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3 years ago

Please Help!!!! I Give Thanks!!!!

Mathematics

1 answer:

svetlana [45]3 years ago

5 0

1) 40 x 85% = 34 games
2) 25 x 80% = 20 new price of the sweater
3) 22/40 = 0.55 x 100% = 55% weight of zinc on the bar
4) 18/40% = 45 exhibits
5) 6/20% = 30 questions
6) 40 x 3% = 12 students passed
7) 6/25% = 24 dog owners
8) 38/40 = 0.95 x 100% = 95% of the painters painted the interior
9) 10,000/40% = 25,000 price of new car
10) 40,000 x 55% = 22,000 money for expenses

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HELLO!! I NEED HELP! Please show full Solutions. I will mark brainliest for the best answer. Thank you and god bless.

Greeley [361]

((x+2)(x+2))-((x-5)(1))=A

x^2+4x+4-x+5=A

x^2+3x+9=A

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3 years ago

Which value is included in the solution set for the inequality graphed on the number line?

Orlov [11]

The answer would be -2 or 0

8 0

3 years ago

5,367➗4= help please

butalik [34]

Answer: 1,341 3/4

Step-by-step explanation:

7 0

3 years ago

Read 2 more answers

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

D is the midpoint of RJ⎯⎯⎯⎯. R has coordinates (−4,7), and D has coordinates (2,−1). Identify the coordinates of J.

Aleksandr [31]

Answer:

$(8,-9)$

Step-by-step explanation:

If the x-coordinate of R is 6 units away from D's x-coordinate, then so will J's x-coordinate. If the y-coordinate of R is 8 units away from D's y-coordinate, then so will J's y-coordinate.

Side note/Extra information:

The line would be $y=\frac{-4}{3} x+\frac{5}{3}$

3 0

2 years ago