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3 years ago

9

If A > B and C > D, then A + C > B + D.

1 answer:

steposvetlana [31]3 years ago

4 0

The answer is Always true. because A is greater than B so put it this way if A=2 and B =1 then 2>1 same goes for C>D so if A+C would have to be greater than B+D for instance 2+2>1+1 or 4>2 so
<span>If A>B and C>D, then A+C>B+D Always true.
</span>

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Maksim231197 [3]

Answer:

The sample standard deviation is 393.99

Step-by-step explanation:

The standard deviation of a sample can be calculated using the following formula:

$s=\sqrt[ ]{\frac{1}{N-1} \sum_{i=1}^{N}(x_{i}-{\displaystyle \textstyle {\bar {x}}}) ^{2} }$

Where:

$s=$ Sample standart deviation

$N=$ Number of observations in the sample

${\displaystyle \textstyle {\bar {x}}}=$ Mean value of the sample

and $\sum_{i=1}^{N}(x_{i}-{\displaystyle \textstyle {\bar {x}}}) ^{2} }$ simbolizes the addition of the square of the difference between each observation and the mean value of the sample.

Let's start calculating the mean value:

$\bar {x}=\frac{1}{N} \sum_{i=1}^{N}x_{i}$

$\bar {x}=\frac{1}{15}*(180+1600+90+140+50+260+400+90+380+110+10+60+20+340+80)$

$\bar {x}=\frac{1}{15}*(3810)$

$\bar {x}=254$

Now, let's calculate the summation:

$\sum_{i=1}^{N}(x_{i}-\bar {x}) ^{2} }=(180-254)^2+(1600-254)^2+(90-254)^2+...+(80-254)^2$

$\sum_{i=1}^{N}(x_{i}-\bar {x}) ^{2} }=2173160$

So, now we can calculate the standart deviation:

$s=\sqrt[ ]{\frac{1}{N-1} \sum_{i=1}^{N}(x_{i}-{\displaystyle \textstyle {\bar {x}}}) ^{2} }$

$s=\sqrt[ ]{\frac{1}{15-1}*(2173160)}$

$s=\sqrt[ ]{\frac{2173160}{14}}$

$s=393.99$

The sample standard deviation is 393.99

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