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Zarrin [17]
3 years ago
15

Which expression is another way to show 9/2?

Mathematics
1 answer:
Mashutka [201]3 years ago
7 0
9 over 2, or 9/2 would also be 9 divided by 2
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Which theorem is not a valid theorem to show that two triangles are congruent?
GenaCL600 [577]

Answer:

B

Step-by-step explanation:

I got 100 on my test and that was my answer

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Please help. NEED THE ANSWER SOON AS POSSIBLE.
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Answer:

y=4x-17

Step-by-step explanation: Distribute the 4(x-5)

4x-20

Then add three to both sides

y=4x-17

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7 (2.268- 2.036) ? answer I need help
Arisa [49]
The answer to this question is 1.624
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Nathan hit 19 homeruns. He scored 3 fewer homeruns than twice the number of homeruns hit by jason. Write and solve an equation t
Vinvika [58]

Jason hit 11 homeruns.

Step-by-step explanation:

Homeruns hit by Nathan = 19

Let,

Homeruns hit by Nathan = x

Homerun hit by Jason = y

According to given statement;

x = 2y - 3   Eqn 1

We know that Nathan hit 19 homeruns, therefore, putting x=19 in Eqn 1

19=2y-3\\19+3=2y\\22=2y\\2y=22

Dividing both sides by 2;

\frac{2y}{2}=\frac{22}{2}\\y=11

Jason hit 11 homeruns.

Keywords: linear equations, addition

Learn more about linear equations at:

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7 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
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