Answer:
x=36 degrees
Step-by-step explanation:
Remember that all angles of a triangle must add up to 180 degrees due to the Angle Addition Postulate. So let's plug it in! x+65+79=180
Now you solve, x=180-79-65
x=36!
x=36 degrees
Answer:
After 45.7 years
Step-by-step explanation:
∵ r = 1/2% per year
∵ P0 = $40
∵ Pn = $50
∵ Pn = P0 ( 1 + r/100)^n-1
∵ P0 is initial value , Pn is the value after n years , is the rate
∴ 50 = 40 (1 + 0.5/100)^n-1
∴ 50/40 = (1.005)^n-1
∴ 1.25 = (1.005)^n-1 ⇒ insert ln in both sides
∴ ln(1.25) = (n-1)ln(1.005)
∴ n - 1 = ln(1.25)/ln(1.005)
∴ n = ln(1.25)/ln(1.005) + 1 = 45.7 years
Answer:
1/4
Step-by-step explanation:
First, add up the marbles:
2 + 9 + 7 = 18
Multiply by 2 because you pick two:
18 × 2 = 36
Simplify:
9/36 = 3/12 = 1/4
Hope this helped.
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)
Answer:
Step-by-step explanation:
first part: 3/25
second part: 1/25
third part: 4/25
