The possibility is 3 : 1/2 * 6/1 = 3/1
Answer:
The approximate difference in the half-lives of the isotopes is 66 days.
Step-by-step explanation:
The decay of an isotope is represented by the following differential equation:
Where:
- Current mass of the isotope, measured in kilograms.
- Time, measured in days.
- Time constant, measured in days.
The solution of the differential equation is:
Where 
is the initial mass of the isotope, measure in kilograms.
Now, the time constant is cleared:
The half-life of a isotope (
) as a function of time constant is:
The half-life difference between isotope B and isotope A is:
If 
, 
and 
, the difference in the half-lives of the isotopes is:
The approximate difference in the half-lives of the isotopes is 66 days.
Answer: 0.0241
Step-by-step explanation:
This is solved using the probability distribution formula for random variables where the combination formula for selection is used to determine the probability of these random variables occurring. This formula is denoted by:
P(X=r) = nCr × p^r × q^n-r
Where:
n = number of sampled variable which in this case = 21
r = variable outcome being determined which in this case = 5
p = probability of success of the variable which in this case = 0.31
q= 1- p = 1 - 0.31 = 0.69
P(X=5) = 21C5 × 0.31^5 × 0.69^16
P(X=5) = 0.0241
<span>You can probably just work it out.
You need non-negative integer solutions to p+5n+10d+25q = 82.
If p = leftovers, then you simply need 5n + 10d + 25q ≤ 80.
So this is the same as n + 2d + 5q ≤ 16
So now you simply have to "crank out" the cases.
Case q=0 [ n + 2d ≤ 16 ]
Case (q=0,d=0) → n = 0 through 16 [17 possibilities]
Case (q=0,d=1) → n = 0 through 14 [15 possibilities]
...
Case (q=0,d=7) → n = 0 through 2 [3 possibilities]
Case (q=0,d=8) → n = 0 [1 possibility]
Total from q=0 case: 1 + 3 + ... + 15 + 17 = 81
Case q=1 [ n + 2d ≤ 11 ]
Case (q=1,d=0) → n = 0 through 11 [12]
Case (q=1,d=1) → n = 0 through 9 [10]
...
Case (q=1,d=5) → n = 0 through 1 [2]
Total from q=1 case: 2 + 4 + ... + 10 + 12 = 42
Case q=2 [ n + 2 ≤ 6 ]
Case (q=2,d=0) → n = 0 through 6 [7]
Case (q=2,d=1) → n = 0 through 4 [5]
Case (q=2,d=2) → n = 0 through 2 [3]
Case (q=2,d=3) → n = 0 [1]
Total from case q=2: 1 + 3 + 5 + 7 = 16
Case q=3 [ n + 2d ≤ 1 ]
Here d must be 0, so there is only the case:
Case (q=3,d=0) → n = 0 through 1 [2]
So the case q=3 only has 2.
Grand total: 2 + 16 + 42 + 81 = 141 </span>
The volume of a cylinder changes when you adjust the height or radius as the volume either increases or reduces.
<h3>How to illustrate the volume?</h3>
Let's assume that the height and radius are 14cm and 7cm. The volume will be:
= πr²h
= 3.14 × 7² × 14
= 2154.04cm³
When the radius and height are reduced to 5cm and 9cm, the volume will be:
= πr²h
= 3.14 × 5² × 9
= 706.5cm³
This illustrates that the volume reduces when the height and radius reduces.
Learn more about volume on:
brainly.com/question/1972490
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