3.5*5.5=19.25
The formula to find the area is the length times width
1) vertical line
2) neither
3) horizontal line
4) neither- these two overlap.
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5 through 6 equation: a²+b²=c²
5) 5²+.8²=c²
25+.64=c²
√25.64=√c²
√25.64=C or 5.063=C
6) 2²+8²=c²
4+64=c²
√68=√c²
√68=C or 8.246=C
I hope this was of help
Yes. If you have very high or very low outliers in your data set, it is generally preferred to use the median - the mid-point when all data points are arranged from least to greatest.
<span>A good example for when to avoid the mean and prefer the median is salary. The mean is less good here as there are a few very high salaries which skew the distribution to the right. This drags the mean higher to the point where it is disproportionately affected by the few higher salaries. In this case, the median would only be slightly affected by the few high salaries and is a better representation of the whole of the data. </span>
<span>In general, if the distribution is not normal, the mean is less appropriate than the median.</span>
The answer would be x>7/2
