An ordinary deck of playing cards contains 52 cards, 26 red and 26 black. If a card is dealt to each of 2 players,. Find in how many different ways this can be done if the following occur.
a. both cards are red. ways
b. both cards are black ways
c. one card is black and the other is red. way

(a)650 ways

(b)650 ways

(c)676 ways

Step-by-step explanation:

There are 26 red and 26 black cards.

If a card is dealt to each of 2 players, we want to find out how many different ways this can be done.

(a)Both cards are red

If both cards are red:

The first red card can be dealt in 26 ways.

The second red card can be dealt in 25 ways.

Therefore: Both Red cards can be dealt in: 26 X 25 = 650 ways

(b)Both cards are black

If both cards are black:

The first black card can be dealt in 26 ways.

The second black card can be dealt in 25 ways.

Therefore: Both black cards can be dealt in: 26 X 25 = 650 ways

(c)One card is black and the other is red.

The red card can be dealt in 26 ways.

The black card can be dealt in 26 ways.

Therefore: Both cards can be dealt in: 26 X 26 = 676 ways

75

step-by-step explanation:

you do x-29=46 (the number of cantaloupes sold)

in order to find x, you have to find what minus 29 equals 46, meaning you'd have to add 29 to 46.

29+46=75

if you replace x with 75, the equation looks like:

75-29=46

x=75

solution:

the given function is ,

      =

the three roots of this quadratic function is , 0 ,8 and 10.all the roots are real and rational.

if a quadratic equation is of the form ,

sum of the roots = a + b +c= 0+8+10=18=  

product of roots taken two at a time = a b + b c+ c a = 0+80+0=80 =  

product of roots = a b c = 0=  

there are 11 doors in a hall starting from 0 to 11.the entry in the hall is permitted through gate number 0 and exit is allowed through only gate number 8 and 10.so, three roots(0,8,10) of quadratic function is given by entry and exit route of this hall.