Answer:
x = 3.3 and -0.3
Step-by-step explanation:
Given the expression
1/x-4 + x/x-2=2/x^2-6x+8
Find the LCM
x-2+x(x-4)/(x-4)(x-2) = 2/x^2-6x+8
Since the denominator, are the same they cancels out
x +1 + x(x-4) = 2
x+1 + x²-4x = 2
x²-3x + 1 - 2 = 0
x² - 3x - 1 = 0
Factorize
x =3±√9+4/2
x =3±√13/2
x = 3±3.6/2
x= 6.6/2 and -0.6/2
x = 3.3 and -0.3
Yes it is,
it is reflection like a mirror, but, it copy's the position and moves it upward
A:b=c:d
7.2:x=1:25. If u see two ratios equal each other, the product of a and d is equal to the product of b and c. This means that (7.2 times 25) is equal to x. This equals 180=x. The actual distance of the trail of 180 km. Hope this helps!
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.
1
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)
What's the probability of being dealt a royal flush in a five card hand from a standard deck of cards? (Note:
A royal flush is a 10, Jack, Queen, King, and Ace of the same suit. A standard deck has 4 suits, each with
13 distinct cards, including these five above.) (NB: The order in which the cards are dealt is unimportant,
and you keep each card as it is dealt -- it's not returned to the deck.)
The probability of drawing any card which could fit into some royal flush is 5/13. Once that card is taken
from the pack, there are 4 possible cards which are useful for making a royal flush with that first card, and
there are 51 cards left in the pack. therefore the probability of drawing a useful second card (given that the
first one was useful) is 4/51. By similar logic you can calculate the probabilities of drawing useful cards for
the other three. The probability of the royal flush is therefore the product of these numbers, or
5/13 * 4/51 * 3/50 * 2/49 * 1/48 = .00000154
1,500(1.02) means 1500×1.02
