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3 years ago

10

Day 4 please help i dont understand it

1 answer:

NNADVOKAT [17]3 years ago

7 0

3200 divided by 8 (the number of boxes) =

H. 400

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A bag contains 28 blue marbles and 12 red marbles. If Milton removes two marbles without replacing them, what is the probability

Daniel [21]

The probability the first marble will be blue is 28/40
The probability the second marble will be red is 12/39
The reason is because 28 + 12 = 40, so this makes the denominator for our first outcome, but then a marble is removed making our denominator 39 because 40-1 = 39.

4 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

3 years ago

How do you solve this compound inequality???? Please show work!!! Thanks!!

svet-max [94.6K]

5n - 1 < -16

Add 1

5n < -15

Divide

n < -3

n is less than -3

-3n - 1 < 8

add 1

-3n < 9

divide by -3

n < -3

n is also less than -3

3 0

3 years ago

Write an equation in slope intercept form for the graph below

labwork [276]

Answer:

$y = 10x$

Step-by-step explanation:

Finding the slope (m) of the line the following points in the line, (0, 0) and (2, 20):

$slope (m) = \frac{y_2 - y_1}{x_2 - x_1} = \frac{20 - 0}{2 - 0} = \frac{20}{2} = 10$

y-intercept (b) is 0. The line intercepts the y-axis at 0. Thus, b = 0.

Substitute m = 10 and b = 0 in $y = mx + b$

The equation in slope-intercept form for the graph would be:

✅ $y = 1x + 0$

$y = 10x$

8 0

3 years ago

HELP ME PLEASEEEEEEEEEEEE

olga nikolaevna [1]

Answer:

L

Step-by-step explanation:

The first number is always x-coordinate. And amongst all only J and L has positive x-coordinates. With J y-coordinate is also positive so it should be L

5 0

3 years ago