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2 years ago

Yall im taking a exit ticket pls help me for question 1

Mathematics

2 answers:

grin007 [14]2 years ago

5 0

Answer:

the answer is 434843144649

Korolek [52]2 years ago

5 0

The answer I don’t Knoxville. O

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-4 x f (8) + 3 x g(-7) HELP ASAP

lubasha [3.4K]

Answer:

-32

Step-by-step explanation:

-4 + f(8) + 3 x g(-7) = ?

To find f(8), look at each of the f(x) function and find x = 8. The y value is the answer you will plug in for f(8). The same goes for g(-7), but you will be looking at the g(x) function at x = -7.

f(8) = -4

g(-7) = -8

Make sure you solve in the correct order using PEMDAS.

-4 + (-4) + 3 x (-8) = -32

8 0

2 years ago

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. Solve for x. Then find the angle measures of the triangle? (4x) (6x+55) (9x+30)

nalin [4]

Answer:

x = 5.

First angle = 20°, Second angle = 85°, Third angle = 75°

Step-by-step explanation:

The first angle of the triangle = (4 x)

The second angle of the triangle = (6x + 55)

The third angle of the triangle = (9x + 30)

By ANGLE SUM PROPERTY of a triangle:

First angle + Second angle + Third angle = 180°

⇒ (4 x) + (6x+55) + (9x+30) = 180°

or, (4x + 6x + 9x) + ( 55 + 30) = 180°

or, 19x = 180 - 85

or, 19 x = 95 ⇒ x = 95 /19 = 5

or, x = 5

Hence, first angle of the triangle = (4 x) = 4 x 5 = 20°

Second angle = ( 6x + 30) = 6(5) + 55 = 85°

Third angle = (9x + 30 = 9(5) + 30 = 75°

8 0

3 years ago

A rectangular prism has a width of 51 ft and a volume of 144 ft3. Find the volume of a similar prism with a width of 17 ft. Roun

meriva

Answer:

$5.3\ ft^3$

Step-by-step explanation:

Similar shapes with scale factor $k$ have proportional volumes with coefficient $k^3$ i.e.

$\dfrac{V_{large}}{V_{small}}=k^3,$

where

$k=\dfrac{\text{width of large prism}}{\text{width of small prism}}.$

Since

$\text{width of large prism}=51\ ft,\\ \\\text{width of small prism}=17\ ft,$

then

$k=\dfrac{51}{17}=3.$

Hence,

$\dfrac{144}{V_{small}}=3^3\Rightarrow V_{small}=\dfrac{144}{27}=\dfrac{16}{3}=5\dfrac{1}{3}\approx 5.3\ ft^3.$

7 0

3 years ago

Int he figure below find AC

Ahat [919]

Given :-

Two triangles with their two angles equal .

To Find :-

Measure of AC .

Answer :-

Here in ∆ABC and ∆EDC ,

$\angle$ CAB = $\angle$ CED (given )
$\angle$ CBA = $\angle$ CDE (given)

Therefore by AA similarity criterion , we can say that ∆ABC ~ ∆EDC . Also we know that corresponding sides of similar triangles are proportional . So ,

→ AC/EC = BC/DC

→ AC/9 = 8/12

→ AC = 8/12 * 9

→ AC = 6

Hence the measure of side AC is 6 .

I hope this helps.

7 0

2 years ago