Answer:
Step-by-step explanation:
This is one minus the probability that all the girls audition for different roles. The total number of ways of assigning roles to the girls is 12^7, because to each of the 7 girls, you have a choice of 12 roles.
Then if each girl is to receive a different role, then there are 12!/5! possibilities for that. If you start assigning roles to the girls, then for the first girl, there are 12 choices, but for the next you have to choose one of the 11 different ones, so 11 for the next, and then one of the 10 remaining for the next etc. etc., and this is 12*11*10*...*6 = 12!/(12-7)! =12!/5!
The probability that a random assignment of one of the 12^7 roles would happen to be one of the 12!/5! roles where each girl has a different role, is
(12!/5!)/12^7 = 12!/(12^7 5!)
Then the probability that two or more girls addition for the same part is the probability that not all the girls are assigned different roles, this is thus:
1 - 12!/(12^7 5!)
Treat this as if the inequality sign was the same thing as an equal sign. Most of the equality rules apply to inequalities (with a few exceptions)

Subtract both sides by 4


Multiply both sides by 2


We want the unknown variable on the left side (because it looks nicer)

Answer:
57
Step-by-step explanation:
parentheses first then multiply by 3
Answer:
V = ⅓πr²h = ⅓(3.14)(7.5/2)²(4.25) = 62.5546875 ≈ 62.6 ft³
Step-by-step explanation:
Answer:
<em>t = 1.51</em>
Step-by-step explanation:
<u>Exponential Model</u>
The exponential model is often used to simulate the behavior of a magnitude that either grow or decay in proportion to the existing amount of that magnitude.
The model can be expressed as

In this case, Mo is the initial mass of the radioactive substance and k is a constant which value is positive if the mass is growing or negative if the mass is decaying.
The value of k is not precisely given in the question, we are assuming 
The model is now

We are required to compute the time it takes the mass to reach one-half of its initial value:

Simplifying

Taking logarithms

Solving for t
