Which distribution is less variable? How do you
know?
This is on my test and I’m pretty stuck on how to answer, thank you for answering if you did:)

Thank you ☺️☺️

Answer: X=11

(2x+19)=(3x+8)

-2x -2x

19=(x+8)

-8 -8

11=x

Answer:

9

Step-by-step explanation:

81^1/2 is actually √81

square root of √81 is 9

Pls mark me as brainliest

The answer is three

Step-by-step explanation:

75 plus 160 is 235 and 235+235+235= 705

Answer:

775/25

Step-by-step explanation:

Answer:

v: the answer is 43/48

Step-by-step explanation:

Answer:

152%

Step-by-step explanation:

There is one whole grid shaded = 100%

The other grid has 13/25 shaded.  This is the same as 52/100=52%

100+52=152%

Answer: 152%

Step-by-step explanation: To easy please like

Answer:

.19m+70=c

Step-by-step explanation:

.19m represents $0.19 per minute

+70 represents the fixed monthly fee

so you'd multiply the minutes by 19cents and add the $70 to find cost per month

Answer:

z=-3

Step-by-step explanation:

combine like terms then solve

Answer:

x = -5

Step-by-step explanation:

I hope this helps!

Answer:

if you need to find the value of x then I can help

First the equation looks like this:

2x + 3 = -7

first find the value of "x" which will be -5 then the equation will look like this:

2(-5) then multiply 2 and -5 which is equal to -10

this will make the equation look like this:

-10 + 3 = -7

then evaluate

-10 + 3 = -7

-7 = -7

Therefore the value of "x" is -5

Answer:

9/14

Step-by-step explanation:

3/6 / 7/9

3/6 * 9/7

3/2 * 3/7

9/14

Answer:

Question 1)

[tex]\text{ Distance from A to the house is approximately 78.34 meters}\\\text{The height of the house is approximately 50.77 meters}[/tex]

Question 2)

[tex]\text{ Part A) \cos(x)=\frac{12}{13}} \\ \text{ Part B) \sin(90-x)=\frac{12}{13} }[/tex][tex]\text{ Part A) }\\\cos(x)=\frac{12}{13}\\\\\text{ Part B)}\\\sin(90-x)=\frac{12}{13}[/tex]

Step-by-step explanation:

Question 1)

Please first reference below for the image.

Also recall the basic trigonometric functions with SohCahToa.

Part 1)

In order to find the distance, we will need to find the value of [tex]x[/tex]. However, to do so, we must find the value of [tex]c[/tex] first.

First, we will need to find BM. We can use the smaller triangle. BM is opposite of the smaller triangle, and we already know that the hypotenuse is 20 meters. So, we can use sine:

[tex]\sin(A)=\frac{\text{opp}}{\text{hyp}}[/tex]

Substitute 32° for A and 20 for the hypotenuse. Our BM is our opposite, so substitute BM for opposite. This yields:

[tex]\sin(32)=\frac{BM}{20}[/tex]

Solve for BM. Multiply both sides by 20:

[tex]BM=20\sin(32)[/tex]

Don't approximate just yet. In order to make our answer to most precise, we will approximate at the very end.

So, BM is 20sin(32) meters. BM of the smaller triangle is the same as BM of the larger triangle, so BM of the larger triangle is also 20sin(32). Now, we can find the value of [tex]c[/tex]. However, we will first need to find the value of [tex]\angle MBT[/tex]. To do so, notice that [tex]\angle ABM, \angle MBT, \text{ and } 40\textdegree[/tex] form a straight angle. Therefore:

[tex]\angle ABM+\angle MBT+40^\circ=180[/tex]

Simplify:

[tex]\angle ABM+\angle MBT=140[/tex]

So, we will need to find [tex]\angle ABM[/tex], which is [tex]\angle B[/tex] of the smaller triangle. So, we can use the trig functions again. Since we now know BM, we can use cosine. So:

[tex]\cos(B)=\frac{\text{adj}}{\text{opp}}[/tex]

Substitute 20sin(32) for the adjacent and 20 for the opposite. So:

