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nevsk [136]
2 years ago
9

Feeling anxious about another pandemic induced run-on toilet paper, Mrs. Phillips is making room in a closet for hording Angel S

oft toilet paper. Using the Fermi process, she wants to estimate the number of rolls of toilet paper she can fit into a rectangular section of a closet with dimensions of length 36 inches, width 36 inches, height 108 inches.
One Angel Soft MEGA roll has a diameter 5 inches, height 3.75 inches.

About how many Angel Soft MEGA rolls can be fit into the closet space?
Show all math work needed to complete this problem.
Mathematics
2 answers:
nevsk [136]2 years ago
5 0

Answer:

1372

Step-by-step explanation:

Rectangular closet dimensions:

  • length = 36 in
  • width = 36 in
  • height = 108 in

Modelling the roll of toilet paper as a cylinder with dimensions:

  • diameter = 5 in
  • height = 3.75 in

If we sit the rolls of toilet paper in the closet with their circular ends as their bases, then we can calculate the number of rolls that cover the base of the closet by dividing the length (and width) of the closet by the diameter of the toilet roll.

⇒ 36 in ÷ 5 in = 7.2 in

Therefore, we can fit 7 toilet rolls along the length and 7 toilet rolls along the width of the closet, meaning that we can fit 7 × 7 = 49 toilet rolls over the base of the closet.

Now calculate how many toilet rolls we can stack on top of each other.

To do this, divide the height of the closet by the height of the toilet roll:

⇒ 108 in ÷ 3.75 in = 28.8

Therefore, we can stack 28 layers of 49 toilet rolls in the closet.

So the total number of toilet rolls = 28 × 49 = 1372

There are other ways of stacking the toilet rolls in the closet (for example, turning them on their side), however, we cannot calculate the number of rolls the closet fits by simply dividing the volume of the closet space by the volume of a toilet roll, since the toilet rolls are cylinders and so there will always be some space between them.

romanna [79]2 years ago
3 0

For closet

It's a rectangle

  • L=36in
  • B=36in
  • H=108in

Volume

  • LBH
  • 36²(108)
  • (36)³(3)
  • 139968in³

For rolls:-

Cylindrical body

  • radius=r=5/2=2.5in
  • Height=h=3.75in

Volume:-

  • πr²h
  • π(2.5)²(3.75)
  • π(6.25)(3.75)
  • 73.6in³

Total rolls:-

  • Volume of closet /Volume of rolls
  • 139968/73.6
  • 1901.7
  • 1902 rolls
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Answer:

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Step-by-step explanation:

First, we need to find the slope. We can do this by using the formula:

1) y₂-y₁/x₂-x₁

2) 9-5/5-3

3) 4/2

4) The slope is 2

Next, we have to first plug the slope (2) and one of the coordinate pairs (I chose (3,5)) into point-slope format.

5) y-y₁=m(x-x₁)

6) y-5 =4(x-3)

7) Now we just simplify this equation to make it slope intercept form (y=mx+b)

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6 0
2 years ago
The accompanying data came from a study of collusion in bidding within the construction industry. No. Bidders No. Contracts 2 6
viktelen [127]

Answer:

a)

The proportion of contracts involved at most five bidders is 0.667.

The proportion of contracts involved at least five bidders is 0.51.

b)

The  proportion of contracts involved  between five and 10 inclusive bidders is 0.5.

The  proportion of contracts involved  strictly between five and 10 bidders is 0.304.

Step-by-step explanation:

No. bidders    No. contracts     Relative frequency of contracts

2                         6                       6/102=0.0588

3                         20                     20/102=0.1961

4                         24                     24/102=0.2353

5                         18                       18/102=0.1765

6                         13                        13/102=0.1275

7                          7                          7/102=0.0686

8                          5                          5/102=0.049

9                          6                          6/102=0.0588

10                         2                           2/102=0.0196

11                          1                            1/102=0.0098

Total                  102

a)

We have to find proportion of contracts involved at most five bidders.

Proportion of at most 5= Relative frequency 2+ Relative frequency 3+ Relative frequency 4+ Relative frequency 5

Proportion of at most 5=0.0588+ 0.1961+0.2353+0.1765

Proportion of at most 5=0.6667

The proportion of contracts involved at most five bidders is 0.667.

proportion of at least five bidders= proportion≥5= 1- proportion less than 5

Proportion less than 5=0.0588+ 0.1961+0.2353=0.4902

proportion of at least five bidders=1-0.4902=0.5098

The proportion of contracts involved at least five bidders is 0.51

b)

We have to find proportion of contracts involved  between five and 10 inclusive bidders.

Proportion of contracts between five and 10 inclusive= Relative frequency 5+ Relative frequency 6+ Relative frequency 7+ Relative frequency 8+ Relative frequency 9+ Relative frequency 10

Proportion of between five and 10 inclusive=0.1765 +0.1275 +0.0686 +0.049 +0.0588 +0.0196

Proportion of between five and 10 inclusive=0.5

The  proportion of contracts involved  between five and 10 inclusive bidders is 0.5

We have to find proportion of contracts involved  strictly between five and 10 bidders.

Proportion of contracts strictly between five and 10=  Relative frequency 6+ Relative frequency 7+ Relative frequency 8+ Relative frequency 9

Proportion of strictly between five and 10=0.1275 +0.0686 +0.049 +0.0588

Proportion of strictly between five and 10=0.3039

The  proportion of contracts involved  strictly between five and 10 bidders is 0.304.

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Answer:

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Step-by-step explanation:

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32500 + 500, or 33000

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zhannawk [14.2K]

Answer:

B would be the answer for this question.


Step-by-step explanation:


7 0
3 years ago
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Butoxors [25]
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So, with area 42, dimensions could be: 1 * 42
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Just draw the lines with that measures, two pair of two equal opposite lines.

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