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marshall27 [118]
3 years ago
14

Please help, I will mark brainliest.

Mathematics
1 answer:
Bess [88]3 years ago
7 0

Answer:

Step-by-step explanation:

Total number of students who roller skated = 80

Number of students who roller skated with laser tag = 39

Number of students who roller skated with no laser tag = 80 - 39 = 41

Therefore, percentage of students who roller skated with no laser tag = \frac{41}{154}\times 100 = 27%

Since, number of students who played laser tag = 85

Percentage of students who played laser tag = \frac{85}{154}\times 100 = 55%

Total number of students who played no laser tag = 45% of 154 = 69

Number of students with no roller skates and no laser tag = 69 - 41 = 28

Percentage of students with no roller skates and no laser tag = \frac{28}{154}\times 100        

                                                                                                       = 18%

Percentage of students who played with no rollers skates = (30 + 18)%

                                                                                                 = 48%

                                  Laser tag                 No laser tag                Total

Roller skate                     25%                           27%                      52%

No Roller skate               30%                           18%                      48%

Total                                55%                           45%                      100%

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Your favorite craft store had a weekend sale. On the first day, all $10 items were 70% off. You bought some number of these of t
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8 0
3 years ago
An open top box is to be built with a rectangular base whose length is twice its width and with a volume of 36 ft 3 . Find the d
denpristay [2]

Answer:

The dimensions of the box that minimize the materials used is 6\times 3\times 2\ ft

Step-by-step explanation:

Given : An open top box is to be built with a rectangular base whose length is twice its width and with a volume of 36 ft³.

To find : The dimensions of the box that minimize the materials used ?

Solution :

An open top box is to be built with a rectangular base whose length is twice its width.

Here, width = w

Length = 2w

Height = h

The volume of the box V=36 ft³

i.e. w\times 2w\times h=36

h=\frac{18}{w^2}

The equation form when top is open,

f(w)=2w^2+2wh+2(2w)h

Substitute the value of h,

f(w)=2w^2+2w(\frac{18}{w^2})+2(2w)(\frac{18}{w^2})

f(w)=2w^2+\frac{36}{w}+\frac{72}{w}

f(w)=2w^2+\frac{108}{w}

Derivate w.r.t 'w',

f'(w)=4w-\frac{108}{w^2}

For critical point put it to zero,

4w-\frac{108}{w^2}=0

4w=\frac{108}{w^2}

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w^3=3^3

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Derivate the function again w.r.t 'w',

f''(w)=4+\frac{216}{w^3}

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So, it is minimum at w=3.

Now, the dimensions of the box is

Width = 3 ft.

Length = 2(3)= 6 ft

Height = \frac{18}{3^2}=2\ ft

Therefore, the dimensions of the box that minimize the materials used is 6\times 3\times 2\ ft

4 0
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