100 POINTS WILL MARK BRAINLIEST NEED BY 9:00
Solve the linear equation below. Show ALL your work AND explain the type of solution you found!

8 - 2(4v + 6) = - 4 - 8v

Answers

Answer 1

Answer:

See below ~

Step-by-step explanation:

8 - 2(4v + 6) = -4 - 8v8 - 8v - 12 = -4 - 8v-8v - 4 = -4 - 8vOn placing any value for v, they will be equalIt has infinitely many solutions
Answer 2

Answer:

Hello! The solution to this linear equation is 0 = 0. To find this solution:

Distribute -

8 - 2(4v+6) = -4 - 8v

8 - 8v - 12 = 14 - 8v

Subtract the numbers -

8 - 8v - 12 = -4 - 8v

-4 - 8v =  -4 - 8v

Rearrange terms -

-4 - 8v = -4 - 8v

-8v - 4 = -4 - 8v

Rearrange terms -

-8v - 4 = -4 - 8v

-8v - 4 = -8v - 4

Add 4 to both sides -

-84v - 4 = -8v - 4

-8v - 4 + 4 = -8v - 4 + 4

Simplify -

-8v - 4 + 4 = -8v - 4 + 4

-8v = -8v - 4 + 4

-8v = -8v

Add 8v to both sides -

-8v = -8v

-8v + 8v = -8v + 8v

Combine like terms and simplify -

-8v + 8v = -8v + 8v

0 = -8v + 8v

0 = 0

The solution is 0 = 0. This is an identity: it is true for all values.

Hope this helps! Have a great day.


Related Questions

Enrique estimates that his utilities will cost him $86.00 per month over the courses of the next 3 years. What is his total estimated cost for utilities over the next 3 year period??

Please help thank you !!!

Answers

Answer:

$3096

Step-by-step explanation:

There are 12 months in 1 year.

3 × 12 = 36

There are 36 months in 3 years.

$86 × 36 = $3096

Answer:

3096

Step-by-step explanation:

I just completed the quiz.

Find the area of the shaded region

Answers

Not getting in to too much details but basically
Area of half circle - area of triangle
3.14r^2-1/2bh
r=6 b=12 h=6

Answer:

20.52 cm²

Area of shaded region:

area of semi-circle - area of triangle

[tex]\dashrightarrow \sf \dfrac{1}{2} \ \pi (radius)^2 \ - \ \sf \dfrac{1}{2} *base*height[/tex]

[tex]\rightarrow \sf \dfrac{1}{2} (3.14)(6)^2 - \dfrac{1}{2} *12*6[/tex]

[tex]\hookrightarrow \sf 56.52 \ - \ 36[/tex]

[tex]\hookrightarrow \sf 20.52 \ cm^2[/tex]

Given x^2+y^2=r^2 and the figure of the right triangle with legs x and y and hypotenuse r, prove cos^2θ+sin^2θ=1.

I need assistance filling out the blanks on the attached document.

Answers

By definition of cosine and sine,

cos(θ) = x/r

sin(θ) = y/r

so that

cos²(θ) + sin²(θ) = (x/r)² + (y/r)²

… = x²/r² + y²/r²

… = (x² + y²)/r²

… = r²/r²

… = 1

To that end, I would say

• [blank1] = "Division property of equality"

That is, we divide both sides of an equation by the same number and equality still holds since r ≠ 0

• [blank2] = "Definition of cos"

• [blank3] = "cos²(θ) = x²/r²"

• [blank4] = "Defintion of sin"

• [blank5] = "sin²(θ) = y²/r²"

• [blank6] = "Simplify"

More specifically, x² + y² = r² is given, so

x²/r² + y²/r² = (x² + y²)/r² = r²/r² = 1

bianca's dad was taking everyone out to eat for her birthday. he spent $8 total on the adults and $9 total on the kids. how much did it cost for everyone?

Answers

How many people in total where their

PLEASE HELP!!!!!!!!! ​

Answers

Answer:

With Graph and Without Graph.

