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2 years ago

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A circular rug has a circumference of 12.56 feet. What is the area of the rug? (Use 3.14 for .) A. 6.28 square feet B. 25.12 squ

are feet C. 56.52 square feet D. 12.56 square feet

1 answer:

dedylja [7]2 years ago

4 0

Hey there!
Here's you your answer: D; 12.56
How i got it: *image attached*
Hope this helps!

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After Halloween, a variety store had some costumes regularly priced at $20.50 on sale for 20% off the regular price. A few weeks

Diano4ka-milaya [45]

20.50 * 0.8 = $16.40

16.40 * 0.4 = $6.56

Final selling price = $6.56

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3 years ago

What is the slope of the line?<br><br> A. 3<br><br> B. -3<br><br> C. 1/3 <br><br> D. -1/3

Inessa05 [86]

Answer:

B. -3

Step-by-step explanation:

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I am joyous to assist you anytime.

3 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

2 years ago

How many one-half cubes with dimensions of 1/2 x1x1 fit in a unit cube?<br> 1<br> 2<br> 4<br> 6

Leni [432]

I got 2 but this could be wrong. Hope this helps ;D

4 0

3 years ago

Read 2 more answers

1/6(a+12)= -4 please help because I don’t understand this

iogann1982 [59]

Answer:

a=-36

Step-by-step explanation:

1/6(a+12)=-4

1) Distribute <u>1/6</u> to a and 12:

1/6a+2=-4

2) Subtract 2 from both sides:

1/6a=-6

3) Multiply both sides by the reciprocal of 1/6: (6/1)

a=-36

Let me know if you have any confusion on how I reached the solution :)

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3 years ago