Answer:
Step-by-step explanation:
The problem relates to filling 8 vacant positions by either 0 or 1
each position can be filled by 2 ways so no of permutation
= 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2
= 256
b )
Probability of opening of lock in first arbitrary attempt
= 1 / 256
c ) If first fails , there are remaining 255 permutations , so
probability of opening the lock in second arbitrary attempt
= 1 / 255 .
57. 1/8
58. 1/32
Hope Your Thanksgiving Goes Well, Here's A Turkey
-TheKoolKid1O1
Answer:
464
Step-by-step explanation:
Answer:
{-7,5,18,24,32}
Step-by-step explanation:
Let's verify all cases to determine the solution to the problem.
CASE A) we have
-9,7,15,22,26
The number 22 is not included in the solution set of the compound inequality.
CASE B) we have
-7,5,18,24,32
The number 22 is not included in the solution set of the compound inequality.
CASE C) we have
16,17,22,23,24
The number 22 is not included in the solution set of the compound inequality.
CASE D) we have
18,19,20,21,22
The numbers 19,20,21,22 are not included in the solution set of the compound inequality.
Full Question:
Find the volume of the sphere. Either enter an exact answer in terms of π or use 3.14 for π and round your final answer to the nearest hundredth. with a radius of 10 cm
Answer:
The volume of the sphere is ⅓(4,000π) cm³ or 4186.67cm³
Step-by-step explanation:
Given
Solid Shape: Sphere
Radius = 10 cm
Required
Find the volume of the sphere
To calculate the volume of a sphere, the following formula is used.
V = ⅓(4πr³)
Where V represents the volume and r represents the radius of the sphere.
Given that r = 10cm,.all we need to do is substitute the value of r in the above formula.
V = ⅓(4πr³) becomes
V = ⅓(4π * 10³)
V = ⅓(4π * 10 * 10 * 10)
V = ⅓(4π * 1,000)
V = ⅓(4,000π)
The above is the value of volume of the sphere in terms of π.
Solving further to get the exact value of volume.
We have to substitute 3.14 for π.
This gives us
V = ⅓(4,000 * 3.14)
V = ⅓(12,560)
V = 4186.666667
V = 4186.67 ---- Approximated
Hence, the volume of the sphere is ⅓(4,000π) cm³ or 4186.67cm³
