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2 years ago

Chose the graph that represents the equation y=|x-4|

Mathematics

2 answers:

Luden [163]2 years ago

8 0

Answer:

V shape.

Vertex at (4,0)

Intersect with y axis at (0,4)

Step-by-step explanation:

In order to solve this you just ahve to search fo the graph that grows like a V because when you deal with functions that have absolute values you can have negative and positive values for the independent variable but not for the dependent, in this case x is the independent so x will have negative and positive values while Y will only grow in the postives, now the vertex of the function is found where Y=0 so to calculate that you equal the function to 0

y=|x-4|

0=|x-4|

x=4

So the vertex will be at (4,0) and the intersect with the y axis will be located where x=0, so you evaluate the function for 0:

y=|x-4|

y=|o-4|

y=|-4|

y=4

So the intersect with the y axis will be at (0,4).

evablogger [386]2 years ago

6 0

You have not shared the graphs from which you have to choose your answer.

Why not graph y=|x-4| yourself, to help you choose the correct answer?

First, graph y=|x|. It looks like a "v" opening up, with the vertex at (0,0).
Next, move the whole "v" graph 4 units to the right.

Compare your result to the given answer choices.

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2 years ago

Let the number of chocolate chips in a certain type of cookie have a Poisson distribution. We want the probability that a cookie

ludmilkaskok [199]

Answer:

$\lambda \geq 6.63835$

Step-by-step explanation:

The Poisson Distribution is "a discrete probability distribution that expresses the probability of a given number of events occurring in a fixed interval of time or space if these events occur with a known constant mean rate and independently of the time since the last event".

Let X the random variable that represent the number of chocolate chips in a certain type of cookie. We know that $X \sim Poisson(\lambda)$

The probability mass function for the random variable is given by:

$f(x)=\frac{e^{-\lambda} \lambda^x}{x!} , x=0,1,2,3,4,...$

And f(x)=0 for other case.

For this distribution the expected value is the same parameter $\lambda$

$E(X)=\mu =\lambda$

On this case we are interested on the probability of having at least two chocolate chips, and using the complement rule we have this:

$P(X\geq 2)=1-P(X$

Using the pmf we can find the individual probabilities like this:

$P(X=0)=\frac{e^{-\lambda} \lambda^0}{0!}=e^{-\lambda}$

$P(X=1)=\frac{e^{-\lambda} \lambda^1}{1!}=\lambda e^{-\lambda}$

And replacing we have this:

$P(X\geq 2)=1-[P(X=0)+P(X=1)]=1-[e^{-\lambda} +\lambda e^{-\lambda}[]$

$P(X\geq 2)=1-e^{-\lambda}(1+\lambda)$

And we want this probability that at least of 99%, so we can set upt the following inequality:

$P(X\geq 2)=1-e^{-\lambda}(1+\lambda)\geq 0.99$

And now we can solve for $\lambda$

$0.01 \geq e^{-\lambda}(1+\lambda)$

Applying natural log on both sides we have:

$ln(0.01) \geq ln(e^{-\lambda}+ln(1+\lambda)$

$ln(0.01) \geq -\lambda+ln(1+\lambda)$

$\lambda-ln(1+\lambda)+ln(0.01) \geq 0$

Thats a no linear equation but if we use a numerical method like the Newthon raphson Method or the Jacobi method we find a good point of estimate for the solution.

Using the Newthon Raphson method, we apply this formula:

$x_{n+1}=x_n -\frac{f(x_n)}{f'(x_n)}$