Start circle: πd = (3.14)(19) = 59.7
Move diagonally to the circle with the radius of 6.2.
Second circle: 2πr = 2(3.14)(6.2) = 39
Move upwards to the circle with the radius of 10.5
third circle: 2πr = 2(3.14)(10.5) = 66
Move right to the circle with the diameter of 16.6
Fourth circle: πd = (3.14)(16.6) = 52.2
Move down to the circle with the diameter of 7.7
fifth circle: πd = (3.14)(7.7) = 24.2
Move down to the circle with the diameter of 50
Sixth circle: πd = (3.14)(50) = 157.1
Move left to the circle with the radius of 11.8
Seventh circle: 2πr = 2(3.14)(11.8) = 74.1
Move down to the circle with the radius of 38
Eight circle: 2πr = 2(3.14)(38) = 238.8
Move right to the circle with the diameter of 1.1
ninth circle: πd = (3.14)(1.1) = 3.5
Move right to the circle with the radius of 14.8
10th circle = 2πr = 2(3.14)(14.8) = 93
Move up to the end.
Hope this helps :)
7 <span> more than 3 times a number </span>
<span>more than is code for + </span>
<span>times is multiply </span>
<span>so 7+3n=31 </span>
<span>solve: </span>
<span>3n=24 </span>
<span>n= 8</span>
Answer:
The volume inside the cylinder is 
.
Step-by-step explanation:
Given : Four spheres of radius 6 cm just fit inside a cylinder.
To find : Calculate the volume inside the cylinder that is empty ?
Solution :
The volume of the cylinder is volume of 4 spheres.
The volume of sphere is 
The radius is 6 cm.
The volume of four spheres is
Therefore, the volume inside the cylinder is .
Answer:
9units squared
Step-by-step explanation:
A standard rubix cube is 3 by 3 and therefor 3 x 3 is 9
Answer:
I used the function normCdf(lower bound, upper bound, mean, standard deviation) on the graphing calculator to solve this.
- Lower bound = 1914.8
- Upper bound = 999999
- Mean = 1986.1
- Standard deviation = 27.2
Input in these values and it will result in:
normCdf(1914.8,9999999,1986.1,27.2) = 0.995621
So the probability that the value is greater than 1914.8 is about 99.5621%
<u><em>I'm not sure if this is correct </em></u><em>0_o</em>
