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2 years ago

The radius of a circle is 3cm. What is the area of the circle? Use 3.14 for pie.

Mathematics

1 answer:

vekshin12 years ago

6 0

Answer: 28.26 square cm

Step-by-step explanation:

$A=\pi r^{2}=\pi(3)^{2} \approx 3.14(3)^{2} \approx \boxed{28.26 \text{ cm}^{2}}$

You might be interested in

What are the extremes of the following proportion 3/15 = 12/60

zalisa [80]

In a proportion there are letters assigned to represent various numbers. a, b, c, and d.

a, b, c, and d are 4 non-zero rational numbers that are terms of proportion.

a/b = c/d or a:b = c:d

a and d are the end terms or known as the extremes.
b and c are the middle terms or known as the means.

With this information, we can gather that the extremes of the following proportion are 3 and 60.

3/15 = 12/60 ; a/b = c/d
a = 3 ; b = 15 ; c = 12 ; d = 60

3 and 60 are the extremes
15 and 12 are the means

7 0

3 years ago

Read 2 more answers

Each front tire on a particular type of vehicle is supposed to be filled to a pressure of 26 psi. Suppose the actual air pressur

Ostrovityanka [42]

Answer:

1) K = 7.895 × 10⁻⁶

2) 0.3024

3) 3.6775 × 10⁻²

4) $f(x)= \frac{1}{20} +\frac{3x^{2} }{38000}$

5) X and Y are not independent variables

$h(x\mid y) = \frac{38000x^2+38000y^2}{3y^2+19000}$

7) 0.54967

8) 25.33 psi

σ = 2.875

Step-by-step explanation:

1) Here we have

$f(x, y) =\begin{cases} & \text (x^{2}+y^{2}) \right. 20\leq x\leq 30 & \ 0 \, Otherwise\end{cases}$

$\int_{x}\int_{y} f(x, y)dydx = 1$

$\int_{x}( \right )\int_{y} f(x, y)dy)dx = 1$

$K\int_{x}( \right )\int_{y}(x^{2} +y^{2})dy)dx = 1$

$K\int_{x}( (x^{2}y +\frac{y^{3}}{3})_{20}^{30})dx = 1$

$K\int_{x}( (x^{2}(30-20)) +\frac{30^{3}-20^{3}}{3})_{20}^{30})dx = 1$

$K\int_{x}( (10x^{2})+\frac{19000}{3})_{20}^{30})dx = 1$

$K( (10\frac{x^{3}}{3})+\frac{19000}{3}x)_{20}^{30})= 1$

$K( (10\frac{30^{3}-20^{3}}{3})+\frac{19000}{3}(30-20)))_{20}^{30}) = 1$

$K =\frac{3}{380000}$

2) The probability that both tires are underfilled

P(X≤26,Y≤26) =

$\int_{20}^{26} \int_{20}^{26}K(x^{2}+y^{2})dydx$

$=K\int_{x}( \right )\int_{y}(x^{2} +y^{2})dy)dx$

$= K\int_{x}( (x^{2}y +\frac{y^{3}}{3})_{20}^{26})dx$

$K\int_{x}( (x^{2}(26-20)) +\frac{26^{3}-20^{3}}{3})_{20}^{26})dx$

$K\int_{x}( (6x^{2})+\frac{9576}{3})_{20}^{26})dx$

$K( (6\frac{x^{3}}{3})+\frac{9576}{3}x)_{20}^{26})$

$K( (6\frac{26^{3}-20^{3}}{3})+\frac{9576}{3}(26-20)))_{20}^{26})$

$38304\times K =\frac{3\times38304}{380000}$

= 0.3024

That is P(X≤26,Y≤26) = 0.3024

3) The probability that the difference in air pressure between the two tires is at most 2 psi is given by

{20 ≤ x ≤ 30, 20 ≤ y ≤ 30, $| x-y |$ ≤ 2}

{20 ≤ x ≤ 30, 20 ≤ y ≤ 30, $\sqrt{(x-y)^2}$ ≤ 2}

{20 ≤ x ≤ 30, 20 ≤ y ≤ 30, y ≤ x - 2}

Which gives

20 ≤ x ≤ 22 :: 20 ≤ y ≤ x + 2

22 ≤ x ≤ 28 :: x - 2 ≤ y ≤ x + 2

28 ≤ x ≤ 30 :: x - 2 ≤ y ≤ 30

From which we derive probability as

P( $| x-y |$ ≤2) = $\int_{28}^{30} \int_{x-2}^{30}K(x^{2}+y^{2})dydx$ + $\int_{20}^{22} \int_{20}^{x+2}K(x^{2}+y^{2})dydx$ + $\int_{22}^{28} \int_{x-2}^{x+2}K(x^{2}+y^{2})dydx$

= K ( $\int_{28}^{30} \int_{x-2}^{30}K(x^{2}+y^{2})dydx$ + $\int_{20}^{22} \int_{20}^{x+2}K(x^{2}+y^{2})dydx$ + $\int_{22}^{28} \int_{x-2}^{x+2}K(x^{2}+y^{2})dydx$ )

= $K\left [ \left (\frac{14804}{15} \right )+\left (\frac{8204}{15} \right ) +\left (\frac{46864}{15} \right )\right ] = \frac{3}{380000}\times \frac{69872}{15} =\frac{4367}{118750}$ = 3.6775 × 10⁻²

4) The marginal pressure distribution in the right tire is

$f_{x}\left ( x \right )=\int_{y} f(x ,y)dy$

$=K( \right )\int_{y}(x^{2} +y^{2})dy)$

$= K( (x^{2}y +\frac{y^{3}}{3})_{20}^{30})$

$K( (x^{2}(30-20)) +\frac{30^{3}-20^{3}}{3})_{20}^{30})$

$K(10x^{2}+\frac{19000}{3})}$

$\frac{3}{38000} (10x^{2}+\frac{19000}{3})}$

$= \frac{1}{20} +\frac{3x^{2} }{38000}$

$f(x)= \frac{1}{20} +\frac{3x^{2} }{38000}$

5) Here we have

The product of marginal distribution given by

$f_x(x) f_y(y) = ( \frac{1}{20} +\frac{3x^{2} }{38000})( \frac{1}{20} +\frac{3y^{2} }{38000})$ = $\frac{(3x^2+1900)(3y^2+1900)}{1444000000}$

≠ f(x,y)

X and Y are not independent variables since the product of the marginal distribution is not joint probability distribution function.

