GEOMETRY!!! PLS HELPP!!
Quadrilateral ABCD is a rhombus.

The Expected value of XX is 1.00.

Given that a box contains 8 cameras and that 4 of them are defective and 2 cameras is selected at random with replacement.

The probability distribution of the hypergeometric is as follows:

[tex]P(x,N,n,M)=\frac{\left(\begin{array}{l}M\\ x\end{array}\right)\left(\begin{array}{l}N-M\\ n-x\end{array}\right)}{\left(\begin{array}{l} N\\ n\end{array}\right)}[/tex]

Where x is the success in the sample of n trails, N represents the total population, n represents the random sample from the total population and M represents the success in the population.

The probability distribution for X is obtained as below:

From the given information, let X be a random variable, that denotes the number of defective cameras following hypergeometric distribution.

Here, M = 4, n=2 and N=8

The probability distribution of X is obtained below:

The probability distribution of X is,

[tex]P(X=x)=\frac{\left(\begin{array}{l}5\\ x\end{array}\right)\left(\begin{array}{l}8-5\\ 2-x\end{array}\right)}{\left(\begin{array}{l} 8\\ 2\end{array}\right)}[/tex]

The probability distribution of X when X=0 is

[tex]\begin{aligned}P(X=0)&=\frac{\left(\begin{array}{l}4\\ 0\end{array}\right)\left(\begin{array}{l}8-4\\ 2-0\end{array}\right)}{\left(\begin{array}{l} 8\\ 2\end{array}\right)}\\ &=\frac{\left(\begin{array}{l}4\\ 0\end{array}\right)\left(\begin{array}{l}4\\ 2\end{array}\right)}{\left(\begin{array}{l} 8\\ 2\end{array}\right)}\\ &=\frac{\left[\left(\frac{4!}{(4-0)!0!}\right)\times \left(\frac{4!}{(4-2)!2!}\right)\right]}{\left(\frac{8!}{(8-2)!2!}\right)}\\ &=0.21\end[/tex]

The probability distribution of X when X=1 is

[tex]\begin{aligned}P(X=1)&=\frac{\left(\begin{array}{l}4\\ 1\end{array}\right)\left(\begin{array}{l}8-4\\ 2-1\end{array}\right)}{\left(\begin{array}{l} 8\\ 2\end{array}\right)}\\ &=\frac{\left(\begin{array}{l}4\\ 1\end{array}\right)\left(\begin{array}{l}4\\ 1\end{array}\right)}{\left(\begin{array}{l} 8\\ 2\end{array}\right)}\\ &=\frac{\left[\left(\frac{4!}{(4-1)!1!}\right)\times \left(\frac{4!}{(4-1)!1!}\right)\right]}{\left(\frac{8!}{(8-2)!2!}\right)}\\ &=0.57\end[/tex]

The probability distribution of X when X=2 is

[tex]\begin{aligned}P(X=2)&=\frac{\left(\begin{array}{l}4\\ 2\end{array}\right)\left(\begin{array}{l}8-4\\ 2-2\end{array}\right)}{\left(\begin{array}{l} 8\\ 2\end{array}\right)}\\ &=\frac{\left(\begin{array}{l}4\\ 2\end{array}\right)\left(\begin{array}{l}4\\ 0\end{array}\right)}{\left(\begin{array}{l} 8\\ 2\end{array}\right)}\\ &=\frac{\left[\left(\frac{4!}{(4-2)!2!}\right)\times \left(\frac{4!}{(4-0)!0!}\right)\right]}{\left(\frac{8!}{(8-2)!2!}\right)}\\ &=0.21\end[/tex]

Use E(X)=∑xP(x) to find the expected values of a random variable X.

The expected values of a random variable X is obtained as shown below:

The expected value of X is,

E(X)=∑xP(x-X)

E(X)=[(0×0.21)+(1×0.57)+(2×0.21)]

E(X)=[0+0.57+0.42]

E(X)=0.99≈1

Hence, the binomial probability distribution of XX when X=0 is 0.21, when X=1 is 0.57 and when X=2 is 0.21 and the expected value of XX is 1.00.

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The fractional number given has a quotient of -2x⁹.

Given that (-8x⁶)÷(4x⁻³) is the fractional number.

