Tom and Susan park at different lots.

To see which lot is busier, they count the numbers of cars in the lots each day as they arrive.

Their data are shown in the box plots.


the questions to compare the variabilities of the data sets.

1. What is the interquartile range for Tom's data? Explain how you found the interquartile range.

Write your answer in the space below.

1

Answer:

Step-by-step explanation:

68 frefq4gtrhrsahhrhshrr

Step-by-step explanation:

1)

7x+20°+3x=180°{straight angle}

10x=180-20

x=160/10

x=16°

again,

y=7x+20{vertically opposed angle r equal}

Answer:

1. x = 20

y = 120

2. x = 263

Step-by-step explanation:

•••

As AOB is a line,

( 7x - 20 ) + 3x = 180

7x - 20 + 3x = 180

10x - 20 = 180

10x = 180 + 20 = 200

10x = 200

x = 200/10 = 20

Then, as COD is a line,

y + 3x = 180

y + 3(20) = 180

y + 60 = 180

y = 180 - 60 = 120

•••

2. Draw a parallel line AB and CD.

Then,

Angle BAE + Angle y = 180 [Co-interior angles]

56 + y = 180

y = 180 - 56 = 124

Angle DCE + Angle z = 180

41 + z = 180

z = 180 - 41

z = 139

Now, z + y = 139 + 124 = 263

[tex][/tex]

Answer:

7 24 25

because , according to Pythagorean triplet....

Answer:

supplementary angle

Step-by-step explanation:

angles 6 and 8 are same side exterior angles

Answer:

stop being lazy and do youre own work lol

Step-by-step explanation:

Answer:

option 2 line graph

Step-by-step explanation:

100% on EDGE

Answer:

b

Step-by-step explanation:

Answer:

The 80% confidence interval for the population proportion of tenth graders reading at or below the eighth grade level is (0.149, 0.207).

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of [tex]\pi[/tex], and a confidence level of [tex]1-\alpha[/tex], we have the following confidence interval of proportions.

[tex]\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]

In which

z is the zscore that has a pvalue of [tex]1 - \frac{\alpha}{2}[/tex].

Suppose a sample of 292 tenth graders is drawn. Of the students sampled, 240 read above the eighth grade level.

So 292 - 240 = 52 read below or at eight grade level, and that [tex]n = 292, \pi = \frac{52}{292} = 0.178[/tex]

80% confidence level

So [tex]\alpha = 0.2[/tex], z is the value of Z that has a pvalue of [tex]1 - \frac{0.2}{2} = 0.9[/tex], so [tex]Z = 1.28[/tex].

The lower limit of this interval is:

[tex]\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.178 - 1.28\sqrt{\frac{0.178*0.822}{292}} = 0.149[/tex]

The upper limit of this interval is:

[tex]\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.178 + 1.28\sqrt{\frac{0.178*0.822}{292}} = 0.207[/tex]

The 80% confidence interval for the population proportion of tenth graders reading at or below the eighth grade level is (0.149, 0.207).

Answer:

100

Step-by-step explanation:

Answer:

m∠Q = 109°

m∠QRT = 109°

x = 4

Step-by-step explanation:

1). "Opposite angles of a parallelogram are equal"

By this property,

m∠Q = m∠S = 109°

2). "Opposite sides of a parallelogram are parallel and equal in measure"

By this property,

RQ║ST and diagonal RT is a transversal line.

m∠QRT = ∠SRT = 30° [Alternate interior angles]

3). "Opposite sides of a parallelogram are parallel and equal in measure"

RS = QT

2x = 8

x = 4

Answer:

y intercept is 11 or (0,11)

Step-by-step explanation:

[tex]y = 2 {(x - 2)}^{2} + 3 \\ y = 2 {(0 - 2)}^{2} + 3 \\ y = 2(4) + 3 \\ y = 8 + 3 \\ y = 11[/tex]

Answer:

361,899

hope this helps

Answer:

350,000

Step-by-step explanation:

1,500,000 - 1,138, 101 = 361,899

361,899 - 350,000 = 11,899

400,000 - 361,899 = 38,101

11,899 is less than 38,101

So, 350,000 is the answer.

Answer: there are 3150 boys and 3850 girl

Step-by-step explanation:

45÷100×7000= 3150 and 7000- 3150 = 3850

Hope this helps!!

Answer:

36 and 72

Step-by-step explanation:

Answer:

V≈17974.16

Step-by-step explanation:

Using the formulas

V=4

3πr3

d=2r

Solving for V

V=1/ 6πd^3=1/ 6·π·32.53≈17974.16422

Answer: b is correct

Step-by-step explanation: i did this

The area of rectangle ABCD is 34 square unit.

