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yuradex [85]
3 years ago
13

(26 points)

Mathematics
2 answers:
Fed [463]3 years ago
8 0

Answer:

Given 4x+7=6x-5

addition property so 4x+12=6x

then subtraction property so 2x=12

divison property x=6

All are equality

ella [17]3 years ago
3 0

Answer:

x = 6

Step-by-step explanation:

4x + 7 = 6x - 5

subtract 7 from both sides (Subtraction Property)

4x = 6x -12

subtract 6x from both sides (Subtraction Property)

-2x = -12

divide by -2 (Division Property)

x = 6

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Can you please tell me what is the area of the given triangle?

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AnnZ [28]

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Step-by-step explanation:

7 0
2 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.

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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
What is x+x+x+x+x+x+x+x+x+x+b+b+b+b+3=
xz_007 [3.2K]

Answer:

10x+4b+3

Step-by-step explanation:

x+x+x+x+x+x+x+x+x+x+b+b+b+b+3

There are 10 x

4 b

and one number

10x+4b+3

8 0
2 years ago
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