Answer:
Correct option is
C
36.25
Modal class =30−40
So we have, l=30,f0=12,f1=32,f2=20 and h=10
⇒ Mode=l+2f1−f0f2f1−f0×h
=30+2×32−12−2032−12×10
=30+6.25
=36.25
∴ Mode =36.25
Step 1. Set up long division
_______
7| 1 9 8 6
Step 2. <span>Calculate 19 ÷ 7, which is 2 with a remainder of 5.
2
</span> _______
7| 1 9 8 6
1 4
_________
5
Step 3. Bring down 8, so that 58 is large enough to be divided by 7.
2
_______
7| 1 9 8 6
1 4
_________
5 8
Step 4. <span>Calculate 58 ÷ 7, which is 8 with a remainder of 2.
</span> 2 8
_______
7| 1 9 8 6
1 4
_________
5 8
5 6
_________
2
Step 5. <span>Bring down 6, so that 26 is large enough to be divided by 7.
</span> 2 8
_______
7| 1 9 8 6
1 4
_________
5 8
5 6
_______
2 6
Step 6. Calculate 26 ÷ 7, which is 3 with a remainder of 5.
2 8 3
_______
7| 1 9 8 6
1 4
_________
5 8
5 6
_______
2 6
2 1
______
5
Step 7. <span>Therefore, 1986 ÷ 7 = 283 with a remainder of 5.
823 With a remainder of 5
Done!
</span><span>Decimal Form If Needed: 283.714286</span>
If the line is perpendicular the slope is the reciprocal so it would be
Y=-1/3x
Then if the line has to go through the point (-2,5) the y-intercept must be 5 making the answer
Y=-1/3x+5
