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2 years ago

It is known that y^24 − 256 = 0. Rewrite this equation in terms of x if x = y^4.

Mathematics

2 answers:

vladimir2022 [97]2 years ago

6 0

Answer and step-by-step explanation:

> y²⁴- 256 = 0
> (y⁴)
x ${}^{6}$ -256 = 0

x ${}^{6}$ = 256

qaws [65]2 years ago

5 0

Answer:

> y²⁴- 256 = 0

> (y⁴)

x{}^{6}6 -256 = 0

x{}^{6}6 = 256

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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

-14=-20y-7x 10y+4=2x Solve by Elimination please!

iogann1982 [59]

Answer:

x = 2 y = 0

Step-by-step explanation:

8 0

2 years ago

Which value of x is a solution to x2 = 16?

Helga [31]

Answer:

Step-by-step explanation:

if we take the 2 that is in the R.H.S and put it in L.H.S

it becomes 16÷2=8

8 0

3 years ago

Which equation could be used to solve this problem? The sum of two consecutive integers is 141. Find the integers. A. n + 1= 141

Kamila [148]

D. works because if you work it out that means 2n=140 and then divide by 2 which means n=70 and the other number is 71

7 0

3 years ago

What is the MAD of this data set? 2, 3, 5, 7, 8, 8, 9

Tasya [4]

Answer:

do you mean the mode cause that would be 8 and the mean would be 6 and median would be 7 and the range would also be 7

the mad would be 2.2857 idk what your rounding to but i hope this helps

Step-by-step explanation:

3 0

3 years ago

Read 2 more answers

It is known that y^24 − 256 = 0. Rewrite this equation in terms of x if x = y^4.​

It is known that y^24 − 256 = 0. Rewrite this equation in terms of x if x = y^4.