The resultant number obtained by dividing two numbers is known as the quotient. Let's divide number a by number b. Then, q=a÷b will be these two numbers' quotient.

Here are the actual numbers (a,b).

In this case,the fraction is (-8x⁶)÷(4x⁻³).

Assume that q=(-8x⁶)÷(4x⁻³)

If you write the denominator variable as x⁻¹=1÷x, you obtain

[tex]\begin{aligned}q&=\frac{-8x^6}{4\frac{1}{x^3}}\\ &=\frac{-8x^6\times x^3}{4}\end[/tex]

As we are aware, if two numbers are multiplied together and their bases are the same, their powers can be represented as sums, such as

q=(-8x⁶⁺³)÷(4)

q=(-8x⁹)÷(4)

By dividing the numerator by the denominator, we get

q=-2x⁹

Hence,the given fraction (-8x⁶)÷(4x⁻³) has a quotient of -2x⁹.

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Answer: option D -2x^9 on edge 2023

Step-by-step explanation:

The solutions to the given quadratic equation [2x² - 3x - 4 = 0] are x = 2.35, -8.85.

The solutions to the given quadratic equation [x² + 2x + 2 = 0] are x = -1+i, -1-i.

Given the equation for the general quadratic formular;

[ -b±√( b² - 4(ac)) ]/2a

a) 2x² - 3x - 4 = 0

We substitute into the equation.

x = [ -(-3)±√(-3)² - 4( 2 × -4 )) ] / 2×2

x = [ 3±√9 + 32 ] / 4

x = [ 3±√41 ] / 4

x = [ 3+√41 ] / 4, [ 3-√41 ] / 4

x = 2.35, -8.85

The solutions to the given quadratic equation [2x² - 3x - 4 = 0] are x = 2.35, -8.85

b) x² + 2x + 2 = 0

We substitute into the equation.

x = [ -2±√( 2² - 4( 1×2 )) ] / 2×1

x = [ -2±√( 4 - 8) ] / 2

x = [ -2±√(-4) ] / 2

Lets re-write -4 as -1( 4 )

x = [ -2±√( -1 × 4) ] / 2

x = [ -2±√4 × √-1 ] / 2

Write √-1 as i

x = [ -2±√4 × i] / 2

x = [ -2± 2 × i] / 2

x = 2[ -1 ± i] / 2

x = (-1 ± i)

x = -1+i, -1-i

The solutions to the given quadratic equation [x² + 2x + 2 = 0] are x = -1+i, -1-i.

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Answer:

The answer is 540 mm^3

Step-by-step explanation:

using v= BH

v= 30×18

v= 540mm^3

pls like and follow, hope this helps.:-):-)

Answer: [tex]\frac{1}{3} \sin x^{3}+C[/tex]

Step-by-step explanation:

Let [tex]u=x^3[/tex]. Then, [tex]3x^2 dx = du \longrightarrow x^2 dx =\frac{1}{3}du[/tex]

So, we can rewrite the original integral as

[tex]\frac{1}{3} \int \cos u \text{ } du=\frac{1}{3} \sin u+C=\frac{1}{3} \sin x^{3}+C[/tex]

Answer:

D

Step-by-step explanation:

Answer:

here is what I came up with 26.4

Considering the number of legs of each animal, it is found that there are 56 legs on the ground.

Hence the number of legs on the ground is given by:

N = 2 x 4 + 4 x 9 + 1 x 4 + 1 x 2 + 3 x 2 = 56.

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Using translation concepts, the algebraic equation that matches the graph is given by:

y = |x - 1| - 1

A translation is represented by a change in the function graph, according to operations such as multiplication or sum/subtraction in it's definition.

In this problem, the function y = |x| is:

Thus the algebraic equation that matches the graph is given by:

y = |x - 1| - 1

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The circle theoremss that cyclic quadrilaterals can be related to include:

It should be noted that cyclic quadrilateral simply means a quadrilateral that's drawn inside a circle.

In this case, every corner of the quadrilateral must touch the circumference of the circle.

Also, the product of the diagonals is equal to the sum of the product of its two pairs of opposite sides.