Step-by-step explanation:

Without graphing calculator, you plug in your x-values into the equation

y = 16 -x^2 to solve for y-values.

f(x) = 16 - x^2

f(-4) = 16 - (-4)^2 = 16 - 16 =0

...

f(4) = 16 - (4)^2 = 16 - 16 =0


With Graphing Calculator.

Find an equation for the graph

Answers

Answer:

y=6^x+0.612 - 4

Step-by-step explanation:

x = -4 ---> horizontal asymptote

m = 6 ---> use points (0, -1) and (1, 5)

Parent function of the graph: [tex]y=b^x[/tex]

Our equation: [tex]y=6^x[/tex]

Add alterations:

Reflections = N/AVertical & horizontal shifts = down 4Vertical & horizontal stretches = left approx 0.612

Final equation: y=6^x+0.612 - 4


7. A sector of a circle has are length 2cm and central angle 0.4 radians. Find its
radius and and area?

Answers

12 is the factor n the bro 92 n minus 8

The base of S is the region enclosed by the parabola y = 8 − 8x2 and the x-axis. Cross-sections perpendicular to the x-axis are isosceles triangles with height equal to the base.

Answers

The base of a solid is the region in the first quadrant bounded by the y-axis, the x-axis, the graph of y=ex, and the vertical line x=1. For this solid, each cross section perpendicular to the x-axis is a square. What is the volume of the solid?

Step-by-step explanation:

Answer: Sometimes I dont want to be happy.

Consider a triangle...

Answers

Answer:

1. Triangle: B = 47.0° , C = 103.05° , c = 2.53 cm

2. Lake: c = 1105.31 ft

Step-by-step explanation:

Law of Sine Formula:[tex]\frac{sin(A)}{A} = \frac{sin(B)}{B} = \frac{sin(C)}{C}[/tex]

Given: A = 30° , a = 1.3 cm , b = 1.9 cm

[tex]\frac{sin(30)}{1.3} = \frac{sin(B)}{1.9} = \frac{sin(C)}{C}[/tex]

Solving for sin(B). Cross Multiply.

[tex]\frac{sin(30)}{1.3} = \frac{sin(B)}{1.9}\\1.3*sin(B)=1.9*sin(30)\\sin(B)=\frac{1.9*sin(30)}{1.3} \\[/tex]

B = sin^-1( [tex]\frac{1.9*sin(30)}{1.3}[/tex] )

B ≈ 46.9509202

B = 47.0°

Solve for C°

A° + B° + C° = 180°

30° + 46.95° + C° = 180°

C° = 180° - 30° - 46.95°

C° = 103.05°

Solve for sin(C)

[tex]\frac{sin(30)}{1.3} = \frac{sin(103.05)}{C}\\[/tex]

Cross Multiply

[tex]C*sin(30)=1.3*sin(103.05)\\C=\frac{1.3*sin(103.05)}{sin(30)}[/tex]

C ≈ 2.532850806

C = 2.53 cm

Law of Cosine Formula: [tex]c^2=a^2+b^2-2*a*b*cos(C)[/tex]

Given: a = 850 ft , b = 960 ft ,  C=75°

Solve for c.

[tex]c^2=a^2+b^2-2*a*b*cos(C)\\c^2=(850ft)^2+(960ft)^2-2*(850ft)*(960ft)*cos(75)\\\\c=\sqrt{(850ft)^2+(960ft)^2-2*(850ft)*(960ft)*cos(75)\\} \\[/tex]

c ≈ 1105.308698

c = 1105.31 ft

Will the product of 2 2/5 x 1/6 be larger or smaller than 2 2/5

Answers

Answer:

Smaller

Step-by-step explanation:

2 2/5 * 1/6 = 2/

5

= 0.4

Conversion a mixed number 2 2/

5

to a improper fraction: 2 2/5 = 2 2/

5

= 2 · 5 + 2/

5

= 10 + 2/

5

= 12/

5

To find a new numerator:

a) Multiply the whole number 2 by the denominator 5. Whole number 2 equally 2 * 5/

5

= 10/

5

b) Add the answer from previous step 10 to the numerator 2. New numerator is 10 + 2 = 12

c) Write a previous answer (new numerator 12) over the denominator 5.

