Fourth-grade classrooms in several elementary schools are randomly assigned to different antibullying training programs at the beginning of the school year. The school district keeps track of the number of incidents of bullying in each classroom.

This study is an example of __________ study.

Answer:

c. y = ¼x - 2

Step-by-step explanation:

Find the slope (m) and y-intercept (b) then substitute the values into y = mx + b (slope-intercept form)

Slope = change in y/change in x

Using two points on the graph, (0, -2) and (4, -1):

Slope (m) = (-1 - (-2))/(4 - 0) = 1/4

m = ¼

y-intercept = the point where the line intercepts the y-axis = -2

b = -2

✔️To write the equation, substitute m = ¼ and b = -2 into y = mx + b:

y = ¼x - 2

Answer:

He played 40 matches in total, scoring 32 goals.

Step-by-step explanation:

Since after a certain number of football matches, a footballer averages 1 goal per game, and he only scored 2 goals in the next 10 games and his average dropped to 0.8 goals per game, to determine how many football matches did he play altogether you must perform the following calculation:

10/10 + 2/10 = 12/20 = 0.6

8/8 + 2/10 = 10/18 = 0.55

12/12 + 2/10 = 14/22 = 0.64

20/20 + 2/10 = 22/30 = 0.73

24/24 + 2/10 = 26/34 = 0.76

30/30 + 2/10 = 32/40 = 0.8

Thus, he played 40 matches in total, scoring 32 goals.

Answer:

The number of selections is 49.

Step-by-step explanation:

drama = 7

horror films = 16

Select 3 movies at least two dramas

For 2 drama and 1 horror film

(3 C 2) x (16  C 1) = 48

For 3 drama

(3 C 3) = 1

So, total number of selections is 48 + 1 = 49.  

Answer:

[tex]Cov(X,Y) = -\frac{ 25}{9}[/tex]

Step-by-step explanation:

Given

[tex]Interval =[0,10][/tex]

[tex]X + Y < 10[/tex]

Required

[tex]Cov(X,Y)[/tex]

First, we calculate the joint distribution of X and Y

Plot [tex]X + Y < 10[/tex]

So, the joint pdf is:

[tex]f(X,Y) = \frac{1}{Area}[/tex] --- i.e. the area of the shaded region

The shaded area is a triangle that has: height = 10; width = 10

So, we have:

[tex]f(X,Y) = \frac{1}{0.5 * 10 * 10}[/tex]

[tex]f(X,Y) = \frac{1}{50}[/tex]

[tex]Cov(X,Y)[/tex] is calculated as:

[tex]Cov(X,Y) = E(XY) - E(X) \cdot E(Y)[/tex]

Calculate E(XY)

[tex]E(XY) =\int\limits^X_0 {\int\limits^Y_0 {\frac{XY}{50}} \, dY} \, dX[/tex]

[tex]X + Y < 10[/tex]

Make Y the subject

[tex]Y < 10 - X[/tex]

So, we have:

[tex]E(XY) =\int\limits^{10}_0 {\int\limits^{10 - X}_0 {\frac{XY}{50}} \, dY} \, dX[/tex]

Rewrite as:

[tex]E(XY) =\frac{1}{50}\int\limits^{10}_0 {\int\limits^{10 - X}_0 {XY}} \, dY} \, dX[/tex]

Integrate Y

[tex]E(XY) =\frac{1}{50}\int\limits^{10}_0 {\frac{XY^2}{2}}} }|\limits^{10 - X}_0 \, dX[/tex]

Expand

[tex]E(XY) =\frac{1}{50}\int\limits^{10}_0 {\frac{X(10 - X)^2}{2} - \frac{X(0)^2}{2}}} }\ dX[/tex]

[tex]E(XY) =\frac{1}{50}\int\limits^{10}_0 {\frac{X(10 - X)^2}{2}}} }\ dX[/tex]

Rewrite as:

[tex]E(XY) =\frac{1}{100}\int\limits^{10}_0 X(10 - X)^2\ dX[/tex]

Expand

[tex]E(XY) =\frac{1}{100}\int\limits^{10}_0 X*(100 - 20X + X^2)\ dX[/tex]

