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3 years ago

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Jack works after school. Each day he is paid a set amount, plus an hourly wage. Look at the table. Choose the linear function f

Jack can use to find his pay. Hours 1 1.5 2 2.5 3 Pay 18 23 28 33 38 A. f(x) = 10x + 8 B. f(x) = 15x + 3 C. f(x) = 0.5x + 1 D. f(x) = x + 18

1 answer:

nekit [7.7K]3 years ago

7 0

Answer:

The correct linear equation is;

f(x) = 10x + 8

Step-by-step explanation:

To choose the correct linear function, we just need to take a specific point, then substitute in each of the equations given to get the one right

The correct one will be the one in which when the x-value is substituted, we will get the y-value

Let us pick (1,18)

Let us begin testing;

A. f(x) = 10x + 8

f(1) = 10(1) + 8 = 10 + 8 = 18

Let us use another point to test this

The second point is (1.5,23)

f(1.5) = 10(1.5) + 8

= 15 + 8 = 23

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Steve has $5. Nancy has 250% as much money as Steve. How much money does Nancy have?

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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

1. Which of the following describes the end behavior of the function ƒ(x) = x^4 + 3x^3 – 2x + 7?

snow_tiger [21]

Hello, when x tends to $\infty$ the term with the highest degree will lead the behaviour.

In other words.

$\displaystyle \lim_{x\rightarrow+\infty} {x^4+3x^3-2x+7}\\\\=\lim_{x\rightarrow+\infty} {x^4}\\\\=+\infty\\\\\\\displaystyle \lim_{x\rightarrow-\infty} {x^4+3x^3-2x+7}\\\\=\lim_{x\rightarrow-\infty} {x^4}\\\\=+\infty$

So, the answer B is correct.

Thank you.

6 0

3 years ago