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3 years ago

Solve for x: -3 + x = -22

Mathematics

1 answer:

Tanya [424]3 years ago

3 0

Answer:

x= -19

Step-by-step explanation:

-3 + x = -22

Add 3 to each side

-3+3 +x = -22 +3

x = -19

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Please help im really confused<br><br> Which expression is equivalent to 4(7 + 8)?

Hatshy [7]

Answer:

With the answer choices:

A. 11 + 8

B. 11 + 12

C. 28 + 12

D. 28 + 32

The answer would be 28+32.

Step-by-step explanation:

Use the distributive law.

4 * 7 = 28

4 * 8 = 32

Hope this helps!

8 0

2 years ago

PLEASE HELP!! look at the image below

Sladkaya [172]

Answer:

A right 4 down 5

Step-by-step explanation:

counting the grid spaces from any point for example R and going to the point R' (R prime) will give you the translation

8 0

3 years ago

Suppose that the length of a side of a cube X is uniformly distributed in the interval 9

Nastasia [14]

Answer:

$f(v) = \left \{ {{\frac{1}{3}v^{-\frac{2}{3}}\ 9^3 \le v \le 10^3} \atop {0, elsewhere}} \right.$

Step-by-step explanation:

Given

$9 < x < 10$ --- interval

Required

The probability density of the volume of the cube

The volume of a cube is:

$v = x^3$

For a uniform distribution, we have:

$x \to U(a,b)$

and

$f(x) = \left \{ {{\frac{1}{b-a}\ a \le x \le b} \atop {0\ elsewhere}} \right.$

$9 < x < 10$ implies that:

$(a,b) = (9,10)$

So, we have:

$f(x) = \left \{ {{\frac{1}{10-9}\ 9 \le x \le 10} \atop {0\ elsewhere}} \right.$

Solve

$f(x) = \left \{ {{\frac{1}{1}\ 9 \le x \le 10} \atop {0\ elsewhere}} \right.$

$f(x) = \left \{ {{1\ 9 \le x \le 10} \atop {0\ elsewhere}} \right.$

Recall that:

$v = x^3$

Make x the subject

$x = v^\frac{1}{3}$

So, the cumulative density is:

$F(x) = P(x < v^\frac{1}{3})$

$f(x) = \left \{ {{1\ 9 \le x \le 10} \atop {0\ elsewhere}} \right.$ becomes

$f(x) = \left \{ {{1\ 9 \le x \le v^\frac{1}{3} - 9} \atop {0\ elsewhere}} \right.$

The CDF is:

$F(x) = \int\limits^{v^\frac{1}{3}}_9 1\ dx$

Integrate

$F(x) = [v]\limits^{v^\frac{1}{3}}_9$

Expand

$F(x) = v^\frac{1}{3} - 9$

The density function of the volume F(v) is:

$F(v) = F'(x)$

Differentiate F(x) to give:

$F(x) = v^\frac{1}{3} - 9$

$F'(x) = \frac{1}{3}v^{\frac{1}{3}-1}$

$F'(x) = \frac{1}{3}v^{-\frac{2}{3}}$

$F(v) = \frac{1}{3}v^{-\frac{2}{3}}$

So:

$f(v) = \left \{ {{\frac{1}{3}v^{-\frac{2}{3}}\ 9^3 \le v \le 10^3} \atop {0, elsewhere}} \right.$

8 0

2 years ago

What is 1 divided by 1/8?

3241004551 [841]

the answer is number 8

8 0

3 years ago

Read 2 more answers

If a pair of standard dice are rolled, what is the probability that one of the dice will be a 3 and the other a 4?

eimsori [14]

Answer:

0.055

Step-by-step explanation:

file is attached with explanation

7 0

2 years ago