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3 years ago

How do I know if it's a function or not?

Mathematics

2 answers:

Effectus [21]3 years ago

5 0

In plain and short, a function has no x-coordinate Repeats, if there's an x-coordinate Repeat, then is not a function.

$\bf \begin{array}{ccll}
x&y\\
\text{\textemdash\textemdash\textemdash}&\text{\textemdash\textemdash\textemdash}\\
6&3\\
8&2\\
9&0\\
\stackrel{re peat}{11}&1\\
\stackrel{re peat}{11}&2
\end{array}$

Free_Kalibri [48]3 years ago

3 0

Your domain would be (6,8,9,11,11)
your range would be (3,2,0,1,2)
This is not a function because the domain 11 is paired with both the range of 1 and 2

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Plssss helpppp ASAPP will mark BRAINLIEST!!!

ra1l [238]

Answer:

4th and 5th option

Step-by-step explanation:

you can easily add eqn 1 and 2 to get rid of the y term for 4th option and get rid of x term by addition as well for the 5th option

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3 years ago

Read 2 more answers

What is the greatest common factor of the pair of numbers?

geniusboy [140]

The answer is 66.
66 * 3 = 198
66 * 5 = 330

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3 years ago

A new medical test provides a false positive result for hepatitis 2% of the time that is a perfectly healthy subject being teste

Oxana [17]

Given that X be the number of subjects who test positive for the disease out of the 30 healthy subjects used for the test.

The probability of success, i.e. the probability that a healthy subject tests positive is given as 2% = 0.02

Part A:

The probability that all 30 subjects will appropriately test as not being infected, that is the probability that none of the healthy subjects will test positive is given by:

 $P(X)={ ^nC_xp^x(1-p)^{n-x}} \\ \\ P(0)={ ^{30}C_0(0.02)^0(1-0.02)^{30-0}} \\ \\ =1(1)(0.98)^{30}=0.5455$


Part B:

The mean of a binomial distribution is given by

 $\mu=np \\ \\ =30(0.02) \\ \\ =0.6$

The standard deviation is given by:

 $\sigma=\sqrt{np(1-p)} \\ \\ =\sqrt{30(0.02)(1-0.02)} \\ \\ =\sqrt{30(0.02)(0.98)}=\sqrt{0.588} \\ \\ =0.7668$


Part C:

This test will not be a trusted test in the field of medicine as it has a standard deviation higher than the mean. The testing method will not be consistent in determining the infection of hepatitis.

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3 years ago

A recent study suggested that 70% of all eligible voters will vote in the next presidential election. Suppose 20 eligible voters

natita [175]

Answer:

0.0479 = 4.79% probability that fewer than 11 of them will vote

Step-by-step explanation:

For each voter, there are only two possible outcomes. Either they will vote, or they will not. The probability of a voter voting is independent of any other voter, which means that the binomial probability distribution is used 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.

70% of all eligible voters will vote in the next presidential election.

This means that $p = 0.7$

20 eligible voters were randomly selected from the population of all eligible voters.

This means that $n = 20$

What is the probability that fewer than 11 of them will vote?

This is:

$P(X < 11) = P(X = 10) + P(X = 9) + P(X = 8) + P(X = 7) + P(X = 6) + P(X = 5) + P(X = 4) + P(X = 3) + P(X = 2) + P(X = 1) + P(X = 0)$

In which

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

$P(X = 10) = C_{20,10}.(0.7)^{10}.(0.3)^{10} = 0.0308$

$P(X = 9) = C_{20,9}.(0.7)^{9}.(0.3)^{11} = 0.0120$

$P(X = 8) = C_{20,8}.(0.7)^{8}.(0.3)^{12} = 0.0039$

$P(X = 7) = C_{20,7}.(0.7)^{7}.(0.3)^{13} = 0.0010$

$P(X = 6) = C_{20,10}.(0.7)^{6}.(0.3)^{12} = 0.0002$

$P(X = 5) = C_{20,5}.(0.7)^{5}.(0.3)^{15} \approx 0$

The probability of 5 or less voting is very close to 0, so they will not affect the outcome. Then

$P(X < 11) = P(X = 10) + P(X = 9) + P(X = 8) + P(X = 7) + P(X = 6) + P(X = 5) + P(X = 4) + P(X = 3) + P(X = 2) + P(X = 1) + P(X = 0) = 0.0308 + 0.0120 + 0.0039 + 0.0010 + 0.0002 = 0.0479$

0.0479 = 4.79% probability that fewer than 11 of them will vote

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3 years ago

When adding two fractions do i leave the denominator how it is?

telo118 [61]

Hoi!

If the two fractions have a common denominator, then yes, you do. Only add the numerators. :)

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3 years ago