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algol13
2 years ago
10

Leveled

Mathematics
1 answer:
Marat540 [252]2 years ago
8 0

Answer:

  48.6

Step-by-step explanation:

Put the values in place of the variables and simplify according to the Order of Operations in the usual way.

  10.2x +9.4y

  = 10.2(2) +9.4(3) . . . . . 2 is substituted for x; 3 is substituted for y

  = 20.4 +28.2

  = 48.6

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Which ordered pairs is a solution to −5x + 3y > 12?
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<u>Answer </u><u>:</u><u>-</u>

Given inequality ,

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From the options ,

<u>When </u><u>x </u><u>is </u><u>3</u><u> </u><u>and </u><u>y </u><u>=</u><u> </u><u>9</u>

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<u>When</u><u> </u><u>x </u><u>is </u><u>-</u><u>5</u><u> </u><u>y</u><u> </u><u>is </u><u>5</u><u> </u>

  • y > 5/3*-5 + 4
  • y > -8.3 + 4
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<u>When </u><u>x </u><u>is </u><u>3</u><u> </u><u>y </u><u>is </u><u>-</u><u>6</u><u> </u>

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<u>When </u><u>x </u><u>is </u><u>-</u><u>2</u><u> </u><u>y </u><u>is </u><u>-</u><u>5</u><u> </u>

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<u>When</u><u> </u><u>x </u><u>is </u><u>-</u><u>6</u><u> </u><u>y </u><u>is </u><u>0</u><u> </u>

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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.

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v = x^3

Make x the subject

x = v^\frac{1}{3}

So, the cumulative density is:

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

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F(x) = \int\limits^{v^\frac{1}{3}}_9 1\  dx

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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
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