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mario62 [17]
2 years ago
12

How many positive three-digit integers have the hundreds digit equal to 7 and the units (ones) digit equal to 1?

Mathematics
1 answer:
Lina20 [59]2 years ago
7 0

Using the Fundamental Counting Theorem, it is found that there are 10 positive three-digit integers have the hundreds digit equal to 7 and the units (ones) digit equal to 1.

<h3>What is the Fundamental Counting Theorem?</h3>

It is a theorem that states that if there are n things, each with n_1, n_2, \cdots, n_n ways to be done, each thing independent of the other, the number of ways they can be done is:

N = n_1 \times n_2 \times \cdots \times n_n

The number of options for each selection are given as follows, considering there are 10 possible digits, and that the last two are fixed at 7 and 1, respectively:

n_1 = 10, n_2 = n_3 = 1

Hence, the number of integers is given by:

N = 10 x 1 x 1 = 10.

More can be learned about the Fundamental Counting Theorem at brainly.com/question/24314866

#SPJ1

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Please help me find the answer amongst the options below.
belka [17]

Answer:

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Step-by-step explanation:

we are given

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and

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we have to find gof(x)=?

for gof(x) we will replace value of x from f(x) with g(x)

that is

(gof)(x)=(x+1)^{2}

hence it will be third option that is correct


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3 years ago
What are the solutions
Lapatulllka [165]
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