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3 years ago

P and a are both prime numbers. Find the values of p and q so that 654p/q is a perfect cube.

Mathematics

1 answer:

daser333 [38]3 years ago

3 0

6x54= 324

6x50=300

6x4=24

300+24=324

The previous nearest cube root is 216 and the next cube root would be 343.

p/q could be any prime number which there are a lot of.

im going to say p is 1 and q is 3

next you would divide 1 by 3

1/3=0.666666667

324x0.666666667=216 exact

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$E(X)=16\cdot 0.3+18\cdot 0.1+20\cdot 0.6=18.6$

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From the information given we know the probability mass function (pmf) of random variable X.

$\left|\begin{array}{c|ccc}x&16&18&20\\p(x)&0.3&0.1&0.6\end{array}\right|$

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The Expected value or the mean value of X with set of possible values D, denoted by E(X) or μ is

$E(X) = $\sum_{x\in D} x\cdot p(x)$

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$E(X)=16\cdot 0.3+18\cdot 0.1+20\cdot 0.6=18.6$

If the random variable X has a set of possible values D and a probability mass function, then the expected value of any function h(X), denoted by E[h(X)] is computed by

$E[h(X)] = $\sum_{D} h(x)\cdot p(x)$

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$E[h(X)] = $\sum_{D} h(x)\cdot p(x)\\E[X^2]=$\sum_{D}x^2\cdot p(x)\\ E(X^2)=16^2\cdot 0.3+18^2\cdot 0.1+20^2\cdot 0.6\\E(X^2)=349.2$

The variance of X, denoted by V(X), is

$V(X) = $\sum_{D}E[(X-\mu)^2]=E(X^2)-[E(X)]^2$

Therefore

$V(X) = E(X^2)-[E(X)]^2\\V(X)=349.2-(18.6)^2\\V(X)=3.24$

Point b:

We know that the price of a freezer having capacity X is 60X − 650, to find the expected price paid by the next customer to buy a freezer you need to:

From the rules of expected value this proposition is true:

$E(aX+b)=a\cdot E(x)+b$

We have a = 60, b = -650, and E(X) = 18.6. Therefore

The expected price paid by the next customer is

$60\cdot E(X)-650=60\cdot 18.6-650=466$

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3 years ago

P and a are both prime numbers. Find the values of p and q so that 6*54*p/q is a perfect cube.

P and a are both prime numbers. Find the values of p and q so that 654p/q is a perfect cube.