Answer:
it is a function
Step-by-step explanation:
the input is not repeated
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
We need to determine the radius and diameter of the circle. If the area of the circle is 10 pi in^2, then, according to the formula for the area of a circle,
A = 10 pi in^2 = pi*r^2. Thus, 10 in^2 = r^2, and r = radius of circle = sqrt(10) in.
Thus, the diam. of the circle is 2sqrt(10) in. This diam. has the same length as does the hypotenuse of one of the triangles making up the square.
Thus, [ 2*sqrt(10) ]^2 = x^2 + x^2, where x represents the length of one side of the square. So, 4(10) in^2 = 2x^2. Then:
40 in^2 = 2x^2, or 20 in^2 = x^2, and so the length x of one side of the square is sqrt(20). The area of the square is the square of this result:
Area of the square = x^2 = [ sqrt(20) ]^2 = 20 in^2 (answer). Compare that to the 10 pi sq in area of the circle (31.42 in^2).
Multiply the first digit by 3 then subtract the answer by 2. Rule: x 3 - 2
Answer:
336
Step-by-step explanation:
i hope i got it right
