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Ratling [72]
1 year ago
12

The dimensions of a rectangular bin are consecutive integers. If the volume of the bin is 4896 cubic inches, what are the dimens

ions of the bin?
Mathematics
2 answers:
Rom4ik [11]1 year ago
7 0

cubed root of 4896 = 16.98

round to 17

17-1 = 16

17 +1 = 18

16* 17 *18 = 4896

 bin is 16 in x 17 in x 18 in

Archy [21]1 year ago
5 0

Answer:

The dimension of the rectangle is 16\times 17\times 18

Step-by-step explanation:

Given : The dimensions of a rectangular bin are consecutive integers. If the volume of the bin is 4896 cubic inches.

To find : What are the dimensions of the bin?

Solution :

The dimensions of a rectangular bin are consecutive integers.

Let the consecutive integers are x,x+1,x+2

So let, length, l=x

Breadth b=x+1

Height h=x+2

The volume of the bin is

V=l\times b\times h

4896=x\times (x+1)\times (x+2)

4896=(x^2+x)(x+2)

4896=x^3+2x^2+x^2+2x

x^3+3x^2+2x-4896=0

Plot the equation graphically to find the value of x.

Refer the attached figure below.

The value of x is 16.

So, Length is l=16

Breadth is b=x+1=16+1=17

Height is h=x+2=16+2=18

The dimension of the rectangle is 16\times 17\times 18

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Answer:

\epsilon = 3.958\times 10^{-3}

Step-by-step explanation:

The Young's Module of Aluminium is E = 69\times 10^{9}\,Pa. The axial stress on the specimen is:

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\sigma = \frac{35500\,N}{(0.01\,m)\cdot (0.013\,m)}

\sigma = 2.731\times 10^{8}\,Pa

The strain is derived of following expression:

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4 0
1 year ago
Find the standard deviation of the following data. Answers are rounded to the nearest tenth. 5, 5, 6, 12, 13, 26, 37, 49, 51, 56
kogti [31]

Answer:

24.2(to the nearest tenth)

Step-by-step explanation:

The question is ungrouped data type

standard deviation =√ [∑ (x-μ)² / n]

mean (μ)=∑x/n

           = \frac{5+5+6+12+13+26+37+49+51+56+56+84}{12}

          =33.3

x-μ   for data 5, 5, 6, 12, 13, 26, 37, 49, 51, 56, 56, 84 will be

                   -28.3, -28.3, -27.3, -21.3, -7.3, 3.7, 15.7,17.7, 22.7,22.7,50.7

(x-μ)² will be 800.89, 800.89,745.29,453.69,53.29,13.69,246.49,313.29,515.29,515.29,2570.49

∑ (x-μ)² will be = 7028.59

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6 0
2 years ago
A cone-shaped pile of gravel has a diameter of 30 m and a height of 9.1 m.
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d=30
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6 0
1 year ago
Read 2 more answers
You just discovered that you have 100 feet of fencing and you have decided to make a rectangular garden. Assume the lengths of t
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3 0
1 year ago
The aquarium at Sea Critters Depot contains 140 fish. Eighty of these fish are green swordtails (44 female and 36 male) and 60 a
maxonik [38]

Given that:

Total number of fish = 140

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Solution:

A. We have to find the probability that the selected fish is a green swordtail.

\text{P(green swordtail)}=\dfrac{\text{Total green swordtail fish}}{\text{Total fish}}

\text{P(green swordtail)}=\dfrac{80}{140}

\text{P(green swordtail)}=\dfrac{4}{7}

Therefore, the probability that the selected fish is a green swordtail is \dfrac{4}{7}.

B.  We have to find the probability that the selected fish is male.

\text{P(Male fish)}=\dfrac{\text{Total male fish}}{\text{Total fish}}

\text{P(Male fish)}=\dfrac{36+24}{140}

\text{P(Male fish)}=\dfrac{60}{140}

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Therefore, the probability that the selected fish is a male, is \dfrac{3}{7}.

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\text{P(Male green swordtail)}=\dfrac{\text{Total male green swordtail fish}}{\text{Total fish}}

\text{P(Male green swordtail)}=\dfrac{36}{140}

\text{P(Male green swordtail)}=\dfrac{9}{35}

Therefore, probability that the selected fish is a male green swordtail is \dfrac{9}{35}.

D.

We have to find the probability that the selected fish is either a male or a green swordtail.

\text{P(Male or green swordtail)}=\dfrac{\text{Total male or green swordtail fish}}{\text{Total fish}}

\text{P(Male or green swordtail)}=\dfrac{44+36+24}{140}

\text{P(Male or green swordtail)}=\dfrac{96}{140}

\text{P(Male or green swordtail)}=\dfrac{24}{35}

Therefore, the probability the selected fish is either a male or a green swordtail is \dfrac{24}{35}.

4 0
1 year ago
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