[tex]\cos(B)=\frac{20\sin(32)}{20}[/tex]

To solve for B, take the inverse cosine of both sides. Evaluate:

[tex]B=\cos^{-1}(\frac{20\sin(32)}{20})=58^\circ[/tex]

Therefore, we can substitute 58 for [tex]\angle ABM[/tex] and solve. This yields:

[tex]58+\angle MBT=140\\\Rightarrow\angle MBT=82^\circ[/tex]

So, we can now find c. We know BM and that [tex]\angle B[/tex] (of the larger triangle) is 82°. To find c, then, we will use cosine. So:

[tex]\cos(B)=\frac{\text{adj}}{\text{hyp}}[/tex]

Substitute 82 for B, 20sin(32) for the adjacent BM, and c for the hypotenuse. So:

[tex]\cos(82)=\frac{20\sin(32)}{c}[/tex]

Solve for c by cross-multiplying:

[tex]c=\frac{20\sin(32)}{\cos(82)}[/tex]

Now, finally, we can solve for [tex]x[/tex]. We know [tex]c[/tex] (the hypotenuse) and we want to find [tex]x[/tex]. We also have an angle of 40°. So, we can use cosine:

[tex]\cos(40)=\frac{x}{c}[/tex]

Multiply both sides by [tex]c[/tex]:

[tex]x=c\cos(40)[/tex]

Substitute [tex]c[/tex]:

[tex]x=\frac{20\sin(32)\cos(40)}{\cos(82)}[/tex]

Approximate:

[tex]x\approx58.34\text{ meters}[/tex]

However, this is only [tex]x[/tex]. The distance from the house is [tex]x+20[/tex]. Therefore, the distance from the house is:

[tex]d\approx58.34+20\approx78.34\text{ meters}[/tex]

Part 2)

We want to find the height h. Since we already determined [tex]c[/tex], we can use it to find [tex]h[/tex]. We will use the sine function this time:

[tex]\sin(40)=\frac{h}{c}\\\Rightarrow h=c\sin(40)\\\Rightarrow h=\frac{20\sin(32)\sin(40)}{\cos(82)}\\\Rightarrow h\approx48.95[/tex]

However, we will still need to add the eye level of 182 centimeters or 1.82 meters. So:

[tex]h\approx48.95+1.82\approx50.77\text{ meters}[/tex]

The height of the house is approximately 50.77 meters.

Question 2)

We know that: [tex]\tan(x)=\frac{5}{12}[/tex]

Remember that tangent is the opposite side over the adjacent side. So, let's find the hypotenuse. We can us the Pythagorean Theorem:

[tex]a^2+b^2=c^2[/tex]

Substitute 5 for a and 12 for b. Solve for c:

[tex]c^2=(5)^2+(12)^2\\\Rightarrow c^2=25+144\\\Rightarrow c=\sqrt{169}=13[/tex]

So, our hypotenuse is 13, our adjacent is 12, and our opposite is 5. We can now solve for the others:

A)

[tex]\cos(x)=\frac{\text{adj}}{\text{hyp}}\\\Rightarrow\cos(x)=\frac{12}{13}[/tex]

B)

[tex]\sin(90-x)=\cos(x)\\\text{Note: This is a cofunction identity}\\\Rightarrow\sin(90-x)=\frac{12}{13}[/tex]

Note: The document I found said [tex]\sin^2(90-x)[/tex]. Anyways, it will just be: [tex]\sin^2(90-x)=\frac{144}{169}[/tex]

Answer:

The angle is 90, my friend gave the answer as 10, so he is incorrect

The complement of the angle is not 10 but 0

The measure of the supplement is 90 degrees

Step-by-step explanation:

The complement of an angle is simply the value of 90 minus the angle

The supplement of an angle is simply the value of 180 minus the angle

So let’s say the angle we are looking at is x

The complement of the angle will be 90 - x

The supplement of the angle will be 180-x

so from the question;

x = 90-x + 180-x

x = 270 - 2x

2x + x = 270

3x = 270

x = 270/3

x = 90

The measure of the angle is 90 degrees

the measure of the complement of this angle is 90-90 = 0 degrees

The measure of the complement is 180-90 = 90 degrees

Answer:

the set of possible values of the independent variable or variables of a function.

