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3 years ago

Is y=x^2 - 1 a linear?

Mathematics

2 answers:

seropon [69]3 years ago

8 0

Not fully sure what you're asking, but the equation is a basic parabola lowered on the y-axis by one.

NARA [144]3 years ago

8 0

No, because when there is an exponent in the equation, the x value is increasing exponentially, which means it's not linear.

Hope this helped!

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tatuchka [14]

Answer:

It is p = 1

Step-by-step explanation:

p⁰=1 because it is basically saying and when simplifying fractional exponents we can flip the bottom exponent up to the top if we change the sign, leaving us with : (p⁵-p⁵)=p⁰

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3 years ago

Solve the inequality 2x - 3

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That's not an inequality.

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3 years ago

Construct the discrete probability distribution for the random variable described. Express the probabilities as simplified fract

lisov135 [29]

Answer:

$P(X = 0) = 0.03125$

$P(X = 1) = 0.15625$

$P(X = 2) = 0.3125$

$P(X = 3) = 0.3125$

$P(X = 4) = 0.15625$

$P(X = 5) = 0.03125$

Step-by-step explanation:

For each toss, there are only two possible outcomes. Either it is tails, or it is not. The probability of a toss resulting in tails is independent of any other toss, 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.

Fair coin:

Equally as likely to be heads or tails, so $p = 0.5$

5 tosses:

This means that $n = 5$

Probability distribution:

Probability of each outcome, so:

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

$P(X = 0) = C_{5,0}.(0.5)^{0}.(0.5)^{5} = 0.03125$

$P(X = 1) = C_{5,1}.(0.5)^{1}.(0.5)^{4} = 0.15625$

$P(X = 2) = C_{5,2}.(0.5)^{2}.(0.5)^{3} = 0.3125$

$P(X = 3) = C_{5,3}.(0.5)^{3}.(0.5)^{2} = 0.3125$

$P(X = 4) = C_{5,4}.(0.5)^{4}.(0.5)^{1} = 0.15625$

$P(X = 5) = C_{5,5}.(0.5)^{5}.(0.5)^{0} = 0.03125$

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2 years ago

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Nikitich [7]

Answer:

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Romashka [77]

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