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
858 i believe...
546/7
78 x 4 (compound a)
312 + original compound b (546)
Hope I helped!
Giving me brainliest is much appreciated! =)
First you would divide everything by bc on both sides to get a by it self. Then you answer will be a=c/bc
Answer:
The required point is P(-7,
) where, x-coordinate twice the y-coordinate on line 2x - 6y = 7
Step-by-step explanation:
Given equation of line is 2x-6y=7
To find point on graph whose x-coordinate twice the y-coordinate:
Let y be y-coordinate of point
Hence, x-coordinate of point will be 2y
The required point is P(2y,y) on equation of line 2x-6y=7
Now,
2x-6y=7
2(2y)-6y=7
-2y=7
y=
Thus,
The required point is P(-7,
)
Note: Figure show equation of line red line and point P as blue dot.
I got 5 I’m not for sure tho