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3 years ago

explain how you can rename 5,400 as hundreds. Include a quick picture or a place- value chart in your explanation

1 answer:

7 0

5400 = 5000 + 400

5000 ⇒ 50 lots of one hundred
5000 ⇒ 50 hundred

400 ⇒ 4 lots of one hundred
400 ⇒ 4 hundred

5400 ⇒ 54 lots of one hundred
5400 ⇒ 54 hundred

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Law of sines: How many distinct triangles can be formed for which m∠A = 75°, a = 2, and b = 3?

butalik [34]

According to the law of sines:

$\frac{a}{sin(A)} = \frac{b}{sin(B)}$

Using the given values, we can find the angle B and find the number of possible triangles that can be formed.

$\frac{2}{sin(75)}= \frac{3}{sin(B)} \\ \\ 
sin(B)=3* \frac{sin75}{2} \\ \\ 
sin(B)=1.45 \\ \\ 
B=sin^{-1} (1.45)$

The range of sin is from -1 to 1. The above expression does not yield any possible value of B, as sin of no angle can be equal to 1.45.

Therefore, we can conclude that no triangle exists with the given conditions.

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3 years ago

If f(x) = 4x^2and g(x) = x+1, find (f•g)(x).

vazorg [7]

Answer:

(f.g)(x)=4x^3+4x^2

Step-by-step explanation:

f(x) = 4x^2and g(x) = x+1, (f•g)(x)=?

(f.g)(x)=f(x)*g(x)

(f.g)(x)=4x^2*(x+1)

(f.g)(x)=4x^3+4x^2

hope it helps. .brainliest please

4 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:

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