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2 years ago

Look at the figure. Find TW

Mathematics

1 answer:

Dmitrij [34]2 years ago

6 0

Answer: NUMBER 9 I'm on gradpoint too

Step-by-step explanation:

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Four runners formed a relay team at Jan’s high school. The team completed the relay in 3.86 minutes. Each runner ran exactly the

xeze [42]

Answer: All four runners ran .96 seconds

Step-by-step explanation: The total time of the four runners, divided by the amount of runners.

3.86/4= .965

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2 years ago

Do all triangles have a hypotenuse, or only right triangles have a hypotenuse?

Nadusha1986 [10]

Only Right triangles

8 0

2 years ago

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Factor the difference of the perfect square. If the binomial will not factor, write prime.

KATRIN_1 [288]

Easy peasy, this is a representation of differnce of 2 perfect squares
a^2-b^2=(a-b)(a+b)

3 0

3 years ago

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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

A glacier moves 152 ft a year how many inches a day does it move (hint 356 days in a year)

matrenka [14]

It moves roughly 5 inches a day. (p.s you made a mistake, there is 365 days in a year not 356)

6 0

2 years ago

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