step-by-step explanation:
<u><em>Verify each statement</em></u>
case A) The radius of a sphere is 8 millimeters
The statement is false
we know that
The volume of the sphere is equal to
we have
substitute and solve for r
case B) The radius of a sphere is 2 millimeters
The statement is True
(see the case A)
case C) The circumference of the great circle of a sphere is 9.42 square millimeters
The statement is false
The units of the circumference is millimeters not square millimeters
The circumference is equal to
we have
substitute
case D) The surface area of a sphere is 50.24 square millimeters.
The statement is True
Because
The surface area of the sphere is equal to
we have
substitute
case E) The circumference of the great circle of a sphere is 12.56 millimeters.
The statement is true
see the case C
case F) The surface area of a sphere is 25.12 square millimeters
The statement is false
because the surface area of the sphere is
see the case D
Cards are drawn, one at a time, from a standard deck; each card is replaced before the next one is drawn. Let X be the number of draws necessary to get an ace. Find E(X) is given in the following way
Step-by-step explanation:
- From a standard deck of cards, one card is drawn. What is the probability that the card is black and a
jack? P(Black and Jack) P(Black) = 26/52 or ½ , P(Jack) is 4/52 or 1/13 so P(Black and Jack) = ½ * 1/13 = 1/26
- A standard deck of cards is shuffled and one card is drawn. Find the probability that the card is a queen
or an ace.
P(Q or A) = P(Q) = 4/52 or 1/13 + P(A) = 4/52 or 1/13 = 1/13 + 1/13 = 2/13
- WITHOUT REPLACEMENT: If you draw two cards from the deck without replacement, what is the probability that they will both be aces?
P(AA) = (4/52)(3/51) = 1/221.
- WITHOUT REPLACEMENT: What is the probability that the second card will be an ace if the first card is a king?
P(A|K) = 4/51 since there are four aces in the deck but only 51 cards left after the king has been removed.
- WITH REPLACEMENT: Find the probability of drawing three queens in a row, with replacement. We pick a card, write down what it is, then put it back in the deck and draw again. To find the P(QQQ), we find the
probability of drawing the first queen which is 4/52.
- The probability of drawing the second queen is also 4/52 and the third is 4/52.
- We multiply these three individual probabilities together to get P(QQQ) =
- P(Q)P(Q)P(Q) = (4/52)(4/52)(4/52) = .00004 which is very small but not impossible.
- Probability of getting a royal flush = P(10 and Jack and Queen and King and Ace of the same suit)
