Adeck of 5 cards (each carrying a distinct number from 1 to 5) is shuffled thoroughly. two cards are then removed one at time from the deck. what is the probability that the two cards are selected with the number on the first card being one higher than the number on the second card?

P= 1/5

Step-by-step explanation:

If we have  5 different cards (1,2,3,4,5) the number of favorable outcomes (that is: The first card, is bigger than the second only by one) is 4, and the total number of outcomes is 20 according to:

Variations of 5 elements in group of 2

V=5!/(5-2)! = 5*4*3!/3!=20

So if we have 4 possible outcomes, out of 20 (total outcomes) the probability will be 4/20 = 1/5

x =±sqrt(1/3)

step-by-step explanation:

x+4/6x=1/x

multiply each side by 6x to get rid of the fractions

6x*(x+4/6x)=1/x*6x

distribute

6x^2 +4 = 6

subtract 4 from each side

6x^2 +4-4 = 6-4

6x^2 =2

divide by 6

6x^2/6 = 2/6

x^2 = 1/3

take the square root of each side

sqrt(x^2) = ±sqrt(1/3)

x =±sqrt(1/3)

111

step-by-step explanation: