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3 years ago

The perimeter of a rectangle is 26 meters. The difference between the length and the width is 5 meters. Find the width

Mathematics

1 answer:

Scilla [17]3 years ago

3 0

Try making a substitution:

You might be interested in

20 points

Bezzdna [24]

Answer:

32%

Step-by-step explanation:

P(G/S) = P(G&S)/P(S)

= (0.32×0.5)/0.5

= 0.32

32%

5 0

3 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

In B E C, F is the centroid, and AC = 12. Find A F and F C

klemol [59]

bearing in mind that the centroid in a triangle cuts each of the three medians in a 2:1 ratio.

since we know that A C = 12, let's split it in a 2:1 ratio then, cleary from the picture the larger is F C, so F C : A F is on a 2:1 ratio.

$\bf A C=A F+F C\qquad \qquad \cfrac{F C}{A F}=\cfrac{2}{1}\qquad \qquad \cfrac{2\cdot \frac{12}{2+1}}{1\cdot \frac{12}{2+1}}\implies \cfrac{2\cdot 4}{1\cdot 4}\implies \cfrac{8}{4}$

4 0

3 years ago

13. If the following fractions were converted to decimals, which one would result in a repeating decimal? A. 3/4 B. 1/9 C. 5/11

Vladimir [108]

Hey there!

In order to find if a fraction would result in a repeating decimal, recall that a fraction is a division problem written vertically. All that you have to do is divide the numerator by the denominator. Also, remember that a repeating decimal will result in the same number after the decimal point as long as the calculator can handle.

3 ÷ 4 = 0.75

1 ÷ 9 = 0.11111111...

5 ÷ 11 = 0.45454545...

3 ÷ 0.42857143...

As you can see, two out of your four answer choices give you a repeating decimal. B gives you a repeated number of "1" while C gives you "45". D doesn't count since there is no pattern of repeated numbers that it follows.

Both B and C fall into the category of repeating decimal. If you're only able to choose one answer, I would ask your teacher, a parent, or a peer what they think.

Hope this helped you out! :-)

5 0

3 years ago

URGENT

Nutka1998 [239]

Answer: Picking the first option and constantly adding on and on you will eventually make way more than what the second option proposes, therefore picking the first option is the obvious choice.

Step-by-step explanation:

Well let’s see it starts off slow for instance 1, 2, 4, 8, 16 and so on adding that on for a month will eventually leave you with more.

4 0

2 years ago