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dusya [7]
3 years ago
13

H e l p !!!!!! D;will give brainlies t

Mathematics
2 answers:
SCORPION-xisa [38]3 years ago
8 0

Answer:

26.778

Step-by-step explanation:

shepuryov [24]3 years ago
4 0

Answer:21,000

Step-by-step explanation:

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Please help! All in the image:
charle [14.2K]

Answer:

66 cm^2

Step-by-step explanation:

By equating the area we get x = 6.

8 0
3 years ago
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Help, I don’t know how to solve this
Inessa [10]

Answer:

a=A/8

Step-by-step explanation:

This MIGHT be the answer, I am 99% sure!

7 0
2 years ago
Mrs. Vance decided to buy dress shirts for her two sons. She bought her older son a blue shirt for $9.91, and she bought her you
AveGali [126]

Hope the picture will help you

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2 years ago
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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
4 years ago
HELP What is the area of this figure?
Mademuasel [1]
The correct answer is:  [B]:  "40 yd² " .
_____________________________________________________
First, find the area of the triangle:

The formula of the area of a triangle, "A":

A = (1/2) * b * h ; 

in which:  " A = area (in units 'squared') ;  in our case, " yd² " ; 

                 " b = base length" = 6 yd.  

                 " h = perpendicular height" = "(4 yd + 4 yd)" = 8 yd.
___________________________________________________
→  A = (1/2) * b * h = (1/2) * (6 yd) * (8 yd) = (1/2) * (6) * (8) * (yd²) ; 
                                                                 
                                                                   =  " 24 yd² " .
___________________________________________________
Now, find the area, "A", of the square:

The formula for the area, "A" of a square:

   A = s² ;

in which:  "A = area (in "units squared") ; in our case, " yd² " ;
                 
                 "s = side length (since a 'square' has all FOUR (4) equal side lengths);

 A = s²  = (4 yd)² = 4² * yd² =  "16 yd² "
_________________________________________________
Now, we add the areas of BOTH the triangle AND the square:
_________________________________________________
        →  " 24 yd²  +  16 yd² " ; 

to get:  " 40 yd² " ;  which is:  Answer choice:  [B]:  " 40 yd² " .
_________________________________________________
4 0
3 years ago
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