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4 years ago

11

Statement as a conditional statement. “Yellow and blue make green." * 1 point geometry

1 answer:

Vinil7 [7]4 years ago

8 0

Answer:

if green without blue is yellow then yellow and blue make green

Step-by-step explanation:

You might be interested in

Find the percent of change from 18 to 12.

melomori [17]

Answer:

18 is the old value and 12 is the new value. In this case we have a negative change (decrease) of -33.33333333 percent because the new value is smaller than the old value.

8 0

3 years ago

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

Point P is located at (-3,5)

Naya [18.7K]

Answer:

P' (- 7, - 5 )

Step-by-step explanation:

A translation of 3 units to the left means subtractin 3 from the x- coordinate with no change to the y- coordinate, thus

P(- 3, 5 ) → (- 3 - 4, 5 ) → (- 7, 5 )

Under a reflection in the x- axis

a point (x, y ) → (x, - y ), thus

(- 7, 5 ) → P'(- 7, - 5 )

5 0

3 years ago

<img src="https://tex.z-dn.net/?f=%5Csqrt%5C%5Cx%5E%7B2%7D%20-%203%5C%5C" id="TexFormula1" title="\sqrt\\x^{2} - 3\\" alt="\sqrt

Liono4ka [1.6K]

You can’t simplify it more than it is

3 0

3 years ago

Use the long division method to find the result when 6x3 + 19x2 + 10x + 1 is<br> divided by 2x + 1.

galina1969 [7]

Answer:

Hope this answer helps you !

3 0

2 years ago