3 years ago

If two cards are randomly chosen from a standard deck, what is the probability that a 6 is chosen, replaced, and another 6 is ch

osen?
A. 16/2652
B. 16/2704
C. 36/2704
D. 1/16

Mathematics

1 answer:

lord [1]3 years ago

7 0

$|\Omega|=52^2=2704\\ |A|=4^2=16\\\\ P(A)=\dfrac{16}{2704}\Rightarrow{\text{B}}$

You might be interested in

1. Determine the zeros of f(x) = x3 – 3x2 – 16x + 48? Somebody explain it please?:/

inn [45]

$x^{2} - 3x^{2} - 16x + 48 = 0$

$x^{2} (x - 3) - 16(x - 3) = 0$

$(x - 3) ( x^{2} - 16) = 0$

$(x - 3) (x - 4) (x + 4) = 0$

$x = -4, 3, 4$

The zeros are: -4, 3, 4

6 0

3 years ago

What is 2÷8 , what will be the answer

AleksAgata [21]

The correct answer is .25

4 0

3 years ago

Read 2 more answers

A rectangular pool is twice as

dedylja [7]

Step-by-step explanation:

L=2B

P=2(L+B)

48=2(2B+B)

48=2(3B)

48=6B

B=8yd

L=2*8

=16yd

8 0

3 years ago

Read 2 more answers

Explain how you got the answer

Gnoma [55]

Answer:

$\huge\boxed{Center = (3,3),Radius = 2\sqrt{15} }$

Step-by-step explanation:

<u><em>Given equation is</em></u>

$x^2 + y^2 -6x-6y -42 = 0$

Adding 42 to both sides

$x^2 + y^2 -6x-6y = 42\\$

Completing squares

$x^2 -6x+y^2 -6x = 42\\(x)^2 - 2(x)(3) +y^2 - 2(y)(3) = 42$

Adding (3)² => 9 and (3)² => 9 to both sides

$(x-3)^2+(y-3)^2 = 42+9+9\\(x-3)^2 + (y-3)^2 = 60\\(x-3)^2 (y-3)^2 = (2{\sqrt{15})^2}$

Comparing it with $(x-h)^2+(y-k)^2 = r^2$ where Center = (h,k) and Radius = r

We get:

Center = (3,3)

Radius = $2\sqrt{15}$

7 0

3 years ago

C(x)=4x-2 and d(x)=x^(2)+5, what is (c*d)(x)

lawyer [7]

Answer:4x^3-2x^2+20x-10

Step-by-step explanation:

5 0

3 years ago