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3 years ago

7

Determine, to one decimal place, the length, width & height of the rectangular prism that would have the greatest volume, wi

th a surface area of 200 cm^2.

1 answer:

Leya [2.2K]3 years ago

7 0

Answer:

The length = The width = The height ≈ 5.8 cm

Step-by-step explanation:

The volume of a rectangular pyramid, V = l × w × h

The surface area of the pyramid = 2 × l × h + 2 × w × h + 2 × l × w = 200

∴ l × h + w × h + l × w = 200/2 = 100

We have that the maximum volume is given when the length, width, and height are equal and one length is not a fraction of the other. Therefore, we get;

At maximum volume, l = w = h

∴ l × h + w × h + l × w = 3·l² = 100

l² = 100/3

l = 10/√3

Therefore, the volume, v = l³ = (10/√3)³

The length = The width = The height = 10/√3 cm ≈ 5.8 cm

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a) $P(X=14)=(20C14)(0.8)^{14} (1-0.8)^{20-14}=0.109$

b) $P(X\geq 10) = 1-P(X \leq 9) = 0.9994$

c) $P(X \leq 16)= 1-P(X>16) =1-P(X \geq 17)= 1- [P(X=17) +...+P(X=20)]=0.589$

d) $E(X)= np = 20 *0.8 = 16$

$Var(X) = np(1-p) = 20*0.8*(1-0.8) = 3.2$

Step-by-step explanation:

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".

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

$X \sim Binom(n=20, p=1-0.2=0.8)$

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

$P(X)=(nCx)(p)^x (1-p)^{n-x}$

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

$nCx=\frac{n!}{(n-x)! x!}$

Part a

For this case we want this probability:

$P(X=14)$

And using the mass function we have this:

$P(X=14)=(20C14)(0.8)^{14} (1-0.8)^{20-14}=0.109$

Part b

For this case we want this probability:

$P(X \geq 10)$

And we can find this using the complement rule:

$P(X\geq 10) = 1-P(X$

$P(X=0)=(20C0)(0.8)^{0} (1-0.8)^{20-0}=1.05x10^{-14}$

$P(X=1)=(20C1)(0.8)^{1} (1-0.8)^{20-1}=8.39x10^{-13}$

$P(X=2)=(20C2)(0.8)^{2} (1-0.8)^{20-2}=3.19x10^{-11}$

$P(X=3)=(20C3)(0.8)^{3} (1-0.8)^{20-3}=7.65x10^{-10}$

$P(X=4)=(20C4)(0.8)^{4} (1-0.8)^{20-4}=1.30x10^{-8}$

$P(X=5)=(20C5)(0.8)^{5} (1-0.8)^{20-5}=1.66x10^{-7}$

$P(X=6)=(20C6)(0.8)^{6} (1-0.8)^{20-6}=1.66x10^{-6}$

$P(X=7)=(20C7)(0.8)^{7} (1-0.8)^{20-7}=1.33x10^{-5}$

$P(X=8)=(20C8)(0.8)^{8} (1-0.8)^{20-8}=8.65x10^{-5}$

$P(X=9)=(20C9)(0.8)^{9} (1-0.8)^{20-9}=0.00046$

And if we replace we got:

$P(X\geq 10) = 1-P(X \leq 9) = 0.9994$

Part c

For this case we want this probability:

$P(X \leq 16)$

And we can use the complement rule like this:

$P(X \leq 16)= 1-P(X>16) =1-P(X \geq 17)= 1- [P(X=17) +...+P(X=20)]$

$P(X=17)=(20C17)(0.8)^{17} (1-0.8)^{20-17}=0.205$

$P(X=18)=(20C18)(0.8)^{18} (1-0.8)^{20-18}=0.137$

$P(X=19)=(20C19)(0.8)^{19} (1-0.8)^{20-19}=0.0576$

$P(X=20)=(20C20)(0.8)^{20} (1-0.8)^{20-20}=0.0115$

And if we replace we got:

$P(X \leq 16)= 1-P(X>16) =1-P(X \geq 17)= 1- [P(X=17) +...+P(X=20)]=0.589$

Part d

The expected value is given by:

$E(X)= np = 20 *0.8 = 16$

$Var(X) = np(1-p) = 20*0.8*(1-0.8) = 3.2$

3 0

3 years ago