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dimaraw [331]
2 years ago
9

Find the sum or difference. -3(h-4) -2(-6h + 5)

Mathematics
1 answer:
KIM [24]2 years ago
3 0

Answer:

sum = -15h+2

Step-by-step explanation:

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Write a decimal for >1.45 to make the comparison true
Softa [21]

Answer:

> this sign means greater than so any number that is greater than 1.45 would work.

2.34 or 1.47, any number above 1.45 will be an excellent answer

Hope this helps ;)

5 0
3 years ago
Mrs. Masinick is interested in buying a notebook computer. If she buys the computer this week, she will save $150.00 while it is
hichkok12 [17]

Answer:

selling

Step-by-step explanation:

selling" (and any subsequent words) was ignored because we limit queries to 32 words.

5 0
3 years ago
Can you help me with this
kifflom [539]

Answer:

the anwer is d

I did the math

8 0
2 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)

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
Blanche bought a 5-year cd for $7100 with an apr of 2.8%, compounded quarterly, but she wants to take all her money out 9 months
lesantik [10]

Answer:

The money she will end up earning in interest on the cd = $11,352.90

Step-by-step explanation:

The formula for getting the accumulated amount(compounded) is;

A =P(1+\frac{r}{n})^n*t

Where

A = Accumulated amount  

P = principle (deposit)

r = interest rate and

n = no of times interest applied per time period.  

The interest is compounded quarterly so in one year it will be 4 times

In 5 years

n = (5×4)-3 = 17  (as she will withdraw 3 month before the completion of five years)

A = 7100(1+\frac{2.8}{100} )^17

  = 7100( 1 + 0.028)^17

  =  7100(1.028)^17  

   = 7100 * 1.599

  = 11,352.90

Therefore the money she will end up earning in interest on the cd = $11,352.90

8 0
3 years ago
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