How many different ways may a set of 4 books be arraged side by side on a shell?

Step-by-step explanation:

we can simplify the stuff inside the parentheses to

[tex] \frac{ {x}^{3} }{2y} [/tex]

now we need to multiply it with itself, giving us

[tex] \frac{ {x}^{6} }{4 {y}^{2} } [/tex]

so yeah, D is the correct answer

Answer:

y - 2 = 1/3(x - 3)

Step-by-step explanation:

Point slope form is y - y1 = m(x - x1), where m is the slope and (x1, y1) is a point on the line.

Plug in the slope:

y - y1 = m(x - x1)

y - y1 = 1/3(x - x1)

Plug in the given point:

y - y1 = 1/3(x - x1)

y - 2 = 1/3(x - 3)

So, the correct answer is y - 2 = 1/3(x - 3)

Answer:

11.7

Step-by-step explanation:

39.1 - 9.2 = 29.9

29.9 - 18.2 = 11.7

Answer:

[tex]Pr =\frac{1}{3}[/tex]

Step-by-step explanation:

Given

[tex]S = \{(1, 1) (1, 2) (1, 3) (1, 4) (1, 5) (1, 6)(2, 1) (2, 2) (2, 3) (2, 4) (2, 5) (2, 6)[/tex]

[tex](3, 1) (3, 2) (3, 3) (3, 4) (3, 5) (3, 6)(4, 1) (4, 2) (4, 3) (4, 4) (4, 5) (4, 6)[/tex]

[tex](5, 1) (5, 2) (5, 3) (5, 4) (5, 5) (5, 6)(6, 1) (6, 2) (6, 3) (6, 4) (6, 5) (6, 6)\}[/tex] --- sample space

First, list out all outcomes whose sum is divisible by 5

[tex]A = \{(4,6), (5,5),(6,4)\}[/tex]

So, we have:

[tex]n(A) = 3[/tex]

Next, list out all outcomes that has an outcome of 5 in both rolls

[tex]B = \{(5,5)\}[/tex]

[tex]n(B) =1[/tex]

The required conditional probability is:

[tex]Pr =\frac{n(B)}{n(A)}[/tex]

[tex]Pr =\frac{1}{3}[/tex]

Answer:

1

Step-by-step explanation:

The average rate of change is the gradient _

Gradient = Rise / Run = (y2 - y1) / (x2 - x1)

From the table ;

x2 = 4 ; x1 = - 1 ; y2 = 5 ; y1 = 0

Gradient = (y2 - y1) / (x2 - x1) = (5-0) / (4 - - 1) = 5/5 = 1