Two and two fifths is twelve fifths

Multiple: 12/

5

* 1/

6

= 12 · 1/

5 · 6

= 12/

30

= 2 · 6/

5 · 6

= 2/

5

Multiply both numerators and denominators. Result fraction keep to lowest possible denominator GCD(12, 30) = 6. In the following intermediate step, cancel by a common factor of 6 gives 2/

5

.

In other words - twelve fifths multiplied by one sixth = two fifths.

How many different three person relay teams can be chosen from six students?

Answers

There can be 20 different three person relay teams chosen from 6 students. Hope this helps!

Bruce wants to make 50 ml of an alcohol solution with a 12% concentration. He has a 10% alcohol solution and a 15% alcohol solution. The equation 0.10x + 0.15(50 – x) = 0.12(50) can be used to find the amount of 10% alcohol solution Bruce should use.
How much of the 10% alcohol solution should Bruce use

Answers

Answer:

  30 mL

Step-by-step explanation:

You are being asked to solve the given equation for the value of x.

__

  0.10x +0.15(50 -x) = 0.12(50) . . . . given

  -0.05x = -0.03(50) . . . . . subtract 0.15(50), combine terms

  x = 30 . . . . . . . . . . . divide by -0.05

Bruce should use 30 mL of the 10% alcohol solution.

5 Four children are measuring their height.
Aisha
1.39 metres
Teddy
1.37 metres
Scott
1.4 metres
Kim
1.43 metres
Order the children from tallest to shortest.

Answers

Kim, Scott, Aisha, Teddy is the answer

At a local school,940 students each wrote 40 letters to students in another country.How many letters were written in all?

Answers

Answer:

37,600

Step-by-step explanation:

940 X 40 = 37,600.

simple multiplication

Answer:

, 37,000

Step-by-step explanation:

940 x 40= 37,000

Solve for an angle in right triangles. Round to the nearest hundredths

Answers

Answer:

Use   SOH  CAH TOA  to rember how the trig function fit on the triangle

Step-by-step explanation:

we are given  the Hypotenuse and the Opposite  or   H and O   look for the trig function with each of those,  SOH is good

Sin Ф =  Opp /  Hyp,    or   SOH

now plug in what you are given

Sin Ф = 5 / 8

use inverse trig fuction to find the angle

arcSin ( Sin Ф) = arcSin ( 5/8)

trig functions cancel out

Ф = arcSin(5/8)

I'm using my calculator to find the arcSin(5/8)

Ф=38.6821...°

also make sure you know if your calculator is in degrees or radians.  

Ф=38.68°   to the nearest hundredth  :)

Answer:

∠A = 38.68°

Step-by-step explanation:

The side opposing ∠A and the hypotenuse are given.

Therefore, take the inverse sin function of ∠A.

sin∠A = 5/8∠A = sin⁻¹ (0.625)∠A = 38.6821875∠A = 38.68° (nearest hundredth)

Solve the inequality įx - 2 Ž. Which number line represents the graph of the solution?
7 8 9 10 11 12 13 14 15
th
-5-4-3-2-1 0 1 2 3 4 5
-1
3
4
niwa
1
4
0
H
8
-
7
9
0
-1
1
9
-
No
2 5 4 1 2
3 9939

Answers

Answer:

2 ND option

solve and write

When yara and her sister came home there was 3/4 of a pan of brownies left over yara and her sister ate 2/3 of the 3/4 pan of brownies how much of the entire pan of brownies did yara and her sister eat

Answers

Answer:

3/4_2/3 find the LCM= 1/12

The committee spent $372 on costumes for 20 people each costume cost the same amount of money, how much did each costume cost, in dollars ? PLEASE ANSWER I'LL GIVE 68 POINTS TO THE FIRST ONE THAT MAKES SINCE, AND I'LL MAKE UU BRAINLIEST, also I'm gonna ask 4 questions on my page and whoever answers them all first get 100 points I promise!

