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2 years ago

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two swimmers swam lobstering practice together the swim 175 laps swmmier b swam 25 more than twice as many as swimmer a how many

laps did swimmer a swim

1 answer:

qwelly [4]2 years ago

6 0

Answer:

swimmer b swims 100 yards. approximately how many more feet did swimmer a swim than swimmer b? 2.

Step-by-step explanation:

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Which American Indian group formed an alliance with the British as the French and Indian War began?

goblinko [34]

I think the correct answer is A

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3 years ago

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1.13 Thank you! (Sorry it’s so blurry)

VikaD [51]

$\sqrt[8]{48^4}$ can be rewritten as $48^4$ to the $\frac{1}{8}$ power:

$(48^4)^ \frac{1}{8}$

Now, applying exponent rules (multiply exponent inside parentheses by the one outside parentheses), we get:

$(48^ \frac{1}{2})$

This is equivalent to $\sqrt{48}$ , so now, we just simplify:

$\sqrt{48}= \sqrt{16*3}= \sqrt{16}* \sqrt{3}=4 \sqrt{3}$

So:

$\sqrt[8]{48^4} =4 \sqrt{3}$

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3 years ago

Math help please <br><br> Thanks <br> A lot

babunello [35]

Answer:

2 hours: 3968 <u>[I don't understand the $ sign in the answer box]</u>

At midnight: 12137

Step-by-step explanation:

The bacteria are increasing by 15% every hour. So for every hour we will have what we started with, plus 15% more.

The "15% more" can be represented mathematically with (1 + 0.15) or 1.15. Let's call this the "growth factor" and assign it the variable b. b is (1 + percent increase).

Since this per hour, in 1 hour we'll have (3000)*(1.15) = 3450

At the end of the second hour we're increased by 15% again:

(3450)*(1.15) = 3968.

Each additional hour add another (1.15) factor, If we assign a to be the starting population, this can be represented by:

P = a(1.15)^t for this sample that increase 15% per hour. t is time, in hours.

If a represents the growth factor, and P is the total population, the general expression is

P = ab^t

Using this for a = 3000 and b = 1.15, we can find the total population at midnight after starting at 2PM. That is a 10 hour period, so t = 10

P = (3000)*(1.15)^10

P = 12137

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2 years ago

Cathy has a nickel, a dime, and a quarter in her purse. She randomly picks a coin, replaces it, and then picks another coin. The

satela [25.4K]

You could 3•2 because your it two times =6 than it’s 2\6 make more smaller 1/3 it will be. Than you get .3333 repeating

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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:

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

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3 years ago