In a cartesian plane there are two coordinates as P(a, b) and Q(c, d) then the distance between these points are

PQ = √(c- a)² + (d- b)²

We have Coordinates of rectangle

A(-1, 4), B(3, 3), C(1, -5) and D(-3, -4)

Using Distance Formula

AB = √ (3-4)² + (3+1)²

AB = √1 + 16

AB = √17 unit

Now, CB = √(-5-3)² + (1-3)²

CB= √ 64 + 4

CB = √68

CB = 2√17 unit

So, the area of Rectangle ABCD is

= l w

= √17 x 2√17

= 2 x 17

= 34 square unit

Learn more about Distance Formula here:

https://brainly.com/question/25841655

#SPJ7

Answer: 2y124√y12

How to: Simplify the radical by breaking the radicand up into a product of known factors, assuming positive real numbers.

Have a great day and stay safe !

9514 1404 393

Answer:

  [tex]2\cdot y^{\frac{5}{8}}[/tex]

Step-by-step explanation:

The applicable rules of exponents are ...

  [tex](a^b)^c=a^{bc}\\\\(ab)^c=a^c\cdot b^c\\\\\sqrt[n]{x^m}=x^{\frac{m}{n}}[/tex]

__

So, the given expression can be rewritten to ...

  [tex](4\sqrt[4]{y^5})^{\frac{1}{2}}=(4^{\frac{1}{2}})(y^{\frac{5}{4}})^{\frac{1}{2}}= 2\cdot y^{\frac{5}{4}\cdot\frac{1}{2}}=\boxed{2\cdot y^{\frac{5}{8}}}[/tex]

_____

k = 2; n = 5/8

Answer:

Adam scored 60, Brandon scored 66, Chen scored 74, Damion scored 75, and Erica scored 75.

Step-by-step explanation:

Since five friends took the maths test, and Adam, Brandon, and Chen together together scored 200 marks; Brandon, Chen and Damion together scored 215; Chen, Damion and Erica together scored 224; and Damion and Erica scored more than Chen; While the five of them together scored 350 marks, to determine what are their individual scores the following calculations must be performed:

Adam + Brandon + Chen = 200

Damion + Erica = 150

Brandon + Chen + Damion = 215

Adam + Erica = 135

Chen + Damion + Erica = 224

Adam + Brandon = 126

Adam + Brandon = 126 + Chen = 200

Chen = 200 - 126

Chen = 74

Damion and Erica scored more than Chen

Chen + Damion + Erica = 224

74 + Damion + Erica = 224

Damion + Erica = 150

Damion = 75

Erica = 75

Brandon + Chen + Damion = 215

Brandon + 74 + 75 = 215

Brandon = 215 - 74 - 75

Brandon = 66

Adam = 350 - 75 - 75 - 74 - 66

Adam = 60

Therefore, Adam scored 60, Brandon scored 66, Chen scored 74, Damion scored 75, and Erica scored 75.

Answer:

[tex]20 > c > 4[/tex]

Step-by-step explanation:

Given

[tex]Sides =8\ and\ 12[/tex]

Required

The third side

To do this, we make use of the triangle inequality theorem which implies that:

[tex]a + b > c[/tex]

[tex]a + c > b[/tex]

[tex]b + c > a[/tex]

Where

[tex]a,b,c\to sides[/tex]

So, we have:

[tex]8 + 12 > c[/tex]

[tex]8 + c > 12[/tex]

[tex]12 + c > 8[/tex]

Solve all inequalities

[tex]8 + 12 > c[/tex]

[tex]20 > c[/tex]

[tex]8 + c > 12[/tex]

[tex]c > 4[/tex]

[tex]12 + c > 8[/tex]

[tex]c > -4[/tex]

Ignore negative inequalities.

So, we have:

[tex]20 > c[/tex] and [tex]c > 4[/tex]

Combine

[tex]20 > c > 4[/tex]

The above implies that the third length is between 5 and 19 (inclusive)

Answer:

1. The surface area of the cube is 324 square units

The volume of the cube 360 unit cube

2. The surface area of the cylinder is approximately 169.65 square units

The volume of the cylinder is approximately 169.65 unit cube

3. The surface area of a square pyramid is 360 square units

The volume of the square pyramid is 400 unit cube

4. The surface area of a cone is approximately 452.39 square units

The volume of the cone is approximately 50.27 unit cube

5. The surface area of the triangular prism is 240 square units

The volume of the triangular prism is 180 unit cube

6. The surface area of the sphere is approximately 804.25 square units

The volume of the sphere is approximately 2.144.66

7. The surface area of the composite figure is approximately 653.46 square units

The volume of the composite figure is approximately 1,474.45 unit cube

Step-by-step explanation:

1. The surface area of the figure, SA = 2 × (w·h + l·w + h·l)

Where;

w = The width of the figure = 6

l = The length of the figure = 12

h = The height of the figure = 5

We get;

SA = 2 × (6 × 5 + 12 × 6 + 5 × 12) = 324

The surface area of the figure, SA = 324

The volume of the figure, V = l × w × h

∴ V = 12 × 6 × 5 = 360

The volume of the figure, V = 360

2. The surface area of a cylinder, SA = 2·π·r² + 2·π·r·h

The radius of the given cylinder, r = 3

The height of the given cylinder, h = 6

∴ SA = 2×π×3² + 2×π×3×6 ≈ 169.65

The surface area of the cylinder, SA ≈ 169.65

The volume of a cylinder, V = π·r²·h

∴ V = π×3²×6 ≈ 169.65

The volume of the cylinder, V ≈ 169.65

3. The surface area of a square pyramid, SA = b² + 4·(1/2)·b·√((b/2)² + h²)

Therefore, for the given square pyramid, we have;

