All categories

3 years ago

The graph of the piecewise function f(x) is shown what is the domain of f(x)

Mathematics

1 answer:

Rom4ik [11]3 years ago

8 0

The domain of a function is where we have a value for x.

Since that's the case the domain of f(x) = {x e R / 1 ≤ x < 5}

We see that we have a value for x = 1 cuz we have a filled circle, but we don't have a value for x = 5, look at the unfilled circle

So, our x can vary between 1 and 5, but can't be 5.

You might be interested in

One number is 6 more than twice the other number. If the sum of the two numbers is 21, find the two numbers

hichkok12 [17]

Answer:

One number, let's call it x, is 6 more than twice the other number, let's say this number is y.

x= 2y+6

x+y=21

We're left with a system of equations.

If we substitute the first equation in the second we find the value of y.

(2y+6)+y=21

3y+6=21

3y=15

y=5

Now that we have found the value of y, we can take that value and substitute it in the second equation (since it's easier) to find the value of x.

x+ (5)=21

x=16

Now to check if our answers are correct, we plug in both values into any equation and see if they equate.

x+y=21

(16)+(5)=21

21=21

Our solution is correct!

4 0

2 years ago

Please Help me solve this. Easy Five Points.

Sveta_85 [38]

Answer:

$\frac{5}{6}$

Step-by-step explanation:

5 = 6y

y = $\frac{5}{6}$

8 0

3 years ago

A bag contains 3 red marbles and 6 black marbles. What is the probability that you select two red marbles if you reach in the ba

Ksju [112]

Answer:

Step-by-step explanation:

3 0

3 years ago

Read 2 more answers

Let X1, X2, ... , Xn be a random sample from N(μ, σ2), where the mean θ = μ is such that −[infinity] < θ < [infinity] and

Sliva [168]

Answer:

$l'(\theta) = \frac{1}{\sigma^2} \sum_{i=1}^n (X_i -\theta)$

And then the maximum occurs when $l'(\theta) = 0$ , and that is only satisfied if and only if:

$\hat \theta = \bar X$

Step-by-step explanation:

For this case we have a random sample $X_1 ,X_2,...,X_n$ where $X_i \sim N(\mu=\theta, \sigma)$ where $\sigma$ is fixed. And we want to show that the maximum likehood estimator for $\theta = \bar X$ .

The first step is obtain the probability distribution function for the random variable X. For this case each $X_i , i=1,...n$ have the following density function:

$f(x_i | \theta,\sigma^2) = \frac{1}{\sqrt{2\pi}\sigma} exp^{-\frac{(x-\theta)^2}{2\sigma^2}} , -\infty \leq x \leq \infty$

The likehood function is given by:

$L(\theta) = \prod_{i=1}^n f(x_i)$

Assuming independence between the random sample, and replacing the density function we have this:

$L(\theta) = (\frac{1}{\sqrt{2\pi \sigma^2}})^n exp (-\frac{1}{2\sigma^2} \sum_{i=1}^n (X_i-\theta)^2)$

Taking the natural log on btoh sides we got:

$l(\theta) = -\frac{n}{2} ln(\sqrt{2\pi\sigma^2}) - \frac{1}{2\sigma^2} \sum_{i=1}^n (X_i -\theta)^2$

Now if we take the derivate respect $\theta$ we will see this:

$l'(\theta) = \frac{1}{\sigma^2} \sum_{i=1}^n (X_i -\theta)$

And then the maximum occurs when $l'(\theta) = 0$ , and that is only satisfied if and only if:

$\hat \theta = \bar X$

6 0

3 years ago

Suppose four students miss an exam. They tell their instructor that they were carpooling to school together on the day of the ex

tia_tia [17]

Answer:

The probability they each indicate the same tire is $\frac{1}{64}=0.02$

Step-by-step explanation:

There will be 4 cases involved as there are 4 tires.

Probability of choosing front left tire is $\frac{1}{4}$ as there are a total of 4 tires and we need to choose 1.

Now, if all the four students chose the front left tire then the probability is the product of individual probabilities.

Therefore, probability that all 4 students chooses front left tire is:

$P(fl)=(\frac{1}{4})^4$

Similarly, the probabilities of choosing the remaining 3 tires by all the students would be:

$P(fr)=(\frac{1}{4})^4$ , $P(rl)=(\frac{1}{4})^4$ , $P(rr)=(\frac{1}{4})^4$

Therefore, the probability that they indicate the same tire is the sum of all these probabilities.

$P(\textrm{same tire by all})=P(fl)+P(fr)+P(rl)+P(rr)\\ P(\textrm{same tire by all})=(\frac{1}{4})^4\times 4\\ P(\textrm{same tire by all})=\frac{4}{4^4}=\frac{1}{4^3}=\frac{1}{64}=0.02$

7 0

4 years ago