(iii) Find the smallest whole number h such that4704/his a cube number.

Mathematics

1 answer:

yan [13]2 years ago

6 0

Answer:

$h=588$

Step-by-step explanation:

Given the fraction $\dfrac{4,704}{h}$

Consider its numerator and factor it:

$4,704=2^5\cdot 3\cdot 7^2$

You cann not take the cube root from $3, 7^2, 2^5,$ you can only take the cube root from $2^3$ , then the smallest whole number $h$ such that $\dfrac{4,704}{h}$ is a cube number is

$h=2^2\cdot 3\cdot 7^2=4\cdot 3\cdot 49=588$

When $h=588,$

$\dfrac{4,704}{588}=8=2^3$

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A="b is in the middle"

B="c is to the right of b"

C="The letter def occur together in that order"

a) b can be in 7 places, but only one is the middle. So, P(A)=1/7

b) X=i, "b is in the i-th position"

Y=j, "c is in the j-th position"

$P(B)=\displaystyle\sum_{i=1}^{6}(P(X=i)\displaystyle\sum_{j=i+1}^{7}P(Y=j))=\displaystyle\sum_{i=1}^{6}\frac{1}{7}(\displaystyle\sum_{j=i+1}^{7}\frac{1}{6})=\frac{1}{42}\displaystyle\sum_{i=1}^{6}(\displaystyle\sum_{j=i+1}^{7}1)=\frac{6+5+4+3+2+1}{42}=\frac{1}{2}$

P(B)=1/2

c) X=i, "d is in the i-th position"

Y=j, "e is in the j-th position"

Z=k, "f is in the i-th position"

$P(C)=\displaystyle\sum_{i=1}^{5}( P(X=i)P(Y=i+1)P(Z=i+2))=\displaystyle\sum_{i=1}^{5}(\frac{1}{7}\times\frac{1}{6}\times\frac{1}{5})=\frac{1}{210}\displaystyle\sum_{i=1}^{5}(1)=\frac{1}{42}$

P(C)=1/42

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