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MAXImum [283]
3 years ago
11

Ellie was still working on her dollhouse. She has boards that are two different lengths. One long board is 54 inches. The length

of the short board is unknown. Ellie put three boards end to end and then added her 12 inch ruler end to end. The total length was still less than the 54 inch board. How do I draw a picture showing how the short and pond boards are related?
Mathematics
1 answer:
svetoff [14.1K]3 years ago
7 0
If it is a trick question, the 2 boards are probably the same type of board like a 1 by 1 or a 2 By 4
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1 gallon (it’s stayed in the question)
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Given f(x) = 15x + 35, what is the domain of f?
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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
3 years ago
Due in 15 please help!!!
Deffense [45]

Answer:

Step-by-step explanation:

its true

4 0
2 years ago
The sum of four consecutive integers is 286. Find the four integers
8_murik_8 [283]
Step 1: Write out the problem.
Step 1: Write out the problem.
Step 1: Write out the problem.
x+(x+1)+(x+2)+(x+3)= 286

Step 2: Take out the parentheses.
x+x+1+x+2+x+3= 286

Step 3: Combine like terms.
4x+6= 286

Step 4: Subtract 6 from both sides.
4x+6 -6= 286 -6

Step 5: Divide 4 on both sides.
4x/4 = 280/4

x=70

Which means the answer is 70, 71, 72, 73

Check Work:
70+71+72+73= 286

So this is correct :)
3 0
3 years ago
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