It can't be A. since if you only look at managers, you are missing all the sales executives.
It may be C. this option is more random but doesn't guarantee that you will represent both groups of employee's. Also, each time you would conduct the survey, you will receive the exact same results since it is the same people.
It isn't D. for the exact same reason as A. but you're missing managers now.
Therefore the answer is B. Some managers and some sales executives selected at random. This way you get a sample from both categories, and within those groups, it is randomly selected.
I hope this helps!
Answer:
Step-by-step explanation:
we know that
The area of the shaded region is equal to the area of the circle minus the area of the two isosceles triangles
so
Find the area of the circle
The area of the circle is equal to
where r is the radius of the circle
In this problem we have
substitute
Find the area of the two triangles
The area of the two isosceles triangles is equal to
![A=2[\frac{1}{2}bh]=bh](https://tex.z-dn.net/?f=A%3D2%5B%5Cfrac%7B1%7D%7B2%7Dbh%5D%3Dbh)
we have
substitute
Find the area of the shaded region
This question is not written correctly
Complete Question
Select all decimals that are equivalent to ( 9 × 100 ) + ( 2 × 10 ) + ( 3 × 1/10 ) + ( 5 × 1/100 ) . A. 92.35 B. 920.350 C. 902.35 D. 92.350 E. 920.35 F. 920.035
Answer:
E. 920.35
Step-by-step explanation:
( 9 × 100 ) + ( 2 × 10 ) + ( 3 × 1/10 ) + ( 5 × 1/100 )
Step 1
(900) + (20) + (3/10) + (5/100)
Step 2
900 + 20 + 0.3 + 0.05
Step 3
920.35
Therefore, the answer = 920.35
I think the correct answer from the choices listed above is option B. Deliberate bias <span> is the source error that can be avoided by locating questions sensitive to bias and changing or dropping them. Hope this answers the question. Have a nice day.</span>
Trapezoidal is involving averageing the heights
the 4 intervals are
[0,4] and [4,7.2] and [7.2,8.6] and [8.6,9]
the area of each trapezoid is (v(t1)+v(t2))/2 times width
for the first interval
the average between 0 and 0.4 is 0.2
the width is 4
4(0.2)=0.8
2nd
average between 0.4 and 1 is 0.7
width is 3.2
3.2 times 0.7=2.24
3rd
average betwen 1.0 and 1.5 is 1.25
width is 1.4
1.4 times 1.25=1.75
4th
average betwen 1.5 and 2 is 1.75
width is 0.4
0.4 times 1.74=0.7
add them all up
0.8+2.24+1.75+0.7=5.49
5.49
t=time
v(t)=speed
so the area under the curve is distance
covered 5.49 meters
