What is the area of the irregular polygon shown below?
6
6
4.1
6
6
10
A. 121.5 sq. units
B. 108 sq. units
C. 56 sq. units
O D. 216 sq. units

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:

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:

12

Step-by-step explanation:

The amount of children that do like ice cream are 14/26 so the children that do not like ice cream 14/26, and the numerator is 12

Answer:

[tex]\displaystyle 8^\bigg{\frac{8}{3}}[/tex]

General Formulas and Concepts:

Pre-Algebra

Order of Operations: BPEMDAS

Algebra I

Step-by-step explanation:

Step 1: Define

Identify

[tex]\displaystyle \bigg(8^\bigg{\frac{2}{3}} \bigg)^4[/tex]

Step 2: Simplify

Answer:

Whe we have two functions, f(x) and g(x), the composite function:

(f°g)(x)

is just the first function evaluated in the second one, or:

f( g(x))

And the domain of a function is the set of inputs that we can use as the variable x, we usually start by thinking that the domain is the set of all real numbers, unless there is a given value of x that causes problems, like a zero in the denominator, for example:

f(x) = 1/(x + 1)

where for x = -1 we have a zero in the denominator, then the domain is the set of all real numbers except x = -1.

Now, we have:

f(x) = x^2

g(x) = x + 9

then:

(f ∘ g)(x) = (x + 9)^2

And there is no value of x that causes problems here, so the domain is the set of all real numbers, that, in interval notation, is written as:

x ∈ (-∞, ∞)

(g ∘ f)(x)

this is g(f(x)) = (x^2) + 9 = x^2 + 9

And again, here we do not have any problem with a given value of x, so the domain is again the set of all real numbers:

x ∈ (-∞, ∞)

(f ∘ f)(x) = f(f(x)) = (f(x))^2 = (x^2)^2 = x^4

And for the domain, again, there is no value of x that causes a given problem, then the domain is the same as in the previous cases:

x ∈ (-∞, ∞)

(g ∘ g)(x) = g( g(x) ) = (g(x) + 9) = (x + 9) +9 = x + 18

And again, there are no values of x that cause a problem here, so the domain is:

x ∈ (-∞, ∞)

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:

a). 106

b). See Explanation

Step-by-step explanation:

According to the Question,

Given, The human resource department at a certain company wants to conduct a survey regarding worker benefits. The department has an alphabetical list of all 4247 employees at the company and wants to conduct a systematic sample of size 40.

a). Thus, the Required Value of K is 4247/40 = 106.175 ≈ 106

b). The individuals who will be administered the survey. Randomly select a number from 1 to k. Suppose that we randomly select 19. Starting with the first individual selected, the individuals in the survey will be

Answer: 10

Step-by-step explanation: 15 - 5 = 10

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:

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:

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]

[tex]m(2m+2n)+3mn+2m^2\implies \stackrel{\textit{distributing}}{2m^2+2mn}+3mn+2m^2 \\\\\\ 2m^2+2m^2+2mn+3mn\implies \stackrel{\textit{adding like-terms}}{4m^2+5mn}[/tex]

Answer:

Number of 10 cents = 18

Number of  20 cents = 10

Step-by-step explanation:

Let number of 10 cents be = x

Let number 20 cents be = y

Total number of coins = x + y = 28  -------- ( 1 )

Total amount in the box = 0.10 x + 0.20y = 3.80 ---------- ( 2 )

Solve the equations to find x and y

( 1 ) => x + y = 28

         x = 28 - y

Substitute x in ( 2 )

( 2 ) => 0.10(28 - y) + 0.20y = 3.80

          2.80 - 0.10y + 0.20y = 3.80

            0.10 y = 3.80 - 2.80

            0.10 y = 1.00

                [tex]y = \frac{1}{0.10} = 10[/tex]

                y = 10

Substitute y in ( 1 ) => x + y = 28

                                x + 10 = 28

                                x = 28 - 10

                                x = 18

Answer:

32

Step-by-step explanation:

52=x+20

52-20=x+20-20

32=x

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]

