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sp2606 [1]
3 years ago
15

Peaches are being sold for $2 per pound. If x represents the number of pounds of peaches bought and y represents the total cost

of the peaches, which best describes the values of x and y?
The values of both x and y can be any real number.
The values of both x and y will be real numbers greater than or equal to 0.
The value of x and y will always be zero
The value of x can be any real number greater than or equal to 0, but the value of y must be an integer greater than or equal to 0.
Mathematics
2 answers:
serious [3.7K]3 years ago
8 0

Answer:

Step-by-step explanation:

The values of both x and y can be any real number.

The values of both x and y will be real numbers greater than or equal to 0.

The value of x and y will always be zero

The value of x can be any real number greater than or equal to 0, but the value of y must be an integer greater than or equal to 0.

maria [59]3 years ago
4 0

Answer:

the answer is #4 -The value of x can be any real number greater than or equal to 0, but the value of y must be an integer greater than or equal to 0.

Step-by-step explanation:

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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
a stack of sheets of tissue paper is about 2.5 inches high. each sheet is about 0.01 inch thick. how many sheets are in the stac
damaskus [11]

Answer:

Step-by-step explanation:

2.5 / 0.01 = 250 sheets in the stack....just a little division problem....thank god for calculators...lol

4 0
3 years ago
4a+(-8), if a= -2 PLZ ANSWER ASAP
Ugo [173]

Answer:

-16

Step-by-step explanation:

4a+(-8)

4(-2)+(-8)

-8+(-8)

= - 16

6 0
3 years ago
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Liono4ka [1.6K]

Answer:

Step-by-step explanation:

4 0
1 year ago
What is the solution to the equation 4w=2/3?
nikdorinn [45]

Answer:

2/12

Step-by-step explanation:

4w=2/3

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6 0
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