All you are basically doing is making ratios.
Basketball to football would be 5 to 11 or 5:11 or 5/11.
What you do is in the order you said it (basketball to football) you find that info and make it a ratio. A ratio must be simplified and can be written 3 ways, as shown above.
Hockey to lacrosse would be 6 to 10 and simplified would be 3 to 5.
Try doing 3 and 4 and 5 on your own to help you understand it more:) if you have any questions tell me.
Answer:416
Step-by-step explanation: Based on the given problem above, it requires a trigonometric solution. So here it goes:
The climb distance is:200 * Sec [ArcTan [10/200]] = 10√ 401 km
and the decent distance is:300 * Sec [ArcTan [10/300]] = 10√901 km
So add the given results above with 500 km and this will be the additional distance that plane moves through the air.
The answer would be in the unit meters.
((10√401 +10√901−500)∗1000)=416 meters
Hope this is the answer that you are looking for.
Answer:
=4x6−x3+x−3
Step-by-step explanation:
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)
