2/3 + _/6 = 1/6 what is the missing numerator options: 3 5 6 9

1 answer:

7 0

Answer:
This question is not possible unless the plus sign is actually a minus sign.

Explanation:
First we need to find the LCM, lowest common multiple , for the faction which is 6. Because 3x2 is 6 and 6x1 is 6. So since we mutiply the bottom of the faction 2/3, we need to muitiply the 2 too, which is 6. But....
6 plus 3 or anyone of your option listed wouldn’t be equaled to 1.
So that’s why I don’t think this question is possible to answer.
But... if that sign was a negative the answer would be 5 because 6-5 is 1.

Fun Fact:
The word Fraction comes from the Latin word fractio which means 'to break'.

I hope this helped and have a nice day

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One hundred items are simultaneously put on a life test. Suppose the lifetimes

romanna [79]

Answer:

a) $\mathrm{E}[\mathrm{T}]=\sum_{\mathrm{H}}^{5} \frac{200}{101-i}$

b) $\mathrm{Var}[\mathrm{T}]=\sum_{k=1}^{5} \frac{(200)^{2}}{(101-i)^{2}}$

Step-by-step explanation:

Given:

The lifetimes of the individual items are independent exponential random variables.

Mean = 200 hours.

Assume, Ti be the time between ( $i-1$ )st and the $ith$ failures. Then, the $T_{i}$ are independent with $\mathrm{T}_{\mathrm{i}}$ being exponential with rate $\frac{(101-i)}{200} .$

Therefore,

a) $E[T]=\sum_{i=1}^{5} E\left[\tau_{i}\right]$

$=\sum_{i=1}^{5} \frac{200}{101-i}$

$\therefore \mathrm{E}[\mathrm{T}]=\sum_{\mathrm{H}}^{5} \frac{200}{101-i}$

$b)$

The variance is given by, $\mathrm{Var}[\mathrm{T}]=\sum_{i=1}^{5} \mathrm{Var}[T]$

$\therefore \mathrm{Var}[\mathrm{T}]=\sum_{k=1}^{5} \frac{(200)^{2}}{(101-i)^{2}}$

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Brrunno [24]

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