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3 years ago

Determine whether the events are independent or dependent: rolling a number cube and tossing a coin

Mathematics

1 answer:

kirill115 [55]3 years ago

5 0

Its independent because it does not depend on the answer before

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How would I start to solve number 15? Would I use distributive property?

natita [175]

$-5-x=2x-(4x+6)$

Using the distributive property is the correct first step. $(4x+6)$ is being multiplied by -1, so distributing gives

$-5-x=2x+(-4x-6)$
$-5-x=2x+((-4x)+(-6))$

Now, using the associative property of addition, you get

$-5-x=(2x+(-4x))+(-6)$
$-5-x=(2x-4x)-6$

Simplify the right hand side:

$-5-x=-2x-6$

Add 5 and $2x$ to both sides.

$(2x+5)-5-x=(2x+5)-2x-6$

On the left side, use the associative property. On the right, commutativity and associativity:

$2x+(5-5)-x=5+(2x-2x)-6$

Simplify.

$2x-x=-1$
$x=-1$

8 0

3 years ago

Factor completely:<br> 21) x2 – 2x - 8

mojhsa [17]

Answer: x = -2 and 4

Step-by-step explanation:

x^2 - 2x - 8 = 0

(x - 4)(x + 2) = 0

4 0

4 years ago

A meatball recipe calls for 2 cups of breadcrumbs for every 5 cups of ground meat. How many cups of breadcrumbs are required for

Ghella [55]

Answer:

1/4 cup for every cup

Step-by-step explanation:

If you divide 2 by 5, you would get this answer. Sorry if it's wrong!

6 0

3 years ago

The total employment for police officers in 2006 was 648,000. If there are 719,000 police officers in 2016, what will the percen

Sever21 [200]

Answer:

The answer is A. 12%

Step-by-step explanation:

6 0

3 years ago

Find the standard equation of a sphere that has diameter with the end points given below. (3,-2,4) (7,12,4)

DiKsa [7]

Answer:

The standard equation of the sphere is $(x-5)^{2} + (y-5)^{2} + (z-4)^{2} = 53$

Step-by-step explanation:

From the question, the end point are (3,-2,4) and (7,12,4)

Since we know the end points of the diameter, we can determine the center (midpoint of the two end points) of the sphere.

The midpoint can be calculated thus

Midpoint = $(\frac{x_{1} + x_{2} }{2}, \frac{y_{1} + y_{2} }{2}, \frac{z_{1} + z_{2} }{2})$

Let the first endpoint be represented as $(x_{1}, y_{1}, z_{1})$ and the second endpoint be $(x_{2}, y_{2}, z_{2})$ .

Hence,

Midpoint = $(\frac{x_{1} + x_{2} }{2}, \frac{y_{1} + y_{2} }{2}, \frac{z_{1} + z_{2} }{2})$

Midpoint = $(\frac{3 + 7 }{2}, \frac{-2+12 }{2}, \frac{4 + 4 }{2})$

Midpoint = $(\frac{10 }{2}, \frac{10}{2}, \frac{8 }{2})\\$

Midpoint = $(5, 5, 4)$

This is the center of the sphere.

Now, we will determine the distance (diameter) of the sphere

The distance is given by

$d = \sqrt{(x_{2} - x_{1})^{2} +(y_{2} - y_{1})^{2} + (z_{2}- z_{1})^{2} }$

$d = \sqrt{(7 - 3)^{2} +(12 - -2)^{2} + (4- 4)^{2}$

$d = \sqrt{(4)^{2} +(14)^{2} + (0)^{2}$

$d = \sqrt{16 +196 + 0$

$d =\sqrt{212}$

$d = 2\sqrt{53}$

This is the diameter

To find the radius, r

From $Radius = \frac{Diameter}{2}$

$Radius = \frac{2\sqrt{53} }{2}$

∴ $Radius = \sqrt{53}$

r = $\sqrt{53}$

Now, we can write the standard equation of the sphere since we know the center and the radius

Center of the sphere is $(5, 5, 4)$

Radius of the sphere is $\sqrt{53}$

The equation of a sphere of radius r and center $(h,k,l)$ is given by

$(x-h)^{2} + (y-k)^{2} + (z-l)^{2} = r^{2}$

Hence, the equation of the sphere of radius $\sqrt{53}$ and center $(5, 5, 4)$ is

$(x-5)^{2} + (y-5)^{2} + (z-4)^{2} = \sqrt{(53} )^{2}$

$(x-5)^{2} + (y-5)^{2} + (z-4)^{2} = 53$

This is the standard equation of the sphere

6 0

3 years ago