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If function has points (1/2,0) and (6,0) then [tex]g(x)=2x^{2} -13x+6\\[/tex]

Given x-intercepts at (1/2,0)(6,0) and we have to find the function g(x)

If the x-intercepts lie at (1/2,0)(6,0) then the function has roots of 6 and 1/2. Function is a relationship between variables having value of y for each value of x. Value of x is domain and value of y is known as codomain.

If we insert x=6 or x=1/2 into the equation of g(x) we will obtain 0 as under:

g(6)=0

In order to get 0 the equation will be x-6.

In order to get 0 for root the equation will be x-1/2.

but x-1/2 does not appear in any of these but its multiples can be there :

[tex]2(x-1/2)=2x-1[/tex]

Therefore the function will be as under:

[tex]g(x)=(x-6)(2x-1)[/tex]

[tex]g(x)=2x^{2} -x-12x+6[/tex]

[tex]g(x)=2x^{2} -13x+6[/tex]

Hence the function g(x) is [tex]g(x)=2x^{2} -13x+6[/tex].

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The point-slope equation of the line is y - 8 = 3(x - 6)

The points are given as:

(x1, y1) = (6, 8)

The slope is given as:

m= 3

The equation is then calculated as:

y - y1 = m(x - x1)

This gives

y - 8 = 3(x - 6)

Hence, the point-slope equation of the line is y - 8 = 3(x - 6)

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Answer:

D

Step-by-step explanation:

D is correct because the sqrt of 91 is 9.5 and 9<9.5<10 makes absolute sense

C is not correct because the sqrt of 33 is 5.7 and 6<5.7<7 does not make sense

B is not correct because the sqrt of 13 is 3.6 and 4<3.6<5 does not make sense

A is not correct because the sqrt of 65 is 8.06 and 7<8.06<8 does not make sense

Answer:

The answer is D(f-g)

Step-by-step explanation:

Profit minus expenses

Answer:

D

Step-by-step explanation:

The profit can be defined as (cost of production) - (revenue) and since the cost is defined as g(x) = x^2 and the revenue is f(x) = 4x+1. The profit can be defined as [tex](f-g) = (4x+1)-x^2 = -x^2+4x+1[/tex]. After 2 months you can calculate the profit by simply plugging in 2 as x. This gives you the equation: [tex](f-g) = (4(2) + 1) - 2^2 = 9-4 = 5\$[/tex]

Answer:

74 yards squared

Step-by-step explanation:

Divide the shape up into 2 3x5 rectangles and one 11x4 rectangle

2(3x5)+11x4=30+44=74

74 yards squared

Answer:

they reached the shore they get 135 miles

Answer:

Yes they are

Step-by-step explanation:

Because AE and FD both make a 50 degree angle with AC they are parallel which means that the two angles are corresponding.

Answer:

17.6 ft

Step-by-step explanation:

tan 37 = 130/x

x = 130/tan 37 = 130/0.7536 = 172.5 feet

tan 40 = 130/y

y = 130/tan 40 = 130/0.8391 = 154.9

172.5 - 154.9 = 17.6 ft

Answer:

17.6 Ft

Step-by-step explanation:

Geometry B E2020!! :3

Answer: 2,464 boxes are made in 7 days.

Step-by-step explanation:

1 day = 352 boxes

7 days = ?

To solve this word problem, you can multiply 352 and 7.

352*7= 2464

Expression (5x² - 2x + 7), ([tex]\frac{1}{5}x^3 -\frac{7}{2}x^2 -2[/tex]) and (y⁵ - √(5y²) + 7) are polynomials.

A polynomial is a type of algebraic expression where exponents of all variables are whole number.

1) 5x² - 2x + 7

Since there are no variables raised to negative or fraction exponents,

the expression is a polynomial.

2) [tex]\frac{1}{5}x^3 -\frac{7}{2}x^2 -2[/tex]

Since there are no variables raised to negative or fraction exponents,

the expression is a polynomial.

3) y⁵ - √(5y²) + 7

This can be re-written as

[tex]y^5-(\sqrt{5})y+7[/tex]

Since there are no variables raised to negative or fraction exponents,

the expression is a polynomial.

4) [tex]\frac{y^3}{3} -\sqrt{y} +2[/tex]

This can be re-written as

[tex]\frac{1}{3}y^2 -y^{1/2}+2[/tex]

Since variable y is raised to a fraction exponent,

the expression is not polynomial.

