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KiRa [710]
3 years ago
15

Cookies are sold singly or in packages of 3 or 9. With this​ packaging, how many ways can you buy 18 ​cookies?

Mathematics
2 answers:
statuscvo [17]3 years ago
7 0
Six ways can be made 
zvonat [6]3 years ago
4 0
6 packs of 3
2 packs of 9
3 packs of 3, 1 pack of 9 

answer: 3 ways
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Help please!!!!!!!!!!!!!
KatRina [158]
Hello!

For this you can plug in the x values in the choices and see if you get the y value

y = -4^{3}  - (-4)

-4^3 is -64

when you subtract a negative you are just adding

-64 + 4 = 60

Then you do the second one

y =  -3^{3}  - (-3)

-3^3 is -27

-27 + 3 is -24

Then the third one

y =  -2^{3}  - (-2)

-2^3 is -8

-8 + 2 is -6

Then the last one

y =  -1^{3}  - (-1)

-1^3 is -1

-1 + 1 = 0

Since we did not get -2 the answer is D

Hope this Helps!
7 0
3 years ago
Savannah planned to evenly disperse 400 chocolate chips among the x pans of brownies. Unfortunately, the last pan of brownies re
bonufazy [111]
400 - 32 which equals 368
4 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
The SAT mathematics scores in the state of Florida for this year are approximately normally distributed with a mean of 500 and a
Greeley [361]
<u></u><u>The correct answer is 47.5%, or 0.475.</u>

Explanation:
The empirical rule states that in any normal distribution:

68% of data will fall within 1 standard deviation of the mean;
95% of data will fall within 2 standard deviations of the mean; and
99.7% of data will fall within 3 standard deviations of the mean.

The mean is 500 and the standard deviation is 100.  This means that 700 is 2 standard deviations away from the mean:

(700-500)/100=200/100=2.

We know that 95% of data will fall within 2 standard deviations from the mean.  However, included in the 95% is data less than the mean and greater than the mean.  Since we are only concerned with the scores from 500 to 700, we only want the half that is greater than the mean:

95/2 = 47.5%, or 0.475.
8 0
3 years ago
Read 2 more answers
-12(r + 4) =72 please help
Ivahew [28]
The answer is r= -10
6 0
3 years ago
Read 2 more answers
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