Answer:
percent (%) of ancient prehistoric thinner than 3 is 0.0013
Step-by-step explanation:
Given data
mean = 5.1 millimeters (mm)
standard deviation = 0.7 mm
to find out
percent (%) of ancient prehistoric thinner than 3
solution
we have given mean and SD
so Z will be = x -mean / SD
Z = ( x - 5.1 ) / 0.7
so P( x < 3 ) = P ( (x - 5.1 ) / 0.7 < ( 3 - 5.1 ) / 0.7 )
P( Z = ( 3 - 5.1 ) / 1.5 )
P( Z = -3 ) = 0.0013
percent (%) of ancient prehistoric thinner than 3 is 0.0013
CXB = 40 degrees
I am not positive but from what I have looked up this is the answer.
Answer:
144 sq. ft.
Step-by-step explanation:
In order to find the area, we have to know the length of the sides.
Since the figure is a cube, the way that the volume was found was by multiplying one side length by itself two times.
s^3 = cube volume.
So, to find the length of one side, we have to find the cube root of the volume.
![\sqrt[3]{1728}](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B1728%7D)
= 12
Now that we know the side length, we just need to multiply it by itself to find the area of the floor space.
12 * 12 = 144
X+x+54=180
2x+54=180
2x=126
x=63
63+54=117
one angle is 63, the other is 117
(Bowls, Height) (1, 2) (5,5)
Slope is (5-2)/(5-1) = 3/4 inch
y = (3/4)x + b
(2) = (3/4)(1) + b
(2)-(3/4) = b
B=1.25. Y= 0.75*x + 1.25.
Part B
X is the number of bowls in the stack and Y is the corresponding height of the stack.
