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3 years ago

I will give BRAINLIEST to whoever is correct.

Mathematics

1 answer:

maw [93]3 years ago

8 0

Remember, you can do anything to an equation as long as you do it to both sides

we hates fractions
times both sides by 3
(we could have added 1/3b to both sides but I'm lazy)

4b+6=12-b
add b to both sides
5b+6=12
minus 6 both sides
5b=6
divide both sides by 5
$b=\frac{6}{5}$

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Help soon eeeeeeeeeeeeeeeeeee

igomit [66]

3/9 because it is the same as 1/3 or 2/6.

6 0

2 years ago

Read 2 more answers

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

15d^3+30d^2+30d/3d3 what is the answer if you divide the following

lawyer [7]

When you separate them
$\frac{15d^{3} }{ {3d}^{3} } + \frac{30d^{2} }{ {3d}^{3} } + \frac{30d }{ {3d}^{3} } = x$
Solve them separately:
$5 + \frac{10}{d} + \frac{10}{ {d}^{2} } = x$
Remove the denominators:
$5 + 10 {d}^{ - 1} + 10 {d}^{ - 2} = x$
Find GCF (Greatest Common Factor):
$5(1 + 2 {d}^{ - 1} + 2 {d}^{ - 2} ) = x$

5 0

3 years ago

2r = 72 please help me

noname [10]

you divide 72 by 2 to get r by itself and get 36

so r=36

6 0

3 years ago

Read 2 more answers

To solve a problem, a student had to find a certain angle θ. The student got to the

stira [4]

Answer: