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3 years ago

Can you help me work this out please?

Mathematics

2 answers:

Readme [11.4K]3 years ago

5 0

Step-by-step explanation:

Always start by writing what you know

Area of a circle

A= 3.142 x r^2

r = 21cm

Nina [5.8K]3 years ago

5 0

Area of a triangle is a=3.14(r)^2
21 is the radius so plug that in for r and solve
hope this helps!!

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WILL GIVE BRAINLIEST IF CORRECT

SCORPION-xisa [38]

9514 1404 393

Answer:

5544 square units

Step-by-step explanation:

Translation does not change the dimensions of the figure. The two figures together tell you the right triangle has a base of 154 units and a height of 72 units.

The area of a triangle is given by ...

A = 1/2bh

A = 1/2(154 units)(72 units) = 5544 square units

The areas of each of the triangles are 5544 square units.

8 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

A restaurant used 231 eggs last week. 46 of them are colored white, w. The remaining eggs are colored brown.Write an equation th

pantera1 [17]

Answer:

46 + W = 231

Step-by-step explanation:

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3 years ago

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AveGali [126]

Answer:

x² = -5 + 8

x² = 3

x = √3 ≈ 1.73

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3 years ago

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Furkat [3]

I can’t see it is too burly I’m Srry

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3 years ago

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