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2 years ago

A children's ride at an amusement park has a rule that a person must be shorter than 4 feet tall to ride it.

Mathematics

1 answer:

Juliette [100K]2 years ago

8 0

Answer:

Step-by-step explanation:

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The product of two consecutive odd integers is 72. Write a quadratic equation to find the integers,

zimovet [89]

<u>Answer:</u>

$x^2 + x - 72 = 0$

<u>Explanation:</u>

We are given product of two consecutive odd integers is equal to 72.

Let one integer be x , then since these are consecutive and odd so the other integer will be equal to x + 1.

Their product is equal to 72. So,

x * (x + 1) = 72

=> $x^2 + x = 72$

=> $x^2 + x - 72 = 0$ is the required quadratic equation.

5 0

3 years ago

Find the volume of the pyramid with a square base, where the area of the base is 16.3ft and the height of the pyramid is 13.7ft

Deffense [45]

Answer:

oooooogg

Step-by-step explanation:

4 0

3 years ago

What number should be added to 77 to make the sum 0?

labwork [276]

You should add -77 to 77 because it will cancel out.

5 0

2 years ago

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Please help me it’s due and i appreciate it

kakasveta [241]

16=16 zzzzzzzzzzzzzzz

3 0

3 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