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wlad13 [49]
3 years ago
13

A ternary digit is either 0, 1, or 2. how many sequences of eight ternary digits containing a single 2 and a single 1 are possib

le?
Mathematics
1 answer:
PIT_PIT [208]3 years ago
8 0

Answer:

56

Step-by-step explanation:

If there can be only one 1 and only one 2, the remaining digits must all be 0. The digits 1 and 2 can be anywhere in the 8 digits, and can be in either order.

There are 8 possible locations in the sequence for the 1, then 7 possible locations for the 2. The total number of possibilities is 8·7 = 56.

12000000, 21000000, 10200000, 20100000, ..., 00000012, 00000021.

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<u>Step-by-step explanation:</u>

Here we have , A manufacturer finds that the revenue generated by selling of cortan commodity is given by function R(x)=80x-0.2x^ 2 , where he maximum reveremany should be manufactured to obtain this maximum units .Let's find out:

We have following function as R(x)=80x-0.2x^ 2 . Let's differentiate this and equate it to zero to find value of x for which the function is maximum!

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⇒ 0=80-2x(0.2)

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