We've hit on a case where a measure<span> of </span>center does<span> not provide all the information spread or variability </span>there<span> is </span>in month-to-month precipitation<span>. based on how busy each </span>month has been in<span> the past, lets managers plan Unit 6: Standard Deviation | Student Guide | Page </span>3<span> </span>If<span> you sum the deviations from the mean, (. ).</span>
<span>In 10 repetitions, there was an instance where 3 or more out of 5 free throw missed. So the probability of her missing 3 or more out of 5 free throw is 0/10=0% or 0.0%. I</span>t appears that your answer of 0% is correct given your experiment.
2/3 would be bigger
I hope that helps
Answer:
<h2>$1380.2</h2>
Step-by-step explanation:
Step one:
given data
weekly earnings= $375
Commission is 4% of sales.= 0.4
supposing sale is x, and the total earning is y
the expression for the total earning will be
y=375+0.4x
Step two:
for x= 2513, y will be
y=375+0.4(2513)
y=375+1005.2
y=1380.2
<em><u>Hence her total earnings for the week will be $1380.2</u></em>
