A standard deck of 52 playing cards has 4 suits. They are clubs, diamonds, spades , and hearts . The deck also has 13 different ranks. They are Ace (A), 2, 3, 4, 5, 6, 7, 8, 9, 10, Jack (J), Queen (Q), and King (K). Johnny draws 5 cards from a standard deck of 52 cards. If the cards are well shuffled, what is the probability of Johnny drawing 5 cards where 3 are one rank and 2 are another rank? (example: a 5 of clubs, a 5 of spades, a 5 of diamonds, a King of hearts, and a King of diamonds)
A. 1/108,290
B. 1/16,660
C. 1/199,920
D. 6/4,165

something that characterises a right triangle is the fact that we may use pythagoras' theorem to find any one of its sides given that we have the length of the other two sides. here, we have the length of the hypotenuse, √(87) yd, and one other side, √(23) yd, thus we can easily find the length of the remaining side using pythagoras' theorem as defined below:

c^2 = a^2 + b^2, where c is the length of the hypotenuse, and a and b are the lengths of the other two sides.

thus, given that c = √(87) and a = √(23), we may substitute these into the formula we defined above to get:

c^2 = a^2 + b^2

(√87)^2 = (√23)^2 + b^2 (substitute c = √87 and a = √23 into the equation)

87 = 23 + b^2 (evaluate (√87)^2 = 87 and (√23)^2 = 23 - remember that the square root of a number, squared will equal that very same number)

64 = b^2 (subtract 23 from both sides)

√64 = b (take the square root of both sides)

8 = b (evaluate √64 = 8)

therefor, the length of the unknown leg in the right triangle is 8 yd (first answer).

sadsadsadsad

step-by-step explanation: