$\bf recall~that\qquad i^2=-1\\\\
-------------------------------\\\\
\cfrac{-5+i}{2i}\cdot \cfrac{i}{i}\implies \cfrac{i(-5+i)}{2i^2}\implies \cfrac{-5i+i^2}{2(-1)}\implies \cfrac{-5i+(-1)}{-2}
\\\\\\
\cfrac{-5i-1}{-2}\implies \cfrac{1+5i}{2}$

6 0

3 years ago

All else being equal, if you cut the sample size in half, how does this affect the margin of error when using the sample to make

gizmo_the_mogwai [7]

Answer:

The margin of error is multiplied by $\sqrt{2}$

Step-by-step explanation:

margin of error (ME) from the mean can be calculated using the formula

ME= $\frac{z*s}{\sqrt{N} }$ where

z is the corresponding statistic of the given confidence level

s is the standard deviation of the sample (or of the population if it is known)

N is the sample size

Since margin of error is proportional with inverse of $\sqrt{N}$ ,

if we cut the sample size in half, the margin of error is multiplied by $\sqrt{2}$ .

8 0

3 years ago

Eorge pays one-fifth of his salary in taxes. If his taxes amount to $217, what is his salary?

Amiraneli [1.4K]

$1,085. Just multiply by 5

5 0

2 years ago