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2 years ago

12

3х – 8y = 19 3х + 5y = — 46

1 answer:

Vadim26 [7]2 years ago

4 0

Answer:

-5

Step-by-step explanation:

multiple the second equation by negative then subtract or add the equations and the answer will be negative 5

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Which is an equation of a circle with center (-10, 6) and radius 8?

kupik [55]

An equation of a circle:

(x - a)² + (y - b)² = r²

(a; b) - a coordinates of a center
r - a radius

r = 8; (-10; 6) ⇒ a = -10 and b = 6

subtitute

(x - (-10))² + (y - 6)² = 8²

Answer: (x + 10)² + (y - 6)² = 64

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3 years ago

Statistics is the science of collecting, organizing, summarizing, and analyzing information to draw a conclusion and answer ques

Elena L [17]

Answer:

4.results should be reported with some measure that represent how convinced we are that our conclusion reflect reality

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2 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

2 years ago

What is the greatest common factor of 14 56 63?

olga55 [171]

14: 1 × 14
      2 × 7

56: 1 × 56
      2 × 28
      4 × 14
      7 × 8

63: 1 × 63
      3 × 21
      7 × 9
GCF(14, 56, 63) = 7

6 0

3 years ago

Given (-2, 4) (1, -2). Find the slope of the line

gavmur [86]

Answer:

-2

Step-by-step explanation:

We can find the slope of a line given two points from

m = (y2-y1)/(x2-x1)

= (-2-4)/(1--2)

= (-6)/(1+2)

=-6/3

=-2

5 0

3 years ago

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