Translate the sentence into an equation.

Two streams flow into a reservoir. Let X and Y be two continuous random variables representing the flow of each stream with join

zlopas [31]

Answer:

c = 0.165

Step-by-step explanation:

Given:

f(x, y) = cx y(1 + y) for 0 ≤ x ≤ 3 and 0 ≤ y ≤ 3,

f(x, y) = 0 otherwise.

Required:

The value of c

To find the value of c, we make use of the property of a joint probability distribution function which states that

$\int\limits^a_b \int\limits^a_b {f(x,y)} \, dy \, dx = 1$

where a and b represent -infinity to +infinity (in other words, the bound of the distribution)

By substituting cx y(1 + y) for f(x, y) and replacing a and b with their respective values, we have

$\int\limits^3_0 \int\limits^3_0 {cxy(1+y)} \, dy \, dx = 1$

Since c is a constant, we can bring it out of the integral sign; to give us

$c\int\limits^3_0 \int\limits^3_0 {xy(1+y)} \, dy \, dx = 1$

Open the bracket

$c\int\limits^3_0 \int\limits^3_0 {xy+xy^{2} } \, dy \, dx = 1$

Integrate with respect to y

$c\int\limits^3_0 {\frac{xy^{2}}{2} +\frac{xy^{3}}{3} } \, dx (0,3}) = 1$

Substitute 0 and 3 for y

$c\int\limits^3_0 {(\frac{x* 3^{2}}{2} +\frac{x * 3^{3}}{3} ) - (\frac{x* 0^{2}}{2} +\frac{x * 0^{3}}{3})} \, dx = 1$

$c\int\limits^3_0 {(\frac{x* 9}{2} +\frac{x * 27}{3} ) - (0 +0) \, dx = 1$

$c\int\limits^3_0 {(\frac{9x}{2} +\frac{27x}{3} ) \, dx = 1$

Add fraction

$c\int\limits^3_0 {(\frac{27x + 54x}{6}) \, dx = 1$

$c\int\limits^3_0 {\frac{81x}{6} \, dx = 1$

Rewrite;

$c\int\limits^3_0 (81x * \frac{1}{6}) \, dx = 1$

The $\frac{1}{6}$ is a constant, so it can be removed from the integral sign to give

$c * \frac{1}{6}\int\limits^3_0 (81x ) \, dx = 1$

$\frac{c}{6}\int\limits^3_0 (81x ) \, dx = 1$

Integrate with respect to x

$\frac{c}{6} * \frac{81x^{2}}{2} (0,3) = 1$

Substitute 0 and 3 for x

$\frac{c}{6} * \frac{81 * 3^{2} - 81 * 0^{2}}{2} = 1$

$\frac{c}{6} * \frac{81 * 9 - 0}{2} = 1$

$\frac{c}{6} * \frac{729}{2} = 1$

$\frac{729c}{12} = 1$

Multiply both sides by $\frac{12}{729}$

$c = \frac{12}{729}$

$c = 0.0165 (Approximately)$

8 0

3 years ago

To edit a manuscript, Miguel charges 12 dollars per hour for the first 4 hours and 15 dollars per hour after the first 4 hours.

aleksley [76]

Answer:

C = 12 + 15(x - 4)

C = 15x - 48

Step-by-step explanation:

Let the number of hours = t

The total amount Miguel earns =

C

To edit a manuscript, Miguel charges 12 dollars per hour for the first 4 hours and 15 dollars per hour after the first 4 hours.

C = 12 + 15(x - 4)

Expand the t

C = 12 + 15x - 60

C = 15x - 48

Therefore, an expression for the amount, in dollars, Miguel charges if it takes him t hours to edit a manuscript, where t>4 is given as:

C = 15x - 48

5 0

2 years ago

The ordered pair (3,6) is the solution of the system of equations represented by which lines

Temka [501]

Answer:

p and n

Step-by-step explanation:

The solution to a system of equations given graphically is at the point of intersection of the 2 lines.

The lines p and n intersect at (3, 6)

Thus (3, 6) is the solution to the system of equations represented by p and n

8 0

2 years ago

A standard roulette wheel has 38 numbered slots for a small ball to land in: 36 are marked from 1 to 36, with half of those blac

Anastasy [175]

Answer:

The expected value of betting $500 on red is $463.7.

Step-by-step explanation:

There is not a fair game. This can be demostrated by the expected value of betting a sum of money on red, for example.

The expected value is calculated as:

$E(G)=\sum p_iG_i$

being G the profit of each possible result.

If we bet $500, the possible outcomes are:

- Winning. We get G_w=$1,000. This happens when the roulette's ball falls in a red place. The probability of this can be calculated dividing the red slots (half of 36) by the total slots (38) of the roulette:

$P(R)=R/T=18/38\approx0.4737$

- Losing. We get G_l=$0. This happens when the ball does not fall in a red place. The probability of this is the complementary of winning, so we have:

$P(not \,R)=1-P(R)=1-18/38=20/38\approx 0.5263$

Then, we can calculate the expected value as:

$E(G)=\sum p_iG_i=p_wG_w+p_lG_l=0.4637*1,000+0.5263*0=463.7$

We expect to win $463.7 for every $500 we bet on red, so we are losing in average $36.3 per $500 bet.

4 0

2 years ago

I need help what times what equals 104

forsale [732]

1 times 104
2 times 52
4 times 26
8 times 13

7 0

3 years ago