Find the slope from<br> 1. A Graph<br> Slope:

Does the graph represent a function?

Lina20 [59]

The graph does not represent a function

4 0

3 years ago

The base change property can help in what type of scenarios? HELP ME!!!!!

Mekhanik [1.2K]

<h3>Base Change Property</h3><h3 />

The Base Change Property is very helpful in scenarios related to simplifying equations where the logarithmic terms have a varying base.

So to solve an equation, which possesses logarithmic functions, all logarithmic terms must have a similar base.

<h3>What is Base Change Property?</h3><h3 />

This refers to the base formula which is used to write a logarithm of a number with a base that is fixed as the ratio of two logarithms both having the same base but different from the base of the initial or original logarithm.

Change of Base Formula is given as:

$log_{b} a = \frac{Log_{c} a }{Log_{c} b}$

See the link below for more about Base Change Property:

brainly.com/question/15318682

6 0

3 years ago

John has 12 marbles of different colors, including one red, one green, and one blue marble. in how many ways can he choose 4 mar

Savatey [412]

3 ways (uncertain)
1 red or blue or green + 3 other colours

7 0

3 years ago

Read 2 more answers

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:

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