[tex]E(XY) =\frac{1}{100}\int\limits^{10}_0 100X - 20X^2 + X^3\ dX[/tex]

Integrate

[tex]E(XY) =\frac{1}{100} [\frac{100X^2}{2} - \frac{20X^3}{3} + \frac{X^4}{4}]|\limits^{10}_0[/tex]

Expand

[tex]E(XY) =\frac{1}{100} ([\frac{100*10^2}{2} - \frac{20*10^3}{3} + \frac{10^4}{4}] - [\frac{100*0^2}{2} - \frac{20*0^3}{3} + \frac{0^4}{4}])[/tex]

[tex]E(XY) =\frac{1}{100} ([\frac{10000}{2} - \frac{20000}{3} + \frac{10000}{4}] - 0)[/tex]

[tex]E(XY) =\frac{1}{100} ([5000 - \frac{20000}{3} + 2500])[/tex]

[tex]E(XY) =50 - \frac{200}{3} + 25[/tex]

Take LCM

[tex]E(XY) = \frac{150-200+75}{3}[/tex]

[tex]E(XY) = \frac{25}{3}[/tex]

Calculate E(X)

[tex]E(X) =\int\limits^{10}_0 {\int\limits^{10 - X}_0 {\frac{X}{50}}} \, dY} \, dX[/tex]

Rewrite as:

[tex]E(X) =\frac{1}{50}\int\limits^{10}_0 {\int\limits^{10 - X}_0 {X}} \, dY} \, dX[/tex]

Integrate Y

[tex]E(X) =\frac{1}{50}\int\limits^{10}_0 { (X*Y)|\limits^{10 - X}_0 \, dX[/tex]

Expand

[tex]E(X) =\frac{1}{50}\int\limits^{10}_0 ( [X*(10 - X)] - [X * 0])\ dX[/tex]

[tex]E(X) =\frac{1}{50}\int\limits^{10}_0 ( [X*(10 - X)]\ dX[/tex]

[tex]E(X) =\frac{1}{50}\int\limits^{10}_0 10X - X^2\ dX[/tex]

Integrate

[tex]E(X) =\frac{1}{50}(5X^2 - \frac{1}{3}X^3)|\limits^{10}_0[/tex]

Expand

[tex]E(X) =\frac{1}{50}[(5*10^2 - \frac{1}{3}*10^3)-(5*0^2 - \frac{1}{3}*0^3)][/tex]

[tex]E(X) =\frac{1}{50}[5*100 - \frac{1}{3}*10^3][/tex]

[tex]E(X) =\frac{1}{50}[500 - \frac{1000}{3}][/tex]

[tex]E(X) = 10- \frac{20}{3}[/tex]

Take LCM

[tex]E(X) = \frac{30-20}{3}[/tex]

[tex]E(X) = \frac{10}{3}[/tex]

Calculate E(Y)

[tex]E(Y) =\int\limits^{10}_0 {\int\limits^{10 - X}_0 {\frac{Y}{50}}} \, dY} \, dX[/tex]

Rewrite as:

[tex]E(Y) =\frac{1}{50}\int\limits^{10}_0 {\int\limits^{10 - X}_0 {Y}} \, dY} \, dX[/tex]

Integrate Y

[tex]E(Y) =\frac{1}{50}\int\limits^{10}_0 { (\frac{Y^2}{2})|\limits^{10 - X}_0 \, dX[/tex]

Expand

[tex]E(Y) =\frac{1}{50}\int\limits^{10}_0 ( [\frac{(10 - X)^2}{2}] - [\frac{(0)^2}{2}])\ dX[/tex]

[tex]E(Y) =\frac{1}{50}\int\limits^{10}_0 ( [\frac{(10 - X)^2}{2}] )\ dX[/tex]

[tex]E(Y) =\frac{1}{50}\int\limits^{10}_0 [\frac{100 - 20X + X^2}{2}] \ dX[/tex]

Rewrite as:

[tex]E(Y) =\frac{1}{100}\int\limits^{10}_0 [100 - 20X + X^2] \ dX[/tex]

Integrate

[tex]E(Y) =\frac{1}{100}( [100X - 10X^2 + \frac{1}{3}X^3]|\limits^{10}_0)[/tex]

