Give the “best” big-oh notation to describe the complexity of the algorithm that prints all bit strings of length n.

Mathematics

1 answer:

artcher [175]2 years ago

4 0

It is correct to state that a bit can be in one of 2 states, either 1 or 0.

A bit is is a single binary digit. A bit can either be "1" or "0".

<h3>What is the explanation to the above answer?</h3>

We stated that It is correct to indicate that a bit can be in one of 2 states, either 1 or 0. This indicates that 2ⁿ strings must be created in total. The string must then be processed twice.

The first time, for generation and
The second for reading.

Hence, Our time is now O(2ⁿ⁺¹) = O(2ⁿ)

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A statistics textbook chapter contains 60 exercises, 6 of which are essay questions. A student is assigned 10 problems. (a) What

Scilla [17]

Answer:

a) P=0.3174

b) P=0.4232

c) P=0.2594

d) The shape of the hypergeometric, in this case, is like a binomial with mean np=1.

Step-by-step explanation:

The appropiate distribution to model this is the hypergeometric distribution:

$P(X=x)=\frac{\binom{s}{x}\binom{N-s}{M-x}}{\binom{N}{M}}=\frac{\binom{6}{x}\binom{54}{10-x}}{\binom{60}{10}}$

a) What is the probability that none of the questions are essay?

$P(X=0)=\frac{\binom{6}{0}\binom{54}{10-0}}{\binom{60}{10}}\\\\P(X=0)=\frac{1*(54!/(10!*44!)}{60!/(10!*50!)} =\frac{2.3931*10^{10}}{7.5394*10^{10}} = 0.3174$

b) What is the probability that at least one is essay?

$P(X=1)=\frac{\binom{6}{1}\binom{54}{9}}{\binom{60}{10}}\\\\P(X=1)=\frac{6*(54!/(9!*43!)}{60!/(10!*50!)} =\frac{3.1908*10^{10}}{7.5394*10^{10}} =0.4232$