I think it's 9.68 but I might be wrong
Let's simplify step-by-step.
<span><span>0.2<span>(<span><span>3b</span>−<span>15c</span></span>)</span></span>+<span>6c
</span></span>Distribute:<span>=<span><span><span><span>(0.2)</span><span>(<span>3b</span>)</span></span>+<span><span>(0.2)</span><span>(<span>−<span>15c</span></span>)</span></span></span>+<span>6c
</span></span></span><span>=<span><span><span><span>0.6b</span>+</span>−<span>3c</span></span>+<span>6c
</span></span></span>Combine Like Terms:
<span>=<span><span><span>0.6b</span>+<span>−<span>3c</span></span></span>+<span>6c
</span></span></span><span>=<span><span>(<span>0.6b</span>)</span>+<span>(<span><span>−<span>3c</span></span>+<span>6c</span></span>)
</span></span></span><span>=<span><span>0.6b</span>+<span>3<span>c</span></span></span></span>
Answer:
it can be redued but not sloved
Step-by-step explanation:
Answer:
k>5/14
Step-by-step explanation:
Answer:
A) Professor A failure proportion = 0.12 ; B) Average = 0.1 ; C) σ = 0.055
Step-by-step explanation:
1) Proportion of failures for Prof A : Failure / Total
= 12 / 100 = 0.12
2) Proportion of failures for Prof B : 1/ 100 = 0.01
Proportion of failures for Prof C : 11 / 100 = 0.11
Proportion of failures for Prof D = 16 / 11 = 0.16
Average proportional failure for all instructors :-
[ Prop (prof A) + Prop (prof B) + Prop (prof C) + Prop (prof D) ] / 4
(0.12 + 0.01+ 0.11 + 0.16 ) / 4 = 0.4 / 4 = 0.1
3) Standard Deviation (σ) = √ Σ (x - x')^2 / n , where x' = mean = 0.1
x x - x' (x - x')^2
0.12 0.02 0.0004
0.01 -0.09 0.0081
0.11 0.01 0.0001
0.16 0.06 <u>0.0036 </u>
<u>Σ = 0.0122</u>
σ = √ 0.0122/ 4 = √0.00305 = 0.055
in common, so we can pull it out. The factored form will be 
left. Notice that both of these terms no longer have a negative sign since the negative was factored out.