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Mrac [35]
2 years ago
11

Farmer Ed has 600 meters of fencing

Mathematics
1 answer:
postnew [5]2 years ago
4 0
Can we have the continuation, this is not even a problem/question
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75% of something = 342<br>Need answer quick
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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 = \pir²h
    (where r is the radius and h is the height)

Volume = 27\pi

Height = 3

⇒ 27\pi = 3\pir²

⇒ 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
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