Which number is not in the solution set of x-6>10?

Mathematics

2 answers:

Allushta [10]3 years ago

6 0

It would be any number that is less than 16 because starting with the original question:

x-6>10

By canceling out the -6:

x-6+6 > 10+6

x > 16

givi [52]3 years ago

4 0

Answer:

options?

Step-by-step explanation:

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What is the ratio of 1/2 to 6? (A) 12 (B) 3 (C)1/3 (D)1/6 (E) 1/12

TiliK225 [7]

Answer:

e, 1/12

Step-by-step explanation:

1) turn the first fraction into a whole number, the closest whole number being 1.

what did you multiply 1/2 by to get 1? the answer would be 2.

whatever you do to oneside, do to the other. when you multiply 6 by 2, you get twelve.

therefore, the answer is 1/12

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3 years ago

Solve for: -5/6 + (-4/9)=

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Answer:

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A box contains four bowling balls numbered 2, 4, 6, and 10 according to their weight. Two are drawn randomly without replacement

Lilit [14]

Answer:

E(M) = 23/3

Var(M) = 53/9.

Step-by-step explanation:

There are three possible values for M: 4, 6, and 10.

$\begin{array}{|l|l|}\cline{1-2}\\[-1em]\text{Larger Number} & \text{Smaller Number}\\ \cline{1-2}\\[-1em]4 & 2 \\\cline{1-2}\\[-1em]6 & 2, ~4\\\cline{1-2}\\[-1em] 10 & 2, ~4, ~6\\\cline{1-2} \end{array}$ .

That's six unique combinations in total. If the balls are drawn randomly, the probability for getting each combination shall be equal. That is:

$\begin{array}{|c|c|}\cline{1-2}\\[-1em]m & P(\text{M} = m)\\\cline{1-2}\\[-1em] 2 & 0\\\cline{1-2}\\[-1em] 4 & 1/6 \\\cline{1-2}\\[-1em] 6 & 2/6\\\cline{1-2}\\[-1em] 10 & 3/6\\\cline{1-2}\end{array}$ .

Consider the formula for the expected value of a discrete random variable:

$\begin{aligned} E({\rm M})& = \sum{[m \cdot P(\mathrm{M} = m)]}\\ &= 4 \times \frac{1}{6} + 6\times \frac{2}{6} + 10 \times \frac{3}{6}\\ &= \frac{23}{3}\end{aligned}$ .

Formula for the variance of a discrete random variable (note that this formula can take many other forms):

$\begin{aligned}Var(\mathrm{M}) &= \sum{[m^{2}\cdot P(\mathrm{M}= m)]} - \mu^{2}\\&= 4^{2}\times \frac{1}{6} + 6^{2} \times \frac{2}{6} + 10^{2} \times \frac{3}{6} - \left(\frac{23}{3}\right)^{2}\\&= \frac{53}{9}\end{aligned}$ .