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Delicious77 [7]
3 years ago
14

Miranda surveyed her classmates about the number of cousins each of them has. Her results are shown in the box plot below. Numbe

r of Cousins 3 4 5 6 7 8 9 Cousins Based on the data in the box plot, which statement could be true? There are more students who have 7 to 9 cousins than 6 to 7 cousins. The range of the number of cousins is 9. There is the same number of students with more than 6 cousins than the number of students with less than 6 cousin. number of cousins is 6.​

Mathematics
1 answer:
lozanna [386]3 years ago
8 0

Answer:

third option

Step-by-step explanation:

the median - of 6 - divides the data set into halves (theoretically, number of cousins is 50% less than 6 and 50% more than 6)

You might be interested in
Approximate π + √70 to the nearest hundredth.<br><br> 8.37<br> 11.14<br> 11.51<br> 12.14
stellarik [79]

Answer:

The answer would be 11.51

Step-by-step explanation:

π is rougroughly 3.14

√70 = 8.37

3.14 + 8.37 = 11.51

8 0
1 year ago
Read 2 more answers
A solid is formed by adjoining two hemispheres to the ends of a right circular cylinder. An industrial tank of this shape must h
mestny [16]

Answer:

Radius =6.518 feet

Height = 26.074 feet

Step-by-step explanation:

The Volume of the Solid formed  = Volume of the two Hemisphere + Volume of the Cylinder

Volume of a Hemisphere  =\frac{2}{3}\pi r^3

Volume of a Cylinder =\pi r^2 h

Therefore:

The Volume of the Solid formed

=2(\frac{2}{3}\pi r^3)+\pi r^2 h\\\frac{4}{3}\pi r^3+\pi r^2 h=4640\\\pi r^2(\frac{4r}{3}+ h)=4640\\\frac{4r}{3}+ h =\frac{4640}{\pi r^2} \\h=\frac{4640}{\pi r^2}-\frac{4r}{3}

Area of the Hemisphere =2\pi r^2

Curved Surface Area of the Cylinder =2\pi rh

Total Surface Area=

2\pi r^2+2\pi r^2+2\pi rh\\=4\pi r^2+2\pi rh

Cost of the Hemispherical Ends  = 2 X  Cost of the surface area of the sides.

Therefore total Cost, C

=2(4\pi r^2)+2\pi rh\\C=8\pi r^2+2\pi rh

Recall: h=\frac{4640}{\pi r^2}-\frac{4r}{3}

Therefore:

C=8\pi r^2+2\pi r(\frac{4640}{\pi r^2}-\frac{4r}{3})\\C=8\pi r^2+\frac{9280}{r}-\frac{8\pi r^2}{3}\\C=\frac{9280}{r}+\frac{24\pi r^2-8\pi r^2}{3}\\C=\frac{9280}{r}+\frac{16\pi r^2}{3}\\C=\frac{27840+16\pi r^3}{3r}

The minimum cost occurs at the point where the derivative equals zero.

C^{'}=\frac{-27840+32\pi r^3}{3r^2}

When \:C^{'}=0

-27840+32\pi r^3=0\\27840=32\pi r^3\\r^3=27840 \div 32\pi=276.9296\\r=\sqrt[3]{276.9296} =6.518

Recall:

h=\frac{4640}{\pi r^2}-\frac{4r}{3}\\h=\frac{4640}{\pi*6.518^2}-\frac{4*6.518}{3}\\h=26.074 feet

Therefore, the dimensions that will minimize the cost are:

Radius =6.518 feet

Height = 26.074 feet

5 0
3 years ago
Evaluate the expression when x =19 and y =7
PtichkaEL [24]

Answer:

Plug in X and Y by replacing the given numbers where you see letters.

Step-by-step explanation:

What I said above.

3 0
3 years ago
How to express each of the following pairs of
vichka [17]

Answer:

<em>a) 3<x<8</em>

<em>b) -4<x<-2</em>

<em>c) -6<x<5</em>

<em>d) -21/4 < x-8/3</em>

<em>e) 0<x<7</em>

Step-by-step explanation:

Given the following inequalities

a)  x > 3, 2x - 3 < 15

Solve 2x - 3 < 15

2x < 15+3

2x<18

x<18/2

x<8

Combine x>3 and x<8

If x>3, then 3<x

On combining, we have:

3<x<8

b) For the inequalities 25 > 1 - 6x, 1 > 3x + 7

25 > 1 - 6x

25-1>-6x

24>-6x

Divide through by -6:

24/-6 > -6x/-6

-4 <x

For the inequality 1 > 3x + 7

1-7>3x

-6>3x

-6/3 > 3x/3

-2 > x

x < -2

Combining both results i.e -4 <x and x < -2, we will have:

-4<x<-2

c) For the inequalities 2x - 7<3< 27 + 4x

On splitting:

2x - 7<3 and 3< 27 + 4x

2x < 3+7

2x<10

x<5

Also for 3< 27 + 4x

3-27<4x

-24<4x

-24/4 < 4x/4

-6<x

Combining both solutions i.e x<5 and -6<x will give;

<em>-6<x<5</em>

d) For the inequalities 3x + 8 <0 < 21 + 4x

3x + 8 <0

3x < -8

x < -8/3

Also for 0 < 21 + 4x

0-21<4x

-21<4x

-21/4 < 4x/4

-21/4 < x

Combining -21/4 < x and x < -8/3 will give;

<em>-21/4 < x-8/3</em>

<em></em>

e) For the inequalities 5x - 36 < -1 < 2x – 1​

Split:

5x - 36 < -1

5x < -1+36

5x<35

5x/5 < 35/5

x < 7

For the expression -1 < 2x – 1​

-1+1 < 2x

0 < 2x

0<x

Combining both inequalities 0<x and x < 7 will give:

<em>0<x<7</em>

7 0
3 years ago
10. A garden 150 m long and 80 m wide has a road 2 m wide all around
Sergio [31]

Answer:

<em>length</em><em> </em><em>l</em><em> </em><em>=</em><em>150</em><em> </em>

<em>breadth</em><em> </em><em>b</em><em> </em><em>=</em><em> </em><em>80</em><em> </em>

<em>lf</em><em> </em><em>2</em><em> </em><em>m</em><em> </em><em>wide</em><em> </em><em> </em><em>of</em><em> </em><em>road</em><em> </em><em>is</em><em> </em><em>inside</em><em> </em><em>the</em><em> </em><em>garden</em>

<em>then</em>

<em>lt's</em><em> </em><em>area</em><em> </em><em>=</em><em> </em><em>l</em><em> </em><em>×</em><em>b</em>

<em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em>=</em><em> </em><em>150</em><em> </em><em>×</em><em> </em><em>2</em>

<em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em>=</em><em> </em><em>300m</em><em>^</em><em>2</em>

6 0
3 years ago
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