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Sunny_sXe [5.5K]
3 years ago
15

The graph shows the volume (y) of a cup of water (in ml). The cup has a small leak and the volume of water is decreasing at a co

nstant rate each second (x). The situation can be expressed as a linear equation. Which values are correct for the slope and the y-intercept of the graph.
Mathematics
1 answer:
love history [14]3 years ago
5 0

Answer:

m = –3.5, b = 22

Step-by-step explanation:

i answered the question.

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Stephen, Gavin and Jim share some sweets in the ratio 4:1:2. Stephen gets 30 more sweets than Jim. How many sweets are there alt
Nutka1998 [239]

Answer:

$17.00

Step-by-step explanation:

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
3 years ago
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Natali5045456 [20]

Answer: 1.6 Ounces.

16/2 is 8.

8/5 is 1.6

You can check this by multiplying 1.6 times 5 then by 2,

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How to solve 8× 10 rasied to the power of 4
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Answer: 80,000

Step-by-step explanation: 8x(10x10x10x10)=8x10,000=80,000

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3 years ago
Center city has a population of twenty one thousand seventy people write the populationin standard form
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It would be 21070

In standard form, 20000+1000+70

5 0
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