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2 years ago

15

ROUND TO THE NEAREST ONES PLACE What is the volume of right circular cone rounded to the nearest whole cubic millimeter? Show wo

rk to support your answer. 4 mm 9 mm

1 answer:

MA_775_DIABLO [31]2 years ago

6 0

V= π r^2 h/3
V= π 4^2 9/3
V= π 16 3
V= 150.72
V=151

Explanation
Formula is the first line. R is the radius with us 4 and h is the height which is 9. Substituting these values into the formula will get you pi times 4 squared times 9 divided by 3. When simplified this gets pi times 16 times 3. Using the approximate value of pi which is 3.14, multiple it with 16 and 3 will get you 150.72, round this to 150.8, which rounds to 151.
Hope this helps

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Suppose that the length of a side of a cube X is uniformly distributed in the interval 9

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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:

$f(v) = \left \{ {{\frac{1}{3}v^{-\frac{2}{3}}\ 9^3 \le v \le 10^3} \atop {0, elsewhere}} \right.$

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Strike441 [17]

Answer:

The 13th term is 81<em>x</em> + 59.

Step-by-step explanation:

We are given the arithmetic sequence:

$\displaystle -3x -1, \, 4x +4, \, 11x + 9 \dots$

And we want to find the 13th term.

Recall that for an arithmetic sequence, each subsequent term only differ by a common difference <em>d</em>. In other words:

$\displaystyle \underbrace{-3x - 1}_{x_1} + d = \underbrace{4x + 4} _ {x_2}$

Find the common difference by subtracting the first term from the second:

$d = (4x+4) - (-3x - 1)$

Distribute:

$d = (4x + 4) + (3x + 1)$

Combine like terms. Hence:

$d = 7x + 5$

The common difference is (7<em>x</em> + 5).

To find the 13th term, we can write a direct formula. The direct formula for an arithmetic sequence has the form:

$\displaystyle x_n = a + d(n-1)$

Where <em>a</em> is the initial term and <em>d</em> is the common difference.

The initial term is (-3<em>x</em> - 1) and the common difference is (7<em>x</em> + 5). Hence:

$\displaystyle x_n = (-3x - 1) + (7x+5)(n-1)$

To find the 13th term, let <em>n</em> = 13. Hence:

$\displaystyle x_{13} = (-3x - 1) + (7x + 5)((13)-1)$

Simplify:

$\displaystyle \begin{aligned}x_{13} &= (-3x-1) + (7x+5)(12) \\ &= (-3x - 1) +(84x + 60) \\ &= 81x + 59 \end{aligned}$

The 13th term is 81<em>x</em> + 59.

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