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frosja888 [35]
3 years ago
8

Please help answer this

Mathematics
1 answer:
Travka [436]3 years ago
4 0

Step-by-step explanation:

Girls - 2 over 5 times 250 over 1 = 100

Boys - 3 over 10 times 250 over 1 = 75

Total number of people - Girls + Boys

250 - 100 + 75

250 - 175 = 75

Therefore,

Adult = 75

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How do you find inverses
MrMuchimi

How To Find Inverses:

1. First, replace f(x) with y . ...

2. Replace every x with a y and replace every y with an x .

3. Solve the equation from Step 2 for y . ...

4. Replace y with f−1(x) f − 1 ( x ) . ...

5. Verify your work by checking that (f∘f−1)(x)=x ( f ∘ f − 1 ) ( x ) = x and (f−1∘f)(x)=x ( f − 1 ∘ f ) ( x ) = x are both true.

5 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
3 years ago
I need help: what is this?
givi [52]

Answer:

9

Step-by-step explanation:

first you have to do the 1+2 to get 3 then you divide 6 by 2 and get 3 then you multiply 3x3 to get 9!

3 0
3 years ago
The area of a square poster is 31 square inches. find the length of one side of the poster. Explain
kobusy [5.1K]

Answer:

5.57

Step-by-step explanation:

In order to find the are of a square you would use the equation <em>a²</em>. However, since you are finding the length of one side and its a square, you use the equation \sqrt{a}. The area of this square is 31 sq in., so you would replace a with 31. The equation then becomes \sqrt{31}<em>. </em>When out into a calculator, the answer is 5.567764... which rounded up to the nearest hundreth is 5.57.

5 0
3 years ago
Help? i'll mark brainliest if correct
Strike441 [17]

Answer:

36,686

Step-by-step explanation:

5.88 x 12 over 10 = 240

240 x 26 = 36,686

5 0
3 years ago
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