Solution :

Given :

mean weight of the dogs, [tex]$\overline x$[/tex] = 69

Number of dogs, n = 37

Standard deviation, σ = 9.2

Confidence interval = 90%

At 90% confidence interval for the true population mean dog weight is given by :

[tex]$= \overline x \pm \frac{\sigma}{\sqrt n} \times z_{0.05}$[/tex]

[tex]$= 69\pm \frac{9.2}{\sqrt_{37}}} \times 1.64485$[/tex]

[tex]$=69 \pm 2.487799$[/tex]

= (66.5122, 71.4878)

Answer:

parrel line never meet

9514 1404 393

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:

60 kg of 40% and 90 kg of 60%

Step-by-step explanation:

Let the amount of 40% chocolate be x.

Let the amount of 60% chocolate be y.

x + y = 150

0.4x + 0.6y = 0.52 * 150

x + y = 150

4x + 6y = 780

     -4x - 4y = -600

(+)   4x + 6y = 780

--------------------------

               2y = 180

y = 90

x + y = 150

x + 90 = 150

x = 60

Answer: 60 kg of 40% and 90 kg of 60%

Answer:

-7

Step-by-step explanation:

We are given the following functions:

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

[tex]h(x) = -3x + 2[/tex]

(g•h) (1)

The multiplication is:

[tex](g \times h)(1) = g(1) \times h(1)[/tex]

So

[tex]g(x) = 1^2 + 4 + 2(1) = 7[/tex]

[tex]h(1) = -3(1) + 2 = -3 + 2 = -1[/tex]

Then

[tex]g(1) \times h(1) = 7(-1) = -7[/tex]

So -7 is the answer.

Answer:

[tex]15,120[/tex]

Step-by-step explanation:

For the first symbol, there are 9 options to choose from. Then 8, then 7, and so on. Since each player chooses 5 symbols, they will have a total of [tex]9\cdot 8 \cdot 7 \cdot 6\cdot 5=\boxed{15,120}[/tex] permutations possible. Since the order of which they choose them matters (as a different order would be a completely different password), it's unnecessary to divide by the number of ways you can rearrange 5 distinct symbols. Therefore, the desired answer is 15,120.

Answer:15,120

Step-by-step explanation:

Answer:

D.80

Step-by-step explanation:

You need to divide thus

16m/0.2m=80m

Answer:

Yes

Step-by-step explanation:

The lengths of this right angle triangle (6, 8, 10) proves that the polygon is indeed a right angle triangle. This is because there are certain ratios to prove that a right angle triangle is indeed a right angle triangle. These are called the Pythagorean Triples . Some examples include; (3, 4, 5), (7, 24, 25) and (28, 45, 53). The Pythagorean Triple 3, 4, 5 can be scaled up to provide the triple 6, 8, 10, where the scale factor is 2.

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:

[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]

Answer:

49.87

Step-by-step explanation:

Normal Probability Distribution:

Problems of normal distributions can be solved using the z-score formula.

In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the z-score of a measure X is given by:

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.

Suppose the mean percentage in Algebra 2B is 70% and the standard deviation is 8%.

This means that [tex]\mu = 70, \sigma = 8[/tex]

What percentage of students receive between a 70% and 94%

The proportion is the p-value of Z when X = 94 subtracted by the p-value of Z when X = 70. So

X = 94

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

[tex]Z = \frac{94 - 70}{8}[/tex]

[tex]Z = 3[/tex]

[tex]Z = 3[/tex] has a p-value of 0.9987.

X = 70

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

[tex]Z = \frac{70 - 70}{8}[/tex]

[tex]Z = 0[/tex]

[tex]Z = 0[/tex] has a p-value of 0.5.

0.9987 - 0.5 = 0.4987.

0.4987*100% = 49.87%.

So the percentage is 49.87%, and the answer, without the percent sign, is 49.87.

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:

y > 0.

Step-by-step explanation:

A.

Answer:

Z +7

Step-by-step explanation:

Z-4/4+8

Divide first

Z -1 +8

Add and subtract

Z +7