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1 year ago

Pls help me with this

Mathematics

1 answer:

Nady [450]1 year ago

6 0

Answer:

Step-by-step explanation:

9. the set containing all objects or elements and of which all other sets are subsets.

10. complement --> the amount in only one of teh sets

intersection --> the amount in both sets

11. 14 or -14

13. 46 1/2

14. 2 31/120

15. 1100% profit

16. 1200 gm

17. cube root of 7?

18. 6000 per year

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You are constructing a concrete landing pad for a helicopter. You have 4704 cubic feet of concrete

Sonja [21]

Answer:

Round it to the nearest hundredth

Step-by-step explanation:

6 0

3 years ago

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:

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

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3 years ago

A TV station has 11 minutes of commercials for each 60 minutes of air time. What percent

Julli [10]

Answer: Approximately 18.33%

Work Shown:

11/60 = 0.1833 approximately

This decimal converts to 18.33% when moving the decimal point over to the right two spots. Round this value however you need to.

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3 years ago

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bagirrra123 [75]

(8x + 2y) - (7x + 2y) = 30 - 24

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3 years ago

Read 2 more answers

Help me, please<br> im serious tho

murzikaleks [220]

Answer:

2/6 or 1/3 so color 2 out of the six squares

Step-by-step explanation:

1/2 - 1/6 is equal to 3/6 - 1/6 so 2/6

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3 years ago