Expand

[tex]E(Y) =\frac{1}{100}( [100*10 - 10*10^2 + \frac{1}{3}*10^3] -[100*0 - 10*0^2 + \frac{1}{3}*0^3] )[/tex]

[tex]E(Y) =\frac{1}{100}[100*10 - 10*10^2 + \frac{1}{3}*10^3][/tex]

[tex]E(Y) =10 - 10 + \frac{1}{3}*10[/tex]

[tex]E(Y) =\frac{10}{3}[/tex]

Recall that:

[tex]Cov(X,Y) = E(XY) - E(X) \cdot E(Y)[/tex]

[tex]Cov(X,Y) = \frac{25}{3} - \frac{10}{3}*\frac{10}{3}[/tex]

[tex]Cov(X,Y) = \frac{25}{3} - \frac{100}{9}[/tex]

Take LCM

[tex]Cov(X,Y) = \frac{75- 100}{9}[/tex]

[tex]Cov(X,Y) = -\frac{ 25}{9}[/tex]

Answer:

parrel line never meet

Answer:

44 años tiene jose . el padre 74 y el hijo 17 años.

Step-by-step explanation:

Answer:

If we define W as the width:

12m  ≤ W  ≤  22m

Step-by-step explanation:

We have a rectangle with length L and width W.

We know that:

"The length of a rectangle should be 9 meters longer than 7 times the width"

Then:

L = 9m + 7*W

We also know that the length must be between 93 and 163 meters long, so:

93m ≤ L ≤ 163m

Now we want to find the restrictions for the width W.

We start with:

93m ≤ L ≤ 163m

Now we know that L = 9m + 7*W, then we can replace that in the above inequality:

93m ≤  9m + 7*W ≤ 163m

Now we need to isolate W.

First, we can subtract 9m in the 3 sides of the inequality

93m - 9m  ≤  9m + 7*W  -9m ≤ 163m -9m

84m ≤ 7*W ≤ 154m

Now we can divide by 7 in the 3 sides, so we get:

84m/7 ≤ 7*W/7 ≤ 154m/7

12m  ≤ W  ≤  22m

Then we can conclude that the width is between 12 and 22 meters long.

Answer:

[tex](g + f)(8) =133[/tex]

Step-by-step explanation:

Given

[tex]g(n) = 3n - 5[/tex]

[tex]f(n) = n^2 + 50[/tex]

Required

[tex](g + f)(8)[/tex]

This is calculated as:

[tex](g + f)(n) =g(n) + f(n)[/tex]

So, we have:

[tex](g + f)(n) =3n - 5 + n^2 +50[/tex]

[tex]Substitute[/tex] 8 for n

[tex](g + f)(8) =3*8 - 5 + 8^2 +50[/tex]

[tex](g + f)(8) =24 - 5 + 64 +50[/tex]

[tex](g + f)(8) =133[/tex]

Answer:

The median weight for shelter A is greater than that for shelter B.

The data for shelter B are a symmetric data set.

The interquartile range of shelter A is greater than the interquartile range of shelter B.

Step-by-step explanation:

The median weight for shelter A is greater than that for shelter B.

The median of A = 21 and the median of B = 18  true

The median weight for shelter B is greater than that for shelter A.

The median of A = 21 and the median of B = 18   false

The data for shelter A are a symmetric data set.

False, looking at the box it is not symmetric

The data for shelter B are a symmetric data set.

true, looking at the box it is  symmetric

The interquartile range of shelter A is greater than the interquartile range of shelter B.

IQR = 28 - 17 = 11 for A

IQR for B = 20 -16 = 4  True

Answer:

[tex]A * B = \left[\begin{array}{ccc}10&-6\\8&5\\19&-10\end{array}\right][/tex]

Step-by-step explanation:

Given

[tex]A =\left[\begin{array}{cc}2&-2\\3&4\\4&-3\end{array}\right][/tex]

[tex]B = \left[\begin{array}{cc}4&-1\\-1&2\end{array}\right][/tex]

Required

[tex]AB[/tex]

To do this, we simply multiply the rows of A by the column of B;

So, we have:

[tex]A * B = \left[\begin{array}{ccc}2*4 + -2*-1&2*-1+-2*2\\3*4+4*-1&3*-1+4*2\\4*4-3*-1&4*-1-3*2\end{array}\right][/tex]

[tex]A * B = \left[\begin{array}{ccc}10&-6\\8&5\\19&-10\end{array}\right][/tex]

9514 1404 393

Answer:

Step-by-step explanation:

First differences are ...

  x -4, 8 -x, y -8, 20 -y

Then second differences are ...

