All categories

3 years ago

I don’t understand the paperso can someone help me llz

Mathematics

1 answer:

V125BC [204]3 years ago

6 0

Dude its easy,you have to count to see how many space there is and when you reach the end you put the number

You might be interested in

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

WILL GIVE BRAINLIEST HELP PLEASE

Tamiku [17]

The relation t relates x to y. To determine if it is a function, see if one x value can give two different y values. Since each x value only has one y value, the relation is a function. To check if the inverse is a function, see if any one y value will give multiple x values. By inversing t, there are two values of x for the value of y = -4 (x = 4 and x = 6), so this is NOT a function.

Answer: Relation t is a function. The inverse of relation t is NOT a function.

3 0

3 years ago

What is the equation in point-slope form of the line that passes through the point (-1, -4) And has a slope of -3?

Reptile [31]

Point-slope form is $y- y_{1} =m(x- x_{1} )$ where y1 and x1 are your points from the coordinate and m is the slope. So yours would look like this: y-(-4)=-3(x-(-1)) or y + 4 = -3(x + 1)

5 0

3 years ago

If a line perpendicular to 2y=x+5 that passes through (2,1) using the slope- intercept form

arlik [135]

Slope of 2y = x + 5:-
This is y = 0.5 x + 2.5 in slope intercept form so its slope = 0.5

Slope of the line perpendicular to it = -1 / 0.5 = -2

it passes through (2 , 1) so we have

y- 1 = -2(x - 2)
y = -2x + 4 + 1

The answer is y = -2x + 5

4 0

3 years ago

A line goes through the points (-4,7) and (1,2) what is the equation of the line

OLga [1]

Answer:

I think it is 4, 14?

Step-by-step explanation:

6 0

3 years ago