K, so
first one, ok, all of them are prime, 5 and 7 and 13 are all prime
2nd one, 24 is not prime
3rd one, 3 and 13 are prime
4th one, 4 is not prime
answe is 1st and 3rd one
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>
Answer:
dividing is always the last step
Step-by-step explanation:
Answer:
y = (x+3)/4
Step-by-step explanation:
add 3 to both sides
so x + 3 = 4y-3+3
x + 3 = 4y
Divide both sides by 4
(x + 3)/4 = 4y/4
(x+3)/4 = y
Answer:
y = -2x - 1
Step-by-step explanation:
Rewrite y + 7 = -2(x - 3) in slope-intercept form (which is y = mx + b):
First, perform the indicated multiplication:
y + 7 = -2x + 6
Combining the constants, we get
y = -2x - 1
This is the desired equation in slope-intercept form. slope is -2 and y-intercept is -1.
