Raise the monomial to a power. -4ax³ to the power of 3.

Mathematics

1 answer:

Nina [5.8K]3 years ago

5 0

The question was hard to follow but I tried my best. Lmk what I can do if it is in incorrect format.

Have a nice day

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Apply each of the four counting/sampling methods (with replacement and with ordering, without replacement and with ordering, wit

dusya [7]

Answer:

Step-by-step explanation:

I will illustrate this solution with a unique birthday party situation between Jane (the celebrant) and her 10 friends.

In the said party , we will assume that Jane only has 5 chocolates to share among her friends

i. Assuming that the 5 chocolates are of the same type, if she doesn't want to give any friend more than one piece of chocolate, the situation here is said to be WITHOUT REPLACEMENT & UN-ORDERED

n = 10 and k = 5

Total number of possible combinations becomes,

$(\left {n} \atop {k}} \right. )=(\left {10} \atop {5}} \right. )=252$

ii. Assuming that the 5 pieces of chocolate are of the same type and she is willing to give a friend more than one piece of chocolate, the situation is said to be WITH REPLACEMENT & UN-ORDERED

n = 10 and k = 5

Total number of possible combinations becomes,

$(\left {n+k-1} \atop {k}} \right. )=(\left {14} \atop {5}} \right. )=2002$

iii. Assuming that the 5 pieces of chocolate are of different types and she isn't willing to give any friend more than one piece of chocolate, the situation is said to be WITHOUT REPLACEMENT & ORDERED

n = 10 & k = 5

Total number of possible combinations becomes,

$P\left {n} \atop {k}} \right=P\left {10} \atop {5}} \right. =30240$

iv. Assuming the the 5 pieces of chocolate are of different types, if she is willing to give a friend more than one piece of chocolate, the situation is said to be WITH REPLACEMENT & ORDERED

n = 10 AND k = 5

Total number of combinations becomes

$n^k=10^5=100,000$