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jolli1 [7]
3 years ago
9

The science club is selling puzzles to raise money. The supplier charges a one-time fee of $55 for each order and $10 for each p

uzzle. Write and solve an equation to find the number of puzzles the science club can buy if they have $1,245.
Mathematics
1 answer:
wariber [46]3 years ago
7 0

I believe you can do 1,245 divided by 65 to find out how many puzzles they can buy I may be wrong :/

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Paraphin [41]
<h3> Hope this attachment helps u</h3>

5 0
3 years ago
Ellen has a bag with 3 red marbles and 2 blue marbles in it. She is going to randomly draw a marble from the bag 300 times, putt
sweet [91]

Answer:

No. of times Ellen draws a blue marble = 120

Step-by-step explanation:

No. of red marbles = 3

No. of blue marbles = 2

\text{Probability of drawing a blue marble = }\frac{\text{No. of possible outcomes}}{\text{Total no. of outcomes}}\\\\P(B)=\frac{2}{5}\\\\No. of draws (n) = 300\\\text{So, total probability = }n\times P(B)\\\\=100\times \frac{2}{5}=120.

So, the required probability is 120.

7 0
3 years ago
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3 1/2 x 2 1/2 as a mixed number in simplest form. if correct you get brainliest
Helga [31]

Answer:

8 3/ 4

Step-by-step explanation:

3 1/ 2

×  

2 1 /2

=      

7 × 5

2 × 2

=  

35 /4

=  

8 3 /4

4 0
3 years ago
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Square of a standard normal: Warmup 1.0 point possible (graded, results hidden) What is the mean ????[????2] and variance ??????
LenaWriter [7]

Answer:

E[X^2]= \frac{2!}{2^1 1!}= 1

Var(X^2)= 3-(1)^2 =2

Step-by-step explanation:

For this case we can use the moment generating function for the normal model given by:

\phi(t) = E[e^{tX}]

And this function is very useful when the distribution analyzed have exponentials and we can write the generating moment function can be write like this:

\phi(t) = C \int_{R} e^{tx} e^{-\frac{x^2}{2}} dx = C \int_R e^{-\frac{x^2}{2} +tx} dx = e^{\frac{t^2}{2}} C \int_R e^{-\frac{(x-t)^2}{2}}dx

And we have that the moment generating function can be write like this:

\phi(t) = e^{\frac{t^2}{2}

And we can write this as an infinite series like this:

\phi(t)= 1 +(\frac{t^2}{2})+\frac{1}{2} (\frac{t^2}{2})^2 +....+\frac{1}{k!}(\frac{t^2}{2})^k+ ...

And since this series converges absolutely for all the possible values of tX as converges the series e^2, we can use this to write this expression:

E[e^{tX}]= E[1+ tX +\frac{1}{2} (tX)^2 +....+\frac{1}{n!}(tX)^n +....]

E[e^{tX}]= 1+ E[X]t +\frac{1}{2}E[X^2]t^2 +....+\frac{1}{n1}E[X^n] t^n+...

and we can use the property that the convergent power series can be equal only if they are equal term by term and then we have:

\frac{1}{(2k)!} E[X^{2k}] t^{2k}=\frac{1}{k!} (\frac{t^2}{2})^k =\frac{1}{2^k k!} t^{2k}

And then we have this:

E[X^{2k}]=\frac{(2k)!}{2^k k!}, k=0,1,2,...

And then we can find the E[X^2]

E[X^2]= \frac{2!}{2^1 1!}= 1

And we can find the variance like this :

Var(X^2) = E[X^4]-[E(X^2)]^2

And first we find:

E[X^4]= \frac{4!}{2^2 2!}= 3

And then the variance is given by:

Var(X^2)= 3-(1)^2 =2

7 0
3 years ago
1/5(x+2)+2x=6x-10 what does x=
Ber [7]
The answer to this problem is x = 52/19 = 2.737... I hope...
8 0
3 years ago
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