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3 years ago

Cecil had 6 ice cubes. He put 1 ice cube in each glass. In how many glasses did Cecil put ice cubes?

Mathematics

2 answers:

Kamila [148]3 years ago

7 0

He only puts 1 in each glass and he has six. The answer is 6.

julsineya [31]3 years ago

6 0

The answer would be 6

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Solve: 2/x+1/3=1/y for y.

Liono4ka [1.6K]

Answer:

y=3x/6+x

Step-by-step explanation:

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3 years ago

If 30+12x represents the area of a partitioned rectangle, what could be expressions for the side lengths? List at least 5 possib

astra-53 [7]

1. 10+4x
2. 15+6x
3. 7.5+3x
4. 6+2.4x
5. 2/5x

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3 years ago

F(x)=x^-2x+4 find f(-1)

otez555 [7]

Hello,

$f(x)=x^{-2}+4\\

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7 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:

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

8 0

3 years ago

3 (x-5) =5<br> what is x?

azamat

Answer: 20/3

So when solving this equation first distribute our 3,we will now have 3x-15=5.

Next we will Add 15 to both sides of our equation. Now we have 3x=20.Then divide our terms,and you will have 3x/3 = 20/3

<u>This will leave you with 1x=20/3. </u>

So the X is 20/3.

Hope that helped!

8 0

3 years ago