Ask question

Ask question

All categories

2 years ago

11

Farmer Ed has 600 meters of fencing

1 answer:

postnew [5]2 years ago

4 0

Can we have the continuation, this is not even a problem/question

You might be interested in

75% of something = 342<br>Need answer quick

Ostrovityanka [42]

The answer is 456

This the answer 456

8 0

3 years ago

An environment engineer measures the amount ( by weight) of particulate pollution in air samples ( of a certain volume ) collect

Serggg [28]

Answer:

$k = 1$

$P(x > 3y) = \frac{2}{3}$

Step-by-step explanation:

Given

$f \left(x,y \right) = \left{ \begin{array} { l l } { k , } & { 0 \leq x} \leq 2,0 \leq y \leq 1,2 y \leq x } & { \text 0, { elsewhere. } } \end{array} \right.$

Solving (a):

Find k

To solve for k, we use the definition of joint probability function:

$\int\limits^a_b \int\limits^a_b {f(x,y)} \, = 1$

Where

${ 0 \leq x} \leq 2,0 \leq y \leq 1,2 y \leq x }$

Substitute values for the interval of x and y respectively

So, we have:

$\int\limits^2_{0} \int\limits^{x/2}_{0} {k\ dy\ dx} \, = 1$

Isolate k

$k \int\limits^2_{0} \int\limits^{x/2}_{0} {dy\ dx} \, = 1$

Integrate y, leave x:

$k \int\limits^2_{0} y {dx} \, [0,x/2]= 1$

Substitute 0 and x/2 for y

$k \int\limits^2_{0} (x/2 - 0) {dx} \,= 1$

$k \int\limits^2_{0} \frac{x}{2} {dx} \,= 1$

Integrate x

$k * \frac{x^2}{2*2} [0,2]= 1$

$k * \frac{x^2}{4} [0,2]= 1$

Substitute 0 and 2 for x

$k *[ \frac{2^2}{4} - \frac{0^2}{4} ]= 1$

$k *[ \frac{4}{4} - \frac{0}{4} ]= 1$

$k *[ 1-0 ]= 1$

$k *[ 1]= 1$

$k = 1$

Solving (b): $P(x > 3y)$

We have:

$f(x,y) = k$

Where $k = 1$

$f(x,y) = 1$

To find $P(x > 3y)$ , we use:

$\int\limits^a_b \int\limits^a_b {f(x,y)}$

So, we have:

$P(x > 3y) = \int\limits^2_0 \int\limits^{y/3}_0 {f(x,y)} dxdy$

$P(x > 3y) = \int\limits^2_0 \int\limits^{y/3}_0 {1} dxdy$

$P(x > 3y) = \int\limits^2_0 \int\limits^{y/3}_0 dxdy$

Integrate x leave y

$P(x > 3y) = \int\limits^2_0 x [0,y/3]dy$

Substitute 0 and y/3 for x

$P(x > 3y) = \int\limits^2_0 [y/3 - 0]dy$

$P(x > 3y) = \int\limits^2_0 y/3\ dy$

Integrate

$P(x > 3y) = \frac{y^2}{2*3} [0,2]$

$P(x > 3y) = \frac{y^2}{6} [0,2]\\$

Substitute 0 and 2 for y

$P(x > 3y) = \frac{2^2}{6} -\frac{0^2}{6}$

$P(x > 3y) = \frac{4}{6} -\frac{0}{6}$

$P(x > 3y) = \frac{4}{6}$

$P(x > 3y) = \frac{2}{3}$

8 0

2 years ago

If y varies directly as x^2 and y=12 when x=2, find y when x=5

Shkiper50 [21]

$y=ax^2\\\\y=12\ \ \ and\ \ \ x=2\ \ \ \Rightarrow\ \ \ 12=a\cdot 2^2\ \ \ \Rightarrow\ \ \ a= \frac{4}{12}= \frac{1}{3} \\\\y=\frac{1}{3} x^2\\\\x=5\ \ \ \Rightarrow\ \ \ y=\frac{1}{3} \cdot5^2=\frac{25}{3} =8\frac{1}{3}$

7 0

3 years ago

Please help me on this question. 20 points guaranteed!!

julsineya [31]

Answer:

A cylinder with a volume of 27 $\pi$ cubic inches and a height of 3 inches

Step-by-step explanation:

Volume of a cylinder = $\pi$ r²h
(where r is the radius and h is the height)

Volume = 27 $\pi$

Height = 3

⇒ 27 $\pi$ = 3 $\pi$ r²

⇒ 9 = r²

⇒ r = √9 = 3

So height = radius

7 0

2 years ago

I need help with this real quick

Leni [432]

Answer:

A

Step-by-step explanation:

4 0

3 years ago

Read 2 more answers