SA = 10² + 4×(1/2)×10×√((10/2)² + 12²) = 360

The surface area of a square pyramid, SA = 360

The volume of a square pyramid, V = (1/3) × Area of Base × Height

Therefore, or the given pyramid we have;

V = (1/3) × 10² × 12 = 400

The volume of the square pyramid, V = 400

4. The surface area of a cone, SA = π·r·(r + l)

Where;

The radius of the cone = r

The slant height of the cone, l = 10

The height of the cone, h = 6

∴ The radius of the cone, r = √(10² - 6²) = 8

∴ SA = π×8×(8 + 10) ≈ 452.39

The surface area of a cone, SA ≈ 452.39

The volume of the cone, V = (1/3) × π·r·h

∴ V = (1/3) × π × 8 × 6 ≈ 50.27

The volume of the cone, V ≈ 50.27

5. The surface area of the triangular prism, SA = 2 × (1/2)× b·h + b·w + h·w + w·l

Where;

b = The base length of the triangular surfaces = 5

h = The height of the triangular surfaces = 12

w = The width of the triangular prism = 6

l = The slant length of the prism = 13

Therefore;

SA = 2 × (1/2)× 5 × 12 + 5 × 6 + 12 × 6 + 6 × 13 = 240

The surface area of the triangular prism, SA = 240

The volume of a triangular prism, V = (1/2)·b·h·w

V = (1/2) × 5 × 12 × 6 = 180

The volume of the triangular prism, V = 180

6. The surface of a sphere, SA = 4·π·r²

Where;

r = The radius of the sphere = 8

∴ SA = 4 × π × 8² ≈ 804.25

The surface area of the sphere, SA ≈ 804.25

The volume of a sphere, V = (4/3)·π·r³

∴ V ≈ (4/3)×π×8³ ≈ 2,144.66

The volume of the given sphere, V ≈ 2.144.66

7. The figure is a composite figure made up of a cone and an hemispher

The surface area of the cone shaped part of the figure, SA = π·r·l

Where;

r = The radius of the cone = 8

l = The slant height of the cone = 10

∴ SA₁ = π × 8 × 10 ≈ 251.34

The surface area of the cone shaped part of the figure, SA₁ ≈ 251.34

The volume of the cone, V₁ = (1/3)·π·r²·h

Where;

h = The height of the cone = √(10² - 8²) = 6

∴ V₁ = (1/3) × π × 8² × 6 ≈ 402.12

The volume of the cone, V₁ ≈ 402.12

The surface area of the hemisphere, SA₂ = 2·π·r²

∴ SA₂ = 2 × π × 8² ≈ 402.12

The surface area of the hemisphere, SA₂ ≈ 402.12

The volume of a hemisphere, V₂ = (2/3)·π·r³

∴ V₂ = (2/3) × π × 8³ ≈ 1072.33

The volume of a hemisphere, V₂ ≈ 1,072.33

The surface area of the composite figure, SA = SA₁ + SA₂

∴ SA = 251.34 + 402.12 = 653.46

The surface area of the composite figure, SA ≈ 653.46

The volume of the composite figure, V = V₁ + V₂

∴ V = 402.12 + 1,072.33 = 1,474.45

The volume of the composite figure, V ≈ 1,474.45.

Answer:

C.18

Step-by-step explanation:

In the chart it shows that 18 boys like to read science fiction. Hope this helps :)

Answer:

A. The curve goes through one cycle and starts over

Step-by-step explanation:

I calculated it logically

Answer:

O.5 cents each

Step-by-step explanation

You have to multiply 75 by 0.15

Answer:

a) (-1/2,3/2)

Step-by-step explanation:

-3x = x + 2

4x + 2 = 0

4x = -2

x = -1/2

y = -3(-1/2) = 3/2

Step-by-step explanation:

3x+9+4x-3+8x-6=180

15x=180

x=12

angle 1= 3x+9=36+9=45°

angle 2=4x-3=48-3=45°

angle 3=8x-6=96-6=90°

1) it's right triangle

2) it's isosceles triangle

Answer:

[tex] M = (-4, -5) [/tex]

Step-by-step explanation:

Given:

C = (-2, -10) and D = (-6, 0)

Required:

Midpoint M of segment CD.

Solution:

[tex] M = (\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}) [/tex]

Where,

[tex] (-2, -10) = (x_1, y_1) [/tex]

[tex] (-6, 0) = (x_2, y_2) [/tex]

Plug in the values into the equation

[tex] M = (\frac{-2 + (-6)}{2}, \frac{-10 + 0}{2}) [/tex]

[tex] M = (\frac{-8}{2}, \frac{-10}{2}) [/tex]

[tex] M = (-4, -5) [/tex]