Answers

Answer:

$18.60

Step-by-step explanation:

The total amount spent is $372, and this can be divided by 20. Each costume would cost $18.60.

Suppose that $3^a = 2$ and $3^b = 5$. If \[3^x = 150,\]then write an expression for $x$ in terms of $a$ and $b$.

Answers

Use logarithms to solve for a.

[tex]3^a = 2 \implies \log_3(3^a) = a\log_3(3) = \log_3(2) \implies a = \log_3(2)[/tex]

Similarly, for b and x.

[tex]3^b = 5 \implies b = \log_3(5)[/tex]

[tex]3^x = 150 \implies x = \log_3(150)[/tex]

Factorize 150:

150 = 2 • 3 • 5²

Then we can expand log₃(150) using the product-to-sum and exponent property,

[tex]\log_3(150) = \log_3(2\times3\times5^2) = \log_3(2) + \log_3(3) + \log_3(5^2)[/tex]

[tex]\implies \log_3(15) = \log_3(2) + 1 + 2 \log_3(5) \iff \boxed{x = 1 + a + 2b}[/tex]


Maitri and Aabhas do a work in 12 hours.

Aabhas and Kavya do the work in 15 hours.

Kavya and Maitri do
work in 20 hours.

In how many hours will they finish it together and separately?

Pls help me

Answers

Answer:

See below ~

Step-by-step explanation:

Given

Maitri and Aabhas do a work in 12 hoursAabhas and Kavya do the work in 15 hoursKavya and Maitri do the work in 20 hours

Solving

Take Maitri, Aabhas, and Kavya to be x, y, z respectivelyx + y = 12 (1)y + z = 15 (2)x + z = 20 (3)

Take Equation 1 and rewrite it so that it is equal to x.

x = 12 - y

Take Equation 2 and rewrite it so that it is equal to z.

z = 15 - y

Now, substitute these values in Equation 3.

x + z = 2012 - y + 15 - y = 20-2y + 27 = 202y = 7y = 7/2 = 3.5 hours [Aabhas]

Substitute the value of y in Equation 1.

x + 3.5 = 12x = 8.5 hours [Maitri]

Substitute the value of y in Equation 2.

3.5 + z = 15z = 11.5 hours [Kavya]

Add the values of x, y, and z together.

x + y + z8.5 + 3.5 + 11.512 + 11.523.5 hours [together]

A
Select the correct answer from each drop-down menu.
The front, back left and right sides of the second floor of the house will be painted. The roof will not be painted. The total surface area to be
painted is 728 square feet. The windows shown each measure 3 feet by 2 feet. There are no other windows on the second floor.
Ich 4) to
hft
2017
25 ft
What is the value of 2
The front and back of the top section of the
The bottom section of the second floor can be modeled by a
second floor can be modeled by the bases of a
The value of hin feet is
Reset
Next

Answers

Answer: rectangular prism, triangular prism, 6

Step-by-step explanation:

The bottom section of second floor can be modeled by a rectangular prism.The front and back of top section of second floor can be modeled by bases of a triangular prism.The value of height is 6 feet.

How to calculate the area of the base of a triangular prism?

The area of base of the triangular prism can be calculated by the half of the product of the height of the triangular base of prism and base of prism.

How to calculate the area of the base of a rectangular prism?

The area of base of the rectangular prism can be calculated by the product of the length of the rectangular base of prism and the breadth of prism.

The area of the bottom section of the second floor = 2( area of the front part of the wall + area of the side of the wall)

= 2( 20*h + 25*h)

=90h

The front and back of top section of second floor = 2* area of the front wall=2*(1/2*20*(h+4))=20h+80

Total area of the second floor=total area of the window+total area to be painted

⇒Total area of the second floor= 728 + (2* area of the window)

⇒Total area of the second floor= 728 + (2*3*2)

area of front and back of top section of second floor + The area of the bottom section of second floor = 728+12

area of front and back of top section of second floor + The area of the bottom section of second floor = 740

⇒(20h+80)+90h=740

⇒110h+80=740

⇒110h=740-80

⇒110h=660

⇒h=660/110

⇒h=6 feet

Therefore the value of h is 6 feet.