Therefore, expression (5x² - 2x + 7), ([tex]\frac{1}{5}x^3 -\frac{7}{2}x^2 -2[/tex]) and (y⁵ - √(5y²) + 7) are polynomials.

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Answer: (1, -5)

x = 1

y = -5

*Hope this helps, enjoy the rest of your day! ♥️*

Answer: (1, -5)

Step-by-step explanation:

I got you...

In solving system of equations there are three ways..

lets use addition

First we need to cancel out one of the variables

Lets cancel out the y

In order to do this lets multiply the top equation by -2 to get 2y and 4y to cancel out.

(-2)(-9x+2y) = -19(-2)

18x - 4y = 38

Now our system looks like this:

18x - 4y = 38

5x  + 4y = -15

lets add these together

18x - 4y = 38

5x  + 4y = -15   +

= 23x = 23

x = 1

We know x = 1, so substitude 1 for x in one of the equations

5(1) + 4y = -15

5 + 4y = -15

4y = -20

y = -5

This is the y value

Our final answer is (1, -5)

As an ordered pair

Answer:

a. a is bigger than -a

b. a is equal to -a

c. a is less than -a

Step-by-step explanation:

Let's say a = 2

If a > 0, 2 > -2

If a = 0, 0 = -0; -0 and 0 are the same

Let's say a = -1

If a < 0, -1 ? -(-1)

-1 < 1

Answer:

[tex] \frac{x}{y} = \frac{10}{5} \\ = \frac{5}{1} [/tex]

1. •°• y is 5 times x

[tex] \frac{x}{y} = \frac{30}{5} \\ = \frac{6}{1} [/tex]

2. •°• y is 6 times x

[tex] \frac{x}{y} = \frac{56}{8} \\ = \frac{7}{1} [/tex]

3. •°• y is 7 times x

[tex] \frac{x}{y} = \frac{88}{11} \\ = \frac{8}{1} [/tex]

4. •°• y is 8 times x

[tex]2 \times \frac{7}{4} [/tex]

•°• y is 2 times x

[tex]5 \times \frac{7}{4} [/tex]

•°• y is 5 times x

[tex]8 \times \frac{7}{4} [/tex]

•°• y is 8 times x

[tex]11 \times \frac{7}{4} [/tex]

•°• y is 11 times x

Step-by-step explanation:

5,000 by dividing 10.000

$5000 of the new deposits is the bank required to keep on hand.

The answer is option A.

Problem-solving is the act of defining a problem; figuring out the purpose of the trouble; identifying, prioritizing, and selecting alternatives for an answer; and imposing an answer.

Problem-solving starts with identifying the issue. As an example, a trainer may need to parent out a way to improve a scholar's overall performance on a writing talent test. To do that, the instructor will overview the writing exams looking for regions for improvement.

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The integer that represents Omar's score is -14.

Integers are whole numbers. It is a number without a fraction or decimal component. Integers can either be positive, negative or zero.

When Omar wins, the points are added and when he losses, the points are subtracted.

0 + 8 - (6 x 4) + 2

0 + 8 - 24 + 2 = -14

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The number of integer values of n for which the value 4000 × (0.4)ⁿ is an integer is; 9

We are given the expression;

4000 × (0.4)ⁿ

This can also be expressed as;

4000 × (²/₅)ⁿ which can further be expressed in exponent form as;

(2⁵ × 5³) × (²/₅)ⁿ = 2⁽⁵ ⁺ ⁿ⁾ × 5⁽³ ⁻ ⁿ⁾

Since this expression is an integer, we need:

1) 5 + n ≥ 0 for which n ≥ -5

2) 3 - n ≥ 0 for which n ≤ 3

Taking the intersection gives;

-5 ≤ n ≤ 3

Thus;

Number of Integer Values = 3 - (-5) + 1 = 9

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Answer:

a = 2

Step-by-step explanation:

[tex] \frac{a - ( - 6)}{ - 8 - a} = - \frac{4}{5} [/tex]

[tex] \frac{a + 6}{a + 8} = \frac{4}{5} [/tex]

[tex]4(a + 8) = 5(a + 6)[/tex]

[tex]4a + 32 = 5a + 30[/tex]

[tex]a = 2[/tex]

Answer:

we will use democracy so that everyone gets a say in what happens to the colony, second the government will make the decisions and the people will provide options and different ideas, and i believe its the best type because everyone has a say in what happens and the government isn't too powerful and has as much power as the people do.

