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2 years ago

Help me help me help me help me

Mathematics

1 answer:

ioda2 years ago

4 0

Answer: B

We aren't subtracting the pages the person reads so it's not A.

It's not C since we're not dividing either

And we aren't multiplying for D.

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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

Pythagorean theorem digital escape can you find the length and area and type the correct code?

lara31 [8.8K]

Answer:

13825

Step-by-step explanation:

4 0

2 years ago

sand falls from an overhead bin and accumulates in a conical pile with a radius that is always four times its height. suppose th

blagie [28]

Answer:

$\frac{dv}{dt} =7239.168 cm/sec$

Step-by-step explanation:

From the question we are told that:

Rate $\frac{dh}{dt}=1cm$

Height $h=12cm$

Radius $r=4h$

Generally the equation for Volume of Cone is mathematically given by

$V=\frac{1}{3}\pi r^2h$

$V=\frac{1}{3}\pi (4h)^2h$

Differentiating

$\frac{dv}{dt} =\frac{16}{3}\pi3h^2\frac{dh}{dt}$

$\frac{dv}{dt} =\frac{16}{3}*3.142*3*12^2*1$

$\frac{dv}{dt} =7239.168 cm/sec$

7 0

3 years ago

I WILL GIVE BRAINLIEST TO THE FIRST TO ANSWER

dimaraw [331]

Answer: Linear

Step-by-step explanation: A linear function can have different rates of change over different intervals.

(I don't know if this is correct, but this is my understanding of it.)

5 0

3 years ago

Solve for n -3n-5=16

Nezavi [6.7K]

Answer:

n = -7

Step-by-step explanation:

Solve for n:

-3 n - 5 = 16

Hint: | Isolate terms with n to the left-hand side.

Add 5 to both sides:

(5 - 5) - 3 n = 5 + 16

Hint: | Look for the difference of two identical terms.

5 - 5 = 0:

-3 n = 16 + 5

Hint: | Evaluate 16 + 5.

16 + 5 = 21:

-3 n = 21

Hint: | Divide both sides by a constant to simplify the equation.

Divide both sides of -3 n = 21 by -3:

(-3 n)/(-3) = 21/(-3)

Hint: | Any nonzero number divided by itself is one.

(-3)/(-3) = 1:

n = 21/(-3)

Hint: | Reduce 21/(-3) to lowest terms. Start by finding the GCD of 21 and -3.

The gcd of 21 and -3 is 3, so 21/(-3) = (3×7)/(3 (-1)) = 3/3×7/(-1) = 7/(-1):

n = 7/(-1)

Hint: | Simplify the sign of 7/(-1).

Multiply numerator and denominator of 7/(-1) by -1:

Answer: n = -7

7 0

3 years ago

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