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3 years ago

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Solve the following system of equations by linear combination:

1 answer:

Sholpan [36]3 years ago

3 0

3d - e =7
d + e = 5
-------------
4d / = 12
d=12:4
d=3

3+e=5
e=5-3
e=2
the solution is (3,2)

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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

There are 32 students in a class. Of the class, 3/8 of the students bring their own lunches. How many students bring their lunch

OverLord2011 [107]

Answer:

The number of students that bring their lunches is 12

Step-by-step explanation:

Let

x -----> the number of students that bring their lunches

y -----> the total number of students in a class

we know that

The number of students that bring their lunches divided by the total number of students in a class must be equal to 3/8

$\frac{x}{y}=\frac{3}{8}$

$x=\frac{3}{8}y$ -----> equation A

$y=32$ -----> equation B

substitute the value of y in equation A and solve for x

$x=\frac{3}{8}(32)$

$x=12$

therefore

The number of students that bring their lunches is 12

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3 years ago

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Nice;-;

does jared need help carrying it

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3 years ago

Two over twenty-two plus six over eleven

kirill [66]

$\frac{2}{22}+\frac{6}{11}$

Simplify $\frac{2}{22}$ to $\frac{1}{11}$

$\frac{1}{11}+\frac{6}{11}$

Both are over 11, so, just sum.

$\frac{1+6}{11}$

$\frac{7}{11}$

4 0

3 years ago

Read 2 more answers

If the hypotenuse of a right triangle is 2 <img src="https://tex.z-dn.net/?f=%5Csqrt%7B3%7D" id="TexFormula1" title="\sqrt{3}" a

Svetradugi [14.3K]

Answer:

3

Step-by-step explanation:

a = the other leg:

$a^{2} =(2\sqrt{3} )^{2} -(\sqrt{3}) ^{2} =(4)(3)-3=12-3$

$a^{2} =9$

$a=\sqrt{9}$

$a=3$

Hope this helps

5 0

1 year ago