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3 years ago

Order the following set of numbers from least to greatest. 0.09, 1%, 11%, 1/8

Mathematics

2 answers:

WITCHER [35]3 years ago

8 0

Answer: 0.09,1%,1/8,11%
Explanation: this is College math pshh

kkurt [141]3 years ago

7 0

1/8=0.125 1%=0.01 11%=0.11
1%, 0.09, 11%, 1/8

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Problem page the sum of two numbers is 67 and the difference is 11 . what are the numbers?

pav-90 [236]

Try this:
1. sum of numbers is x+y=67, the difference between them is x-y=11
2. it is possible to make up and resolve the system:
$\left \{ {{x+y=67} \atop {x-y=11}} \right. \ =\ \textgreater \ \ \left \{ {{x=39} \atop {y=28}} \right.$

8 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

Which expression is equivalent to r to the 9th over r to the 3rd

alisha [4.7K]

Answer:

$r^{6}$

Step-by-step explanation:

To make this a bit easier to see, we'll expand that expression.

R to the 9th is just r times itself 9 times. R to the third is r times itself 3 times.

(excuse my bad formatting)

The expression:

r • r • r • r • r • r • r • r • r • 1

r • r • r • 1

Remember, anything and everything has a coefficient or denominator of 1.

So, we can cancel 3 r's from the numerator and denominator.

r • r • r • r • r • r • 1

Simplify....

$\frac{1r^{6}}{1}$

That just equals $r^{6}$ .

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2 years ago

Elena packed 48 cubes into this box.Each cube has edges that are 1 centimeter.How many layers of cubes did Elena make

solniwko [45]

Elena packed 48 cubes.
Each cube has an edge of 1 centimeter.
The number of layers that Elena can make depends on how each layer is arranged and depends on how many cubes are there in a layer.
Assume that each layer has only 1 cube, then there are 48 layers.

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3 years ago

A field is 3645 feet long. What is the length of the field in yards?

Ilya [14]

Answer:

answer of this question is 12150

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