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3 years ago

Dan earns a commission of 15% on his sales. Last month his sales were: $550.23, $793.23,

Mathematics

1 answer:

elixir [45]3 years ago

5 0

9514 1404 393

Answer:

$533.86

Step-by-step explanation:

The sum of the sales amounts is $3,559.07. Dan's commission is ...

0.15 × $3,559.07 = $533.86

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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

Suppose you have a cache of radium, which has a half-life of approximately 1590 years. how long would you have to wait for 1/5 o

lubasha [3.4K]

<span>You would have to wait 318 years for 1/5 (20%) of the Radium to disappear. 1590 is the point where half of the Radium would decay, so if you take 20% of that, it would equate out to 318 years. Radium has a long half life!</span>

4 0

3 years ago

What is bigger 3/4 or 4/15?

Tems11 [23]

Answer:

3/4 is a bigger

Step-by-step explanation:

and 4/15 is a smaller

3 0

3 years ago

(09.02)

Yuki888 [10]

Answer:

x = - 2, x = 10

Step-by-step explanation:

Given

x² - 8x - 20 = 0

Consider the factors of the constant term (- 20) which sum to give the coefficient of the x- term (- 8)

The factors are - 10 and + 2 , since

- 10 × 2 = - 20 and - 10 + 2 = - 8 , then

(x - 10)(x + 2) = 0

Equate each factor to zero and soilve for x

x + 2 = 0 ⇒ x = - 2

x - 10 = 0 ⇒ x = 10

4 0

3 years ago

6. Use the graph to answer the question below:

aleksandr82 [10.1K]

Answer:

22%

Step-by-step explanation:

total people 300

pistachio 24 people

chocolate 129 people

vanilla 36 people

strawberry 45 people

cookies and cream 66 people

66/300×100=22%

3 0

2 years ago

Read 2 more answers