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3 years ago

15

Express the inequality using interval notation.−4≤x≤0

1 answer:

Aleonysh [2.5K]3 years ago

4 0

Answer:

[-4,0]

Step-by-step explanation:

we use closed bracket to show " it includes" and -4 and 0 are included in the solution set

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Use the net to find the surface area of the regular pyramid. I’ll mark brainliest if correct

Butoxors [25]

Answer: surface area = area of BASE + 1/2 × perimeter of base × slant HEIGHT.

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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

$P(x > 3y) = \frac{2^2}{6} -\frac{0^2}{6}$

$P(x > 3y) = \frac{4}{6} -\frac{0}{6}$

$P(x > 3y) = \frac{4}{6}$

$P(x > 3y) = \frac{2}{3}$

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3 years ago

If you drive 23 miles south, make a turn and drive 39 miles east, how far are you, in a straight line, from

Rom4ik [11]

Answer:

45.27 miles

Step-by-step explanation:

Pythagorean theorem

23^2 + 39^2 = d ^ 2

d = 45.27 miles

6 0

2 years ago

A ladder is leaning up against a house. The ladder is 20 feet long and the base of the ladder is 12 feet away from the house. Ho

MrRa [10]

Answer:

16 feet

Step-by-step explanation:

The length of the ladder=20 feet

Distance from the base of the ladder to the house = 12 feet

You will notice that a wall is vertical and the ladder makes an angle with the horizontal ground(making it the hypotenuse). This is a right triangle problem.

To find the how far up the house can the ladder can reach, we simply find the third side of the right triangle.

From Pythagoras theorem

$Hyp^2=Opp^2+Adj^2\\20^2=12^2+Adj^2\\Adj^2=400-144\\Adj^2=256\\Adj=\sqrt{256}=16$

The third side of the right triangle is 16. Therefore the ladder leans 16 feet from the ground.

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3 years ago

Read 2 more answers

Ill give brainliest, please

Flura [38]

Answer:

1st One goes to the third one

2nd one goes to the 1st one

3rd one goes to the 4th one

and the 4th one goes to the 2nd one

7 0

3 years ago