  (8 -x) -(x -4)

  (y -8) -(8 -x)

  (20 -y) -(y -8)

Each of these is said to be 2, so we have ...

  12 -2x = 2   ⇒   x = 5

  28 -2y = 2   ⇒   y = 13

And an equation we can use to check:

  x +y -16 = 2   ⇒   5 +13 -16 = 2 . . . . true

_____

The explicit formula for the sequence is ...

  an = n^2 -2n +5

Answer:

B. [tex] 4x^2 + \frac{3}{2}x - 7 [/tex]

Step-by-step explanation:

[tex] f(x) = \frac{x}{2} - 3 [/tex]

[tex] g(x) = 4x^2 + x - 4 [/tex]

(f + g)(x) = f(x) - g(x)

= [tex] \frac{x}{2} - 3 + 4x^2 + x - 4 [/tex]

Add like terms

[tex] = 4x^2 + \frac{x}{2} + x - 3 - 4 [/tex]

[tex] = 4x^2 + \frac{3x}{2} - 7 [/tex]

[tex] = 4x^2 + \frac{3}{2}x - 7 [/tex]

Cosine(angle) = adjacent leg/ hypotenuse

Cosine( angle ) = 18/20

Angle = arccos(18/20)

Angle = 26 degrees

Answer:

26°

Step-by-step explanation:

For a right triangle, we can use trigonometry equations :-

In this case we need to use cosine equation .

cos A = adjacent side / hypotenuse

cos A = 18 / 20

A = cos × 18/20

A = arccos × 18/20

A = 26°

Answer:

Step-by-step explanation:

Answer:

24

Step-by-step explanation:

Find the prime factorization of the number 3,600. Factor Tree.

2|3,600.

2|1,800.

2|900.

2|450

5|225

5|45

3|9

3|3

|1

Setup the equation for determining the number of factors or divisors.

3600=2x2x2x2x3x3x5

Sum factors=2+2+2+2+2+3+3+5=24

Answer:

6400

Step-by-step explanation:

Given the profit function ;

P(c) = –20c2 + 320c + 5,120

The maximum value is given by :

f(h) ; where, h = - b /2a

From P(C) ; a = - 20 ; b = 320

h = - b / 2a = - 320 / 2(-20) = - 320 / 40 = 8

c = h

P(8) = –20(8)² + 320(8) + 5,120

P(8) = - 1280 + 2560 + 5120

= 6400

Answer:

B.) $640,000

Step-by-step explanation:

Answer:

The appropriate answer is "21".

Step-by-step explanation:

Given:

Afraid percentage,

p = 21%

or,

  = 0.21

Sample size,

n = 100

As we know,

⇒ [tex]X=np[/tex]

By putting the values, we get

        [tex]=0.21\times 100[/tex]

        [tex]=21[/tex]

Answer:

1386

Step-by-step explanation:

22 × 9 × 7 = 1386 cubic feet

Answer:

Step-by-step explanation:

(x²-6x)+(y²+8y)=-21

(x²-6x+9)+(y²+8y+16)=-21+9+16

(x-3)²+(y+4)²=4

center=(3,-4),radius=√4=2

Answer: 1:55 PM

Step-by-step explanation:

Turn 4:15 to 24-hr clock system which is 1615hrs

16:15 - 02:20 = 1355hrs

Answer:

a

Step-by-step explanation:

because the diagram the length decimal place degree

Answer:

denominator of 12 (first blank) making the numerator 9. equivalent fraction 9/12 (second blank)

Step-by-step explanation:

lowest common denominator is 12. make sure what you do to the denominator (4times3)= 12, you do to the numerator (3times3)=9

Answer:

The first blank is the denominator of 12 so the numerator is 9. Which makes the answer 9/12

Step-by-step explanation:

Hope this helps! :)>

Answer:

Option C.

