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3 years ago

We study proofs in computing because it helps us understand data structures

Mathematics

1 answer:

alisha [4.7K]3 years ago

8 0

Answer:

false

Step-by-step explanation:

we study proofs in computing because it helps us to understand the mathematical computation. With the help of proofs in computing it is very easy to understand all the mathematical computation but in given statement it is given that it helps us to understand data structure so it is a false statement.

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Solve Z-8/4=-1 1. Z= -12 2. Z= 28 3. Z= 4 4. Z= -36

Gnesinka [82]

Answer:

Step-by-step explanation:

8 0

3 years ago

In the united states, voters who are neither democrat nor republican are called independent. it is believed that 12% of voters a

Nitella [24]

Answer:

a) 0.0245 = 2.45%

b) 0.9483 = 94.83%

c) 0.6931 = 69.31%

Step-by-step explanation:

For each person, there are only two possible outcomes. Either they are independent, or they are not. The probability of a person being independent is independent of any other person. So we use the binomial probability distribution to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

It is believed that 12% of voters are independent.

This means that $p = 0.12$

Survey of 29 people

This means that $n = 29$

a. what is the probability that none of the people are independent? probability =

This is P(X = 0). So

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 0) = C_{29,0}.(0.12)^{0}.(0.88)^{29} = 0.0245$

0.0245 = 2.45% probability that none of the people are independent

b. what is the probability that fewer than 7 are independent? probability =

$P(X < 7) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6)$

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 0) = C_{29,0}.(0.12)^{0}.(0.88)^{29} = 0.0245$

$P(X = 1) = C_{29,1}.(0.12)^{1}.(0.88)^{28} = 0.0971$

$P(X = 2) = C_{29,2}.(0.12)^{2}.(0.88)^{27} = 0.1853$

$P(X = 3) = C_{29,3}.(0.12)^{3}.(0.88)^{26} = 0.2274$

$P(X = 4) = C_{29,4}.(0.12)^{4}.(0.88)^{25} = 0.2016$

$P(X = 5) = C_{29,5}.(0.12)^{5}.(0.88)^{24} = 0.1374$

$P(X = 6) = C_{29,6}.(0.12)^{6}.(0.88)^{23} = 0.0750$

$P(X < 7) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6) = 0.0245 + 0.0971 + 0.1853 + 0.2274 + 0.2016 + 0.1374 + 0.0750 = 0.9483$