Mathematics: In the equation $rac{1}{j} + rac{1}{k} = rac{1}{3}$, both $j$ and $k$ are positive integers. What is the sum of all possible values for $k$?

In the equation $\frac{1}{j} + \frac{1}{k} = - id1114045930, explained1256

Ответ:

gennhill14

04.09.2021

Mathematics

Hello,

answer 22

Step-by-step explanation:

$\dfrac{1}{j} +\dfrac{1}{k} =\dfrac{1}{3} \\\\\dfrac{k+j}{j*k} =\dfrac{1}{3} \\\\\\3k+3j=j*k\\\\k(3-j)=-3j\\\\k=\dfrac{3j}{j-3} \\\\k=\dfrac{3j-9+9}{j-3} \\\\k=3+\dfrac{9}{j-3} \\\\\\j-3\ must\ be \ a divisor\ of\ 9 == 1,3,9\\\\j-3=1 == j=4 , k=3+\dfrac{9}{1} =12\\\\j-3=3 == j=6 , k=3+\dfrac{9}{3} =6\\\\j-3=9 == j=12 , k=3+\dfrac{9}{9} =4\\\\\sum\ k=12+6+4=22\\$

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Ответ:

alkaline27

14.04.2021

Mathematics

The confidence interval is 6.6<μ<6.8.

Step-by-step explanation:

We have:

Number of observations = 601

Mean = 6.7

Standard deviation σ = 1.5

The z-score for a 95% confidence interval is 1.96.

The limits of the confidence interval can be calculated as

$X \pm z*\frac{\sigma}{\sqrt{n}}\\\\LL=X-z*\frac{\sigma}{\sqrt{n}}=6.7-1.96*\frac{1.5}{\sqrt{601} } =6.7-0.1199=6.6\\\\UL=X+z*\frac{\sigma}{\sqrt{n}}=6.7+1.96*\frac{1.5}{\sqrt{601} } =6.7+0.1199=6.8$

The confidence interval is 6.6<μ<6.8.

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In the equation $\frac{1}{j} + \frac{1}{k} = \frac{1}{3}$, both $j$ and $k$ are positive integers. What is the sum of all possible values for $k$?

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