Step-by-step explanation:

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Answer:

  see attached

Step-by-step explanation:

The reflection across y=-x swaps the coordinates and negates both of them. The first-quadrant figure becomes a third-quadrant figure.

  (x, y) ⇒ (-y, -x)

Answer:

x = 1 , y = -4

Step-by-step explanation:

2x - 5y = 22 ------- ( 1 )

y = 3x - 7      ------- ( 2 )

Substitute ( 2 ) in ( 1 ) :

2x - 5 (3x - 7) = 22

2x - 15x + 35 = 22

- 13x = 22 - 35

- 13x =  - 13

 x = 1

Substitute x in ( 1 ) :

2x - 5y = 22

2 ( 1 ) - 5y = 22

- 5y = 22 - 2

-5y = 20

y = - 4

Answer:

D.80

Step-by-step explanation:

You need to divide thus

16m/0.2m=80m

Answer:

(3,10)

Step-by-step explanation:

When x is 3

[tex]{ \bf{y = 3x + 2}} \\ { \bf{y = (3 \times 3) + 2}} \\ { \bf{y = 11}}[/tex]

y is 11, and this is invalid because it is not at accord.

The correct coordinates which is NOT a solution to the linear equation

y = 3x+2 is,

⇒ (3, 10)

The equation of line in point-slope form passing through the points

(x₁ , y₁) and (x₂, y₂) with slope m is defined as;

⇒ y - y₁ = m (x - x₁)

Where, m = (y₂ - y₁) / (x₂ - x₁)

Given that;

Linear equation is,

⇒ y = 3x + 2

We know that;

The solution of linear equation is satisfy the eqaution.

Hence, We can check as;

⇒ y = 3x + 2

Put x = 1. y = 5

⇒ 5 = 3 × 1 + 2

⇒ 5 = 5

Thus, It is solution of linear equation.

⇒ y = 3x + 2

Put x = 2. y = 8

⇒ 8 = 3 × 2 + 2

⇒ 8 = 8

Thus, It is solution of linear equation.

⇒ y = 3x + 2

Put x = 3, y = 10

⇒ 10 = 3 × 3 + 2

⇒ 10 ≠ 11

Thus, It is not solution of linear equation.

Thus, The correct coordinates which is NOT a solution to the linear equation y = 3x+2 is,

⇒ (3, 10)

Learn more about the equation of line visit:

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Answer:

2

Step-by-step explanation:

2 is the answer because the circle touches the line only 2 times.

Hope this helps.

9514 1404 393

Answer:

  The triangle is not acute because 2² + 4² < 5²

Step-by-step explanation:

The square of the hypotenuse of a right triangle with the given short sides would be 2² +4² = 20. So, that hypotenuse would be √20, about 4.47. The long side of this triangle is longer than that, so the angle opposite is larger than 90°. The triangle with sides 2, 4, 5 is an obtuse triangle.

  The triangle is not acute because 2² + 4² < 5²

The triangle is not acute because 22 + 42 < 52.

Option C is the correct answer.

A triangle is a 2-D figure with three sides and three angles.

The sum of the angles is 180 degrees.

We can have an obtuse triangle, an acute triangle, or a right triangle.

We have,

To determine if a triangle is acute, we need to check whether all three angles of the triangle are acute angles (less than 90 degrees).

Pythagorean theorem,

- If the square of the length of the hypotenuse is greater than the sum of the squares of the other two sides, then the triangle is acute.

- If the square of the length of the hypotenuse is less than the sum of the squares of the other two sides, then the triangle is obtuse.

Now,

The triangle with side lengths 2 in., 5 in., and 4 in. is not a right triangle.

So we can't use the Pythagorean theorem directly.

Now,

We can check if the sum of the squares of the two shorter sides is greater than the square of the longest side.

2² + 4² = 4 + 16 = 20

5² = 25

Since 20 < 25, we know that the triangle is not acute.

Therefore,

The triangle is not acute because 22 + 42 < 52.

Learn more about triangles here:

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Answer:

y > 0.

Step-by-step explanation:

A.