Select the graph that best represents the solution set of the system of linear inequalities

Which of the following binomials is a factor of x cubed plus 4 x squared plus x minus 6?

dusya [7]

The first thing we are going to do in this case is to rewrite the expression.
We have then:
x ^ 3 + 4x ^ 2 + x - 6
Factoring the expression we have:
(x-1) * (x + 2) * (x + 3)
Therefore, we observe that the following binomial is a common factor:
(x + 2)
Answer:
B. x plus 2

7 0

3 years ago

Gabrielle uses 7.6 pints of blue paint and white paint to paint her bedroom walls. 1 4 of this amount is blue paint, and the res

zysi [14]

Answer:

1.9

Step-by-step explanation:

Gabrielle uses 7.6 pints of blue and white paint to paint her bedroom

1/4 of the paint is blue paint

Therefore the amount of white paint she used to paint her bedroom can he calculated as follows

= 1/4

= 0.25×100

= 25%

25/100×7.6

= 2.5 × 7.6

= 1.9

Hence 1.9 pints of white paint was used to paint the room

6 0

3 years ago

What is the image of point (-6,1) after the reflection over the line y=2

Lynna [10]

The image is (-6, 3)

Hope it helps!

5 0

3 years ago

How do you solve rational expressions

Stells [14]

Answer:

lots of multiplying I've noticed

5 0

2 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:

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

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:

$\lambda \geq 6.63835$

4 0

3 years ago