Step-by-step explanation:

Answer:

a)  u - v

b)  2v - 2

c)  3u + 3

d)  ¹/₂

Step-by-step explanation:

Given:

 [tex]u=\log_{10}3[/tex]

 [tex]v=\log_{10}5[/tex]

Part (a)

Rewrite 0.6 as a fraction:

[tex]\implies \log_{10}(0.6)=\log_{10}\left(\dfrac{3}{5}\right)[/tex]

[tex]\textsf{Apply the quotient log law}: \quad \log_a\frac{x}{y}=\log_ax - \log_ay:[/tex]

[tex]\implies \log_{10}\left(\dfrac{3}{5}\right)=\log_{10}3-\log_{10}5[/tex]

Substitute the values of u and v:

[tex]\implies \log_{10}3-\log_{10}5=u-v[/tex]

Part (b)

Rewrite 0.25 as 25/100:

[tex]\implies \log_{10}(0.25)=\log_{10}\left(\dfrac{25}{100}\right)[/tex]

[tex]\textsf{Apply the quotient log law}: \quad \log_a\frac{x}{y}=\log_ax - \log_ay[/tex]

[tex]\implies \log_{10}\left(\dfrac{25}{100}\right)=\log_{10}(25)-\log_{10}(100)[/tex]

Rewrite 25 as 5² and 100 as 10²:

[tex]\implies \log_{10}(25)-\log_{10}(100)=\log_{10}(5^2)-\log_{10}(10^2)[/tex]

[tex]\textsf{Appy the Power log law}: \quad \log_ax^n=n\log_ax[/tex]

[tex]\implies \log_{10}(5^2)-\log_{10}(10^2)=2\log_{10}5-2\log_{10}10[/tex]

[tex]\textsf{Apply the log law}: \quad \log_aa=1[/tex]

[tex]\implies 2\log_{10}5-2\log_{10}10=2\log_{10}5-2(1)[/tex]

Substitute the value of v:

[tex]\implies 2\log_{10}5-2(1)=2v-2[/tex]

Part (c)

Rewrite 27000 as 30³:

[tex]\implies \log_{10}(27000)=\log_{10}(30^3)[/tex]

[tex]\textsf{Appy the Power log law}: \quad \log_ax^n=n\log_ax[/tex]

[tex]\implies \log_{10}(30^3)=3\log_{10}(30)[/tex]

[tex]\textsf{Apply the log product law}: \quad \log_axy=\log_ax + \log_ay[/tex]

[tex]\implies 3\log_{10}(30)=3\log_{10}(3)+3\log_{10}(10)[/tex]

[tex]\textsf{Apply the log law}: \quad \log_aa=1[/tex]

[tex]\implies 3\log_{10}(3)+3\log_{10}(10)=3\log_{10}(3)+3(1)[/tex]

Substitute the value of u:

[tex]\implies 3\log_{10}(3)+3(1)=3u+3[/tex]

Part (d)

Rewrite √10 as [tex]10^{\frac{1}{2}}[/tex] :

[tex]\implies \log_{10}(\sqrt{10})=\log_{10}(10^{\frac{1}{2}})[/tex]

[tex]\textsf{Appy the Power log law}: \quad \log_ax^n=n\log_ax[/tex]

[tex]\implies \log_{10}(10^{\frac{1}{2}})=\dfrac{1}{2}\log_{10}(10)[/tex]

[tex]\textsf{Apply the log law}: \quad \log_aa=1[/tex]

[tex]\implies \dfrac{1}{2}\log_{10}(10)=\dfrac{1}{2}(1)=\dfrac{1}{2}[/tex]

[tex]y = \sqrt[3]{x + 10} [/tex]

[tex]x = \sqrt[3]{y + 10} [/tex]

[tex]x {}^{3} = y + 10[/tex]

[tex]y = {x}^{3} - 10[/tex]