Step-by-step explanation:

Answer:

All the x-values in the ordered pairs together make up the domain

Step-by-step explanation:

Answer:

m = 9

Step-by-step explanation:

m/4-3=9

m/(4-3)=9

m/1=9

m=9

Answer:

66 candies

Step-by-step explanation:

70-4=66

Answer:

If Eric sells boxes of cookies for $8 each, and Yolanda sells boxes of cookies for $10 each, and they made the same amount of money, then Eric sold 5 boxes.

Step-by-step explanation:

To solve this problem, you must find the lowest common multiple of eight and ten. That would be 40, which is how much they both made. $40 / $8 per box = 5 boxes.

...And you don't need to know this, but Yolanda sold 4.

Answer:

It would be positive, zero, negative, and undefined (the first answer).

Hope this helps!

Answer:

6,500 = 4th year

6,000 = 5th year

5,500 = 6th year.

The amount of depreciation of the piece of machinery in the sixth year is $5000.

A reduction in the value of an asset over time, due in particular to wear and tear.

Given that, a piece of machinery depreciates $8000 the first year, $7500 the second year, and $7000 the third year

The rate of depreciation is $500 per year

For 6 years the reduction is = 6*500 = $3000

Therefore, after 6 years = $8000 - $3000 = $5000

Hence, The amount of depreciation of the piece of machinery in the sixth year is $5000

For more references on  depreciation, click;

https://brainly.com/question/15085226

#SPJ2

Answer:

this is the answer for first question with this i bet you can do the rest

Step-by-step explanation:

first one 3.99 cent for small bag and 3.349 cent in big bag

You can write a ratio in 3 ways… example-7lb/27.93 or 7:27.93 or 7 to 27.93. Or you can actually divide the number and get a smaller version.

Part B is the answer you get when you divide 7/27.93. Equivalent fraction would be 1/?? Whatever the answer is.

Part C is which ever one is the lesser amount per 1/??

better way and help for rest

Answer:

Woah this is hard

Step-by-step explanation:

Answer:

27/20

Step-by-step explanation:

The absolute value of any expression is always positive or zero

3/5 + 3/4

Add the fractions

27/20

Solution

27/20

Alternate form

1 7/20, 1.35

Answer:

1.35

Step-by-step explanation:

I hope this helps!

Answer:

5.5 years

Step-by-step explanation:

24900 times 6.75% = 1,680.75

so this is how much it goes down per year, but we have to find it were the price reaches 15900. so 9000-1680.75=

7,319.25( one year)

7,319.25-1680.75 = 5,638.5 (second year)

5,638.5-1680.75= 3,957.75 (third year)

3,957.75-1680.75 = 2,277 (fourth year)

2,277-1680.75= 596.25 (fifth year)

Answer:

the correct answer would be 5.5

Step-by-step explanation:

I solved it and got it correct on the test

Answer:

green

Step-by-step explanation:

Answer and explanation:

Find attached

Answer:

(4-3i)(-3+6i)-(2-i)=

1i×9i-1i=

9i-1i=

8i

Answer:

A) La traducción algebraica de la oración es [tex]x^{2}-5=220[/tex].

B) La traducción algebraica de la oración es [tex]4\cdot x^{2} = 100[/tex].

C) Ese problema se traduce bajo esta forma lógica como [tex]w\cdot l = 5625\,m^{2}[/tex] [tex]\land[/tex] [tex]w = l[/tex] [tex]\implies[/tex] [tex]w = 75\,m[/tex] [tex]\land[/tex] [tex]l = 75\,m[/tex].

Step-by-step explanation:

Debemos proceder en este ejercicio como sigue:

1) Leer cuidadosamente la frase.

2) Escribir su equivalente matemático a medida que se lee y hasta culminarlo.

