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3 years ago

12

PLEASEEE HELP! 3/8x+y=3 find the intercept and sketch the line.

1 answer:

just olya [345]3 years ago

8 0

X-intercept is 8
Y-intercept is 3

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Which expression is equivalent to 6p + 6p?<br><br> A. 12p<br> B. 36p<br> C. 12 + 2p<br> D. p6

sveticcg [70]

Hello!

6p + 6p = 12p

The answer is A)12p

Hope this helps!

6 0

3 years ago

(x+2)(x-18)=0<br><br>FIND THE SOLUTIONS OF THE EQUATION

vladimir2022 [97]

Either one at any given time could be equal to zero. So
x + 2 = 0
x = - 2

x - 18 = 0
x = 18

I've enclosed a graph of this so you can see that the answers I've given are the ones expected.

6 0

3 years ago

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Debra travels a distance of 2 miles in 1 minute how much distance can she cover in 25 minutes

Marta_Voda [28]

Debra will travel 50 miles in 25 minutes

Step-by-step explanation:

Given

distance = 2 miles

Time = 1 minute

Other time = 25 minutes

We can use the proportions to find the number of miles debra will travel in 25 minutes

So,

$Distance:Time::Distance:Time\\2:1::Distance:25\\\frac{2}{1} = \frac{Distance}{25}\\Distance = 25 *2\\= 50\ miles$

Hence,

Debra will travel 50 miles in 25 minutes

Keywords: Proportion, rate

Learn more about proportion at:

#LearnwithBrainly

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

How do I Factor -12 out of -24k + 120

Juli2301 [7.4K]

Answer:

-12(2k-10)

Step-by-step explanation:

You divide each term by -12, and then put -12 outside of the parenthesis. The parenthesis contain the new equation you just created.

4 0

2 years ago

Read 2 more answers