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Phantasy [73]
3 years ago
14

Three cube-shaped boxes are stacked one above the other. The volumes of two of the boxes are 1,331 cubic meters each, and the vo

lume of the third box is 729 cubic meters. What is the height of the stacked boxes in meters?
Mathematics
2 answers:
Zolol [24]3 years ago
8 0

Answer: The height of the stacked boxes is 31 meters.

Step-by-step explanation:

Since, the Volume of a cube = (side)³

The volume of first box = 1331 cubic meters,

⇒ (side)³ = 1331

⇒

Similarly, the side of second box = 11 meters ( Because, both boxes have the same volume )

Now, the volume of third box = 729 cubic meters

⇒ ⇒ (side)³ = 729

⇒

Thus, the height of the stacked boxes = Side of first box + side of second box + side of third box

= 11 + 11 + 9

= 31 meters.

blsea [12.9K]3 years ago
7 0
The volume of a box with dimensions x by x by x is  x^{3}.

So if we are given the volume V of a cube-shaped box, the side length of thit is \sqrt[3]{V}.

So we calculate the cubic roots of the volumes we have, and we add their heights.

To calculate the cubic root of 729, we can factorize it, and group the perfect cubes together, as follows:

\frac{729}{3}= \frac{600}{3} + \frac{120}{3} + \frac{9}{3}=200+40+3=243

\frac{243}{3}= \frac{240}{3}+ \frac{3}{3}=80+1=81, which we recognize as 3^{4}

so 729=3*3* 3^{4}= 3^{6}= ( 3^{2} )^{3}= 9^{3}

similarly 1,331 can be found to be 11^{3}.

Thus we have 2 boxes with side length equal to 11 m and one with side length equal to 9 m.


Answer: h= 11+11+9 = 31 (meters)



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Step-by-step explanation:

1) Previous concepts

The binomial distribution is a "DISCRETE probability distribution that summarizes the probability that a value will take one of two independent values under a given set of parameters. The assumptions for the binomial distribution are that there is only one outcome for each trial, each trial has the same probability of success, and each trial is mutually exclusive, or independent of each other".

2) Solution to the problem

Let X the random variable of interest, on this case we now that:

X \sim Binom(n=10, p=0.32)

The probability mass function for the Binomial distribution is given as:

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Where (nCx) means combinatory and it's given by this formula:

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Part a

P(X=2)=(10C2)(0.32)^2 (1-0.32)^{10-2}=0.211

Part b

P(X> 2)=1-P(X\leq 2)=1-[P(X=0)+P(X=1)+P(X=2)]

P(X=0)=(10C0)(0.32)^0 (1-0.32)^{10-0}=0.0211  

P(X=1)=(10C1)(0.32)^1 (1-0.32)^{10-1}=0.0995

P(X=2)=(10C2)(0.32)^2 (1-0.32)^{10-2}=0.211

P(X> 2)=1-P(X\leq 2)=1-[0.0211+0.0995+0.211]=0.668

Part c

P(2 \leq x \leq 5)=P(X=2)+P(X=3)+P(X=4)+P(X=5)

P(X=2)=(10C2)(0.32)^2 (1-0.32)^{10-2}=0.211

P(X=3)=(10C3)(0.32)^3 (1-0.32)^{10-3}=0.264

P(X=4)=(10C4)(0.32)^4 (1-0.32)^{10-4}=0.218

P(X=5)=(10C5)(0.32)^5 (1-0.32)^{10-5}=0.123

P(2 \leq x \leq 5)=0.211+0.264+0.218+0.123=0.816

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