A standard deck of 52 cards contains 13 cards with hearts, 13 with diamonds, 13 with clubs, and 13 with spades. How many 7-card hands are possible with 2 hearts, 2 diamonds, 2 clubs, and 1 spade? Assume that the order of the cards
does not matter, and use the formula for combinations to find your answer.
474,552
6,169,176
49,353,408
133,784,560

133,784,560

Step-by-step explanation:

using the combinations formula: n!/(n-r!)r!, the equation would be 52!/52-7! x 7! which is simplified to 52!/45!7! which is equal to 133,784,560

tickets for the harlem globetrotters

show at michigan state university in

2010 cost $16, $23, or, for vip seats, $40.

total revenue = 16x + 23y + 40z

if 9 times as many $16 tickets were sold

as vip tickets,

x = 9z => x – 9z = 0, if 9 times as many $16 tickets were sold

as vip tickets,

x = 9z => x – 9z = 0, sales of all three kinds of tickets would

total $46,575.

16x + 23y + 40z = 46575, how many of each kind of ticket would

have been sold?

solve for x, y and z

x − 9z = 0

x − y − z = 55

16x + 23y + 40z = 46575

x = 1170

y = 985

z = 130

welcome!

well what you have to do is divide 84 by 6 so theanswer

step-by-step explanation:

84 divided by 6 = 14

and that lightbulb changing party must be freaking lit!