Learn more about area of triangular Prism

here: https://brainly.com/question/17111476

#SPJ2

Question #12: A department store has a discount on shoes based on a
percentage of the price. Suppose one pair of shoes is marked down from
$70 to $49. What is the price for a $110 pair of shoes after the discount is
applied?
a
O $89.00
O $77.00
O $73.33
O $33.00

Answers

Answer:

$77

Step-by-step explanation:

The answer is $77 because first you need to find how much money was discounted by doing 70-49 to get 21. Then you need to find how much percent 21 is of 70 by doing 21/70, then you would get 0.3 which is 30% since you have to multiply it by 100. This means that there is a 30% discount. Then you would do 0.3*110=33. This means that the 30% discount takes away $33. So 110-33=77. The answer is $77.

5. Convert the rectangular equation x² + y² - 6y = 0 into a polar equation.
A. r = 6 sin theta
B. r = 6 cos theta
C. r = √6 sin theta
D. r = √6 cos thea​

Answers

Answer:

A. r = 6 sin theta

Step-by-step explanation:

Given equation is: [tex]x^2+y^2-6y=0[/tex]....(1)

Using the formulae that link Cartesian to Polar coordinates.

[tex]x=r\cos\theta \: and \: y = r\sin\theta[/tex]

Plugging the values of x and y in equation (1), we find:

[tex](r\cos\theta)^2+(r\sin\theta)^2-6(r\sin\theta)=0[/tex]

[tex]\implies r^2\cos^2\theta+r^2\sin^2\theta-6r\sin\theta=0[/tex]

[tex]\implies r^2(\cos^2\theta+\sin^2\theta)=6r\sin\theta[/tex]

[tex]\implies r^2(1)=6r\sin\theta[/tex]

[tex](\because \cos^2\theta+\sin^2\theta=1)[/tex]

[tex]\implies\frac{ r^2}{r}=6\sin\theta[/tex]

[tex]\implies\huge{\purple{ {r}=6\sin\theta}}[/tex]

Answernone

none

none

Step-by-step explanation:

Find the length of the arc.
120°
6ft

Answers

Answer:

The length of the arc is 8*pi cm. An arc that subtends a central angle of 120 degrees has a length of 120/360 = 1/3 the length of the total circumference of the circle. We know the entire circumference of the circle is 2πr, which in this case is 2π*12 = 24π.

Step-by-step explanation:

sorry I only have the 8

[tex] \displaystyle \rm\int_{0}^1 { ln }^{2k} \left \lgroup \frac{ ln \left \lgroup \dfrac{1 - \sqrt{1 - {x}^{2} } }{x} \right \rgroup }{ ln \left \lgroup \dfrac{1 + \sqrt{1 - {x}^{2} } }{x} \right \rgroup } \right \rgroup \: dx[/tex]​​

Answers

Substitute [tex]x\mapsto\sqrt{1-x^2}[/tex], which transforms the integral to

[tex]\displaystyle \int_0^1 \ln^{2k} \left(\frac{\ln\left(\frac{1-\sqrt{1-x^2}}x\right)}{\ln\left(\frac{1-\sqrt{1-x^2}}x\right)}\right) \, dx = \int_0^1 \ln^{2k}\left(\frac{\ln\left(\frac{1-x}{\sqrt{1-x^2}}\right)}{\ln\left(\frac{1+x}{\sqrt{1-x^2}}\right)}\right) \frac{x}{\sqrt{1-x^2}} \, dx[/tex]

and factoring [tex]\sqrt{1-x^2}=\sqrt{(1-x)(1+x)}[/tex] reduces this to

[tex]\displaystyle = \int_0^1 \ln^{2k}\left(\frac{\ln\left(\sqrt{\frac{1-x}{1+x}}\right)}{\ln\left(\sqrt{\frac{1+x}{1-x}}\right)}\right) \frac x{\sqrt{1-x^2}} \, dx[/tex]

