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3 years ago

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What is the equation of the line of the best fit for the following data? Round to the slope and y-intercept of the line to three

decimal places

1 answer:

BigorU [14]3 years ago

4 0

Answer:

D. $y=-0.927x+13.634$

Step-by-step explanation:

We graph the points on the graph. The graph is attached.

Let us take two points (1, 14) and (15, 1), and calculate the slope $m$ between them <em>(we choose these points because the line passing through them will be the best fit for all points) </em>

$m=\frac{14-1}{1-15} =\frac{13}{-14} =-0.9286$

Thus we have the equation

$y=-0.9286x+b$

Let us now calculate $b$ from the point $(1,14)$

$14=-0.9286(1)+b$

$b=14.9286$

So the equation we get is

$y=-0.9286x+14.9286$

Let us now turn to the choices given and see which choice is closest to our equation: we see that choice D. $y=-0.927x+13.634$ is the closest one, so we pick it.

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Step-by-step explanation:

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Which expression is equivalent to 3x+10−x+12<br> a.24x<br> b.4x + 22<br> c.26x<br> d.2x + 22

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3 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:

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

Where :

$f(x_n)=\lambda -ln(1+\lambda)+ln(0.01)$

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

Iterating as shown on the figure attached we find a final solution given by:

$\lambda \geq 6.63835$

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Answer:

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Step-by-step explanation:

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