6) Here we have the conditional probability of Y given X = x and the conditional probability of X given that Y = y is given by

$h(y\mid x) =\frac{f(x,y))}{f_{X}\left (x \right )}$ = Here we have

$h(y\mid x) =\frac{x^2+y^2}{\frac{1}{20} +\frac{3x^2}{38000} } = \frac{38000x^2+38000y^2}{3x^2+19000}$

Similarly, the the conditional probability of X given that Y = y is given by

$h(x\mid y) =\frac{x^2+y^2}{\frac{1}{20} +\frac{3y^2}{38000} } = \frac{38000x^2+38000y^2}{3y^2+19000}$

7) Here we have

When the pressure in the left tire is at least 25 psi gives

$K\int\limits^{25}_{20} \frac{38000x^2+38000y^2}{3x^2+19000} {} \, dx$

Since x = 22 psi, we have

$K\int\limits^{25}_{20} \frac{38000\cdot 25^2+38000y^2}{3\cdot 25^2+19000} {} \, dx = K \int\limits^{25}_{20} 10.066y^2+6291.39, dx = 57041.942\times \frac{3}{380000}$ = 0.45033

For P(Y≥25) we have

$K \int\limits^{30}_{25} 10.066y^2+6291.39, dx = 69624.72\times \frac{3}{380000}$ = 0.54967

8) The expected pressure is the conditional mean given by

$E(Y\mid x) = K\int\limits^{30}_{20} yh(y \mid x)\, dy$

$E(Y\mid x) = K\int\limits^{30}_{20} 10.066y^3+6291.39y\, dy = \frac{3}{380000} \times 3208609.27153$

= 25.33 psi

The standard deviation is given by

$Standard \, deviation =\sqrt{Variance}$

Variance = $K\int\limits^{30}_{20} [y-E(Y\mid x) ]^2h(y \mid x)\, dy$

$=K\int\limits^{30}_{20} [y-25.33]^2(10.066y^2+6291.39)\, dy$

= $\frac{3}{380000} \times 1047259.78$ = 8.268

The standard deviation = √8.268 = 2.875.

3 0

3 years ago

A phone company surveys a sample of current customers to determine if they use their phones most often to text or use the Intern

Gre4nikov [31]

Answer: payment B: 23 customers

plan B text: 13 customers

fraction form: 13/23

Step-by-step explanation:

7 0

3 years ago

You deposit $5000 in an account earning 7.5% simple interest . How long will it take for the balance of the account to be $6500

GREYUIT [131]

Since this is a problem of simple interest, then the interest earned will be based always on the principal amount which is 5000. So assuming that, lets also assume that the duration to get interest will be in years since that is the most commonly used duration anyway. So first, let's multiply 5000 with 0.075 to get the interest for the first year. So we will have 375. This means that we will be getting 375 interest every year. We can do trial an error method to get the number of years that will yield us to 6500. Through that, I was able to get 4 years. So 4 times 375 equals 1500 plus the original balance of 5000. It will take 4 years before your balance reaches 6500

7 0

3 years ago

6. Solve by factoring: 2x²-11x+14=0

Mariana [72]

Step-by-step explanation:

Solve 2x^2-11x+14=0 give me 2 ways to get the 2 values of x?

The first way is to make factors by middle-term splitting. You get (x-2)(2x-7)=0, so the solution is x=2,7/2.

The other way is to use the formula for the solution of roots. Roots= (-b+ rootD)/2a, (-b-rootD)/2a, if the equation is of the form ax^2+bx+c=0. Here D is the discriminant. D=b^2-4ac

How do I solve 12x^2+11x+2=0?

Solve for x. 5x3−2x2−47x−14=0?

How can I prove that x^2+2x+2=0?

Is the sequence X^2 + 2x -2 =0?

What are the steps to solve 2^(2x)-3(2^x) +2=0?

Lets assume your equation is of the form ax^2+ bx + c =0

First method :

By breaking 'b' factor of x into two parts using factors of 'ac' ( a *c)

like 2* 14 = 28 = 7* 4 ( b is 11 which could be splitted in values 7 and 4)

equation becomes 2x^2 -4x-7x+14 = 0

take out common factors 2x(x-2) -7 (x-2) = 0

which is (2x-7) (x-2) =0 ; x =2,7/2 are two values

Second method:

make equation in the form of (x+h)^ 2 - k = 0 ; then x+h = +sqrt(k) and -sqrt(k)

which will give x as -h+sqrt(k) , +sqrt(k)

2x^2 - 11x +14 =0 wil become x^2 -11/2 x + 7 = 0

which is (x-11/4)^2 + 7- 121/16 =0

(x-11/4)^2 = (121/16) - 7 = 9/16

x- 11/4 = 3/4 and -3/4

x = 14/4 and 8/4 = 7/2 and 2

3 0

2 years ago

Read 2 more answers

The radius of a circle is 3cm. What is the area of the circle? Use 3.14 for pie.​

The radius of a circle is 3cm. What is the area of the circle? Use 3.14 for pie.