The inner logarithms differ only by a sign, so that

[tex]\displaystyle = \int_0^1 \ln^{2k}(-1) \frac x{\sqrt{1-x^2}} \, dx[/tex]

Using the principal branch of the complex logarithm, we have

[tex]\ln(-1) = \ln|-1| + i\arg(-1) = i\pi[/tex]

and hence

[tex]\displaystyle \int_0^1 \ln^{2k} \left(\frac{\ln\left(\frac{1-\sqrt{1-x^2}}x\right)}{\ln\left(\frac{1-\sqrt{1-x^2}}x\right)}\right) \, dx = (i\pi)^{2k} \underbrace{\int_0^1 \frac x{\sqrt{1-x^2}} \, dx}_{=1} = \boxed{(-\pi^2)^k}[/tex]

where I assume k is an integer.

What is the difference between a regular and irregular quadrilateral?

Answers

Answer:

ir

Step-by-step explanation:

irregular

regular

what do you notice different?

the both have r, e, g, u, l, a, r, but one has ir at the beginning.

. What is the vertical asymptote(s) for y=x-5/x^2-4x-12​

Answers

Answer:

x = -2 and 6

Step-by-step explanation:

To find vertical asymptote, set the denominator equal to 0 and solve for x. See the guidelines below for determining VA

[tex]y=\frac{x-5}{x^{2} -4x-12}[/tex]

[tex]x^{2} -4x-12=0[/tex]

[tex]x^{2} -6x+2x-12=0[/tex]

[tex]x(x-6)+2(x-6)=0[/tex]

[tex](x+2)(x-6)=0[/tex]

[tex]x=-2,6[/tex]

Hope this helps and God bless!

4x - 5y = 6
- 8x +10y = -12
gausse elimination

Answers

Step-by-step explanation:

4x−5y=6

−8x+10y=−12

Isolate x for 4x –5y = 6: x=6-5y/4

Substitute x= 6+5y/4

[-8 (6-5y/4) +10y=-12]

Simplify

[-12= -12 ]

The solutions to the system of equations are:

x=6+5y/4

What key features do the functions f(x) = 12x and g of x equals the square root of x minus 12 end root have in common?

Both f(x) and g(x) include domain values of [-12, ∞) and range values of (-∞, ∞), and both functions have an x-intercept in common.
Both f(x) and g(x) include domain values of [12, ∞) and range values of [0, ∞), and both functions have a y-intercept in common.
Both f(x) and g(x) include domain values of [-12, ∞) and range values of (-∞, ∞), and both functions increase over the interval (-6, 0).
Both f(x) and g(x) include domain values of [12, ∞), and both functions increase over the interval (12, ∞).

Answers

Both f(x) and g(x) include domain values of [12, ∞), and both functions increase over the interval (12, ∞). Then the correct option is D.

What are domain and range?

The domain means all the possible values of the x and the range means all the possible values of the y.

The functions are given below.

[tex]\rm f(x) = 12x \\\\g(x) = \sqrt{x - 12}[/tex]

Then the domain of f(x) is (-∞, ∞) and the domain of g(x) is (12, ∞). And both the functions increases in the interval of (12, ∞).

More about the domain and range link is given below.

https://brainly.com/question/12208715

#SPJ1

solve 1 + cos theta = 2 cos^2 theta

Answers

[tex]~~~~1+ \cos \theta = 2 \cos^2 \theta \\\\\implies 2\cos^2 \theta -\cos \theta -1 = 0\\\\\implies 2 \cos^2 \theta -2\cos \theta + \cos \theta -1 = 0\\\\\implies 2 \cos \theta( \cos \theta -1) +(\cos \theta -1)=0\\\\\implies (\cos \theta -1)(2 \cos \theta +1)=0\\\\\implies \cos \theta = 1, ~~\cos \theta = -\dfrac 12\\\\\implies \theta = 2n\pi,~~~ \theta = 2n\pi \pm \dfrac{2\pi}3[/tex]

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