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3 years ago

12

Help!!! I’m taking a test and it’s due in 5 minutes!!!!!!

1 answer:

katovenus [111]3 years ago

5 0

Answer:7

Step-by-step explanation:

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Just need the answer

riadik2000 [5.3K]

Answer:

nope sorry man

Step-by-step explanation:

7 0

3 years ago

Can someone help me on 6 and 9?

Papessa [141]

6. instead of multiplying by 4/3 it would be 2/3 because it is half a sphere.
9. you should have multiplied by pi first then multiply by 4/3 to make it more easy for you

3 0

3 years ago

For every real number x,y, and z the statement (x-y)z=xz-yz is

VikaD [51]

Answer:

true.

Step-by-step explanation:

(x - y)z = xz - yz

Apply the distributive property of multiplication over addition to the left side:

(x - y)z =

= z(x - y)

= xz - yz

This is the same as the right side, so the statement is true.

5 0

3 years ago

Factor each expression.<br><br> Problem: 2p^3 + 6p^2 + 3p + 9<br> Answer: (p + 3) (2p2 + 3)

oee [108]

Factor by grouping
$2p^3 + 6p^2 + 3p + 9$
$2 p^3+6 p^2+3 p+9=\left(2 p^3+6 p^2\right)+(3 p+9)=\left(2 p^2\right) (p+3)+3 (p+3)$
$2 p^2 (p+3)+3 (p+3)$
$(p+3) \left(2 p^2+3\right)$

4 0

3 years ago

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

4 years ago