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3 years ago

P(x)=x(x^2+4) list the roots

Mathematics

1 answer:

vlada-n [284]3 years ago

6 0

The answer would be x=0, 2i, -2i

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Which expression is equivalent to the radical expression shown below when it is simplified?

babymother [125]

Answer:

$\sqrt{\frac{3}{64} }$ in simplified form is $\frac{\sqrt{3}}{8}$

Step-by-step explanation:

We need to solve the expression

$\sqrt{\frac{3}{64} }$

We know that $\sqrt{\frac{a}{b}}=\frac{\sqrt{a}}{\sqrt{b}}$

and 64 = 8*8

Solving we get

$=\frac{\sqrt{3}}{\sqrt{64}}\\=\frac{\sqrt{3}}{\sqrt{8*8}}\\=\frac{\sqrt{3}}{\sqrt{8^2}}\\=\frac{\sqrt{3}}{8}$

So $\sqrt{\frac{3}{64} }$ in simplified form is $\frac{\sqrt{3}}{8}$

8 0

3 years ago

Give an example of a point that is the same distance from (3, 0) as it is from (7, 0). Find lots of examples. Describe the confi

Artist 52 [7]

Supposse that the distance from the point $(x,y)$ to the point $(3,0)$ is equal to the distance from $(x,y)$ to the point $(7,0)$ . Then, by the formula of the distnace we must have

$\sqrt{(x-3)^2 + (y-0)^2}=\sqrt{(x-7)^2 + (y-0)^2}$

cancel the square root and the $(y-0)^2$ 's, and then expand the parenthesis to obtain

$x^2 - 6x + 9 = x^2 - 14x + 49$

then, simplifying we obtain

$8x = 40$

therfore we must have

$x=5$

this means that the points satisfying the propertie must have first component equal to 5. So we can give a lot of examples of such points: $(5,0), (5,7),(5,1/2), (5,-10),...$ . The set of this points give us a straight line and the points (3,0) and (7,0) are symmetric with respect to this line.

3 0

2 years ago

Kumar is opening a lemonade stand. He spent $4.50 to build the stand and $2.50 for lemons and sugar. He has enough ingredients t

Triss [41]

He would want to charge $0.85 per glass of lemonade to cover his expenses and have $10.00 profit. But in reality he would'nt make $17.00 because people don't carry freaking nickels and dimes.

3 0

3 years ago

Read 2 more answers

An article suggests that a poisson process can be used to represent the occurrence of structural loads over time. suppose the me

kirill115 [55]

Answer:

a) $\lambda_1 = 2*2 = 4$

And let X our random variable who represent the "occurrence of structural loads over time" we know that:

$X(2) \sim Poi (4)$

And the expected value is $E(X) = \lambda =4$

So we expect 4 number of loads in the 2 year period.

b) $P(X(2) >6) = 1-P(X(2)\leq 6)= 1-[P(X(2) =0)+P(X(2) =1)+P(X(2) =2)+...+P(X(2) =6)]$

$P(X(2) >6) = 1- [e^{-4}+ \frac{e^{-4}4^1}{1!}+ \frac{e^{-4}4^2}{2!} +\frac{e^{-4}4^3}{3!} +\frac{e^{-4}4^4}{4!}+\frac{e^{-4}4^5}{5!}+\frac{e^{-4}4^6}{6!}]$

And we got: $P(X(2) >6) =1-0.889=0.111$

c) $e^{-2t} \leq 2$

We can apply natural log in both sides and we got:

$-2t \leq ln(0.2)$

If we multiply by -1 both sides of the inequality we have:

$2t \geq -ln(0.2)$

And if we divide both sides by 2 we got:

$t \geq \frac{-ln(0.2)}{2}$

$t \geq 0.8047$

And then we can conclude that the time period with any load would be 0.8047 years.

Step-by-step explanation:

Previous concepts

The exponential distribution is "the probability distribution of the time between events in a Poisson process (a process in which events occur continuously and independently at a constant average rate). It is a particular case of the gamma distribution". The probability density function is given by:

$P(X=x)=\lambda e^{-\lambda x}$

The exponential distribution is "the probability distribution of the time between events in a Poisson process (a process in which events occur continuously and independently at a constant average rate). It is a particular case of the gamma distribution"

Solution to the problem

Let X our random variable who represent the "occurrence of structural loads over time"

For this case we have the value for the mean given $\mu = 0.5$ and we can solve for the parameter $\lambda$ like this:

$\frac{1}{\lambda} = 0.5$

$\lambda =2$

So then $X(t) \sim Poi (\lambda t)$

X follows a Poisson process

Part a

For this case since we are interested in the number of loads in a 2 year period the new rate would be given by:

$\lambda_1 = 2*2 = 4$

And let X our random variable who represent the "occurrence of structural loads over time" we know that:

$X(2) \sim Poi (4)$

And the expected value is $E(X) = \lambda =4$

So we expect 4 number of loads in the 2 year period.

Part b

For this case we want the following probability:

$P(X(2) >6)$

And we can use the complement rule like this

$P(X(2) >6) = 1-P(X(2)\leq 6)= 1-[P(X(2) =0)+P(X(2) =1)+P(X(2) =2)+...+P(X(2) =6)]$

And we can solve this like this using the masss function:

$P(X(2) >6) = 1- [e^{-4}+ \frac{e^{-4}4^1}{1!}+ \frac{e^{-4}4^2}{2!} +\frac{e^{-4}4^3}{3!} +\frac{e^{-4}4^4}{4!}+\frac{e^{-4}4^5}{5!}+\frac{e^{-4}4^6}{6!}]$

And we got: $P(X(2) >6) =1-0.889=0.111$

Part c

For this case we know that the arrival time follows an exponential distribution and let T the random variable:

$T \sim Exp(\lambda=2)$

The probability of no arrival during a period of duration t is given by:

$f(T) = e^{-\lambda t}$

And we want to find a value of t who satisfy this:

$e^{-2t} \leq 2$

We can apply natural log in both sides and we got:

$-2t \leq ln(0.2)$

If we multiply by -1 both sides of the inequality we have:

$2t \geq -ln(0.2)$

And if we divide both sides by 2 we got:

$t \geq \frac{-ln(0.2)}{2}$

$t \geq 0.8047$

And then we can conclude that the time period with any load would be 0.8047 years.

3 0

3 years ago

PLEASE I AM BEGGING help me all I need to so is show work I have to turn this in tmrw pls

GenaCL600 [577]

Hello and Good Morning/Afternoon:

Let's take this problem step-by-step:

Let's consider all the information given:

Eddie's pension is based on the average of his last four year's salaries

$\hookrightarrow \text{average } = \text {sum of all salaries} / \text{ number of years}\\ \hookrightarrow \text{average}= 392000 / 4 = 98000$

2. The employer will pay 1.2% of that average for each year he worked

$\hookrightarrow \text{1.2 percent of average} = \frac{1.2}{100} *98000=1176$

3. The employer will therefore pay in total

$\hookrightarrow \text{Total pension} = 1176 * 20 \text{ years} = 23520$

Answer: $23520

Hope that helps!

#LearnwithBrainly

6 0

1 year ago