A continuación, presentamos el desarrollo y la conclusión de cada ejercicio:

A) Interpretamos la oración paso a paso conforme al lenguaje matemático conocido:

(i) El cuadrado de un número:

[tex]x^{2}[/tex]

(ii) El cuadrado de un número menos cinco:

[tex]x^{2}-5[/tex]

(iii) El cuadrado de un número menos cinco es igual a:

[tex]x^{2}-5=[/tex]

(iv) El cuadrado de un número menos cinco es igual a doscientos veinte:

[tex]x^{2}-5=220[/tex]

La traducción algebraica de la oración es [tex]x^{2}-5=220[/tex].

B) Interpretamos la oración paso a paso conforme al lenguaje matemático conocido:

(i) Alejandra pensó un número:

[tex]x[/tex]

(ii) Alejandra pensó un número, lo elevó al cuadrado,:

[tex]x^{2}[/tex]

(iii) Alejandra pensó un número, lo elevó al cuadrado, multiplicó el resultado por cuatro:

[tex]4\cdot x^{2}[/tex]

(iv) Alejandra pensó un número, lo elevó al cuadrado, multiplicó el resultado por cuatro y obtuvo cien:

[tex]4\cdot x^{2} = 100[/tex]

La traducción algebraica de la oración es [tex]4\cdot x^{2} = 100[/tex].

C) Interpretamos la oración paso a paso conforme al lenguaje matemático conocido:

(i) Si una pista de baile tiene un área de 5625 m²:

[tex]A = 5625\,m^{2}[/tex]

(ii) Si una pista de baile tiene un área de 5625 m² y se sabe que su ancho mide lo mismo que su ancho:

Lo que se infiere de la afirmación es que la pista de baile sería un cuadrilátero y más precisamente, un cuadrado:

[tex]A = 5625\,m^{2}[/tex]

[tex]A = w\cdot l[/tex] ([tex]l[/tex] - Largo, [tex]w[/tex] - Ancho)

[tex]w = l[/tex]

Entonces la ecuación algebraica sería:

[tex]l^{2} = 5625\,m^{2}[/tex]

(iii) Si una pista de baile tiene un área de 5625 m² y se sabe que su ancho mide lo mismo que su ancho, cuanto medirán las dimensiones de la pista de baile:

Lo que se infiere de la afirmación es que la pista de baile sería un cuadrilátero y más precisamente, un cuadrado:

[tex]A = 5625\,m^{2}[/tex]

[tex]A = w\cdot l[/tex] ([tex]l[/tex] - Largo, [tex]w[/tex] - Ancho)

[tex]w = l[/tex]

Entonces la ecuación algebraica sería:

[tex]l^{2} = 5625\,m^{2}[/tex]

Y obtenemos la longitud por despeje:

[tex]l = \sqrt{5625\,m^{2}}[/tex]

[tex]l = 75\,m[/tex]

[tex]w = 75\,m[/tex]

Ese problema se traduce bajo esta forma lógica como [tex]w\cdot l = 5625\,m^{2}[/tex] [tex]\land[/tex] [tex]w = l[/tex] [tex]\implies[/tex] [tex]w = 75\,m[/tex] [tex]\land[/tex] [tex]l = 75\,m[/tex].

Answer:

Step-by-step explanation:

Area of a rectangle = Length x Breadth.

Area= L x B

We are told the length is twice the breadth.

You can write this as :

Length = 2 x Breadth

L = 2B

If you put that into the area formula at the beginning, we get:

Area = L x B

Area = 2B x B = [tex]2B^2[/tex]

They also told us that the area of the field is 3200 square meters. So,

Area = [tex]2B^2[/tex] = 3200

Divide both sides by 2

[tex]B^2[/tex]= [tex]\frac{3200}{2} = 1600[/tex]

Now to solve for B we need to square root both sides.

[tex]B = \sqrt{1600}[/tex] = 40

So now we know the Breadth is 40meters

Remember earlier we said Length = 2 x Breadth

[tex]L = 80[/tex]

Cost of fencing the field = 2L + 2B

Cost of fencing the field = 2 x 80 + 2 x 40 = $240

Answer:

-4/5

Step-by-step explanation: