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mojhsa [17]
2 years ago
14

PLS HALP-

Mathematics
1 answer:
den301095 [7]2 years ago
8 0

Answer:

d im preety sure if you need the work i will give it to you

Step-by-step explanation:

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Evan wants to buy a digital camera that sells for $199.99 if he uses his dicount card he will save 18%. about how much will he s
ikadub [295]
35.99 ROUND IT OFF TO 36.00 SO THAT MEANS HE PAID 163.99
199.99X 18/100
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7 0
2 years ago
If a 35 year old woman has averaged 8 hours of sleep each 24 hours, how many days of her life has she spent sleeping
irina1246 [14]

Answer:  4258.33

Step-by-step explanation:    365 x 35= 12775/3= 4258.33

8 0
3 years ago
If s is an increasing function, and t is a decreasing function, find Cs(X),t(Y ) in terms of CX,Y .
Sedbober [7]

Answer:

C(X,Y)(a,b)=1−C(s(X),t(Y))(a,1−b).

Step-by-step explanation:

Let's introduce the cumulative distribution of (X,Y), X and Y :

F(X,Y)(x,y)=P(X≤x,Y≤y)

  • FX(x)=P(X≤x)
  • FY(y)=P(Y≤y).

Likewise for (s(X),t(Y)), s(X) and t(Y) :

F(s(X),t(Y))(u,v)=P(s(X)≤u

  • t(Y)≤v)
  • Fs(X)(u)=P(s(X)≤u)
  • Ft(Y)(v)=P(t(Y)≤v).

Now, First establish the relationship between F(X,Y) and F(s(X),t(Y)) :

F(X,Y)(x,y)=P(X≤x,Y≤y)=P(s(X)≤s(x),t(Y)≥t(y))

The last step is obtained by applying the functions s and t since s preserves order and t reverses it.

This can be further transformed into

F(X,Y)(x,y)=1−P(s(X)≤s(x),t(Y)≤t(y))=1−F(s(X),t(Y))(s(x),t(y))

Since our random variables are continuous, we assume that the difference between t(Y)≤t(y) and t(Y)<t(y)) is just a set of zero measure.

Now, to transform this into a statement about copulas, note that

C(X,Y)(a,b)=F(X,Y)(F−1X(a), F−1Y(b))

Thus, plugging x=F−1X(a) and y=F−1Y(b) into our previous formula,

we get

F(X,Y)(F−1X(a),F−1Y(b))=1−F(s(X),t(Y))(s(F−1X(a)),t(F−1Y(b)))

The left hand side is the copula C(X,Y), the right hand side still needs some work.

Note that

Fs(X)(s(F−1X(a)))=P(s(X)≤s(F−1X(a)))=P(X≤F−1X(a))=FX(F−1X(a))=a

and likewise

Ft(Y)(s(F−1Y(b)))=P(t(Y)≤t(F−1Y(b)))=P(Y≥F−1Y(b))=1−FY(F−1Y(b))=1−b

Combining all results we obtain for the relationship between the copulas

C(X,Y)(a,b)=1−C(s(X),t(Y))(a,1−b).

7 0
3 years ago
Use the grouping method to factor the polynomial in the box completely.
kvv77 [185]
5x^4(x^3+8x+6) this is the correct answer
6 0
3 years ago
Average precipitation for the first 7 months of the year, the average precipitation in toledo, ohio, is 19.32 inches. if the ave
Colt1911 [192]
Part A:

The probability that a normally distributed data with a mean, μ and standard deviation, σ is greater than a given value, a is given by:

P(x\ \textgreater \ a)=1-P(x\ \textless \ a)=1-P\left(z\ \textless \  \frac{a-\mu}{\sigma}\right)

Given that the average precipitation in Toledo, Ohio for the past 7 months is 19.32 inches with a standard deviation of 2.44 inches, the probability that <span>a randomly selected year will have precipitation greater than 18 inches for the first 7 months is given by:

P(x\ \textgreater \ 18)=1-P(x\ \textless \ 18) \\  \\ =1-P\left(z\ \textless \ \frac{18-19.32}{2.44}\right) \\  \\ =1-P(z\ \textless \ -0.5410) \\  \\ =1-0.29426=\bold{0.7057}



Part B:

</span>The probability that an n randomly selected samples of a normally distributed data with a mean, μ and standard deviation, σ is greater than a given value, a is given by:

P(x\ \textgreater \ a)=1-P(x\ \textless \ a)=1-P\left(z\ \textless \ \frac{a-\mu}{\frac{\sigma}{\sqrt{n}}}\right)

Given that the average precipitation in Toledo, Ohio for the past 7 months is 19.32 inches with a standard deviation of 2.44 inches, the probability that <span>5 randomly selected years will have precipitation greater than 18 inches for the first 7 months is given by:

</span>P(x\ \textgreater \ 18)=1-P(x\ \textless \ 18) \\ \\ =1-P\left(z\ \textless \ \frac{18-19.32}{\frac{2.44}{\sqrt{5}}}\right) \\ \\ =1-P(z\ \textless \ -1.210) \\ \\ =1-0.1132=\bold{0.8868}
7 0
3 years ago
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