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Black_prince [1.1K]
3 years ago
13

What 2 square numbers have a difference of 55

Mathematics
1 answer:
Mrrafil [7]3 years ago
5 0
a^2-b^2 =55 \\  \\ 28^2 -27^2=55 \\  \\ 28^2 =784 \\  \\27^2 =729 \\  \\ 784-729=55 \\  \\  \\ \sf So,~the~ difference~ of~ the~ two~ squares~ of~ 28~ and~ 27~ equals~ to~ 55.
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Answer:

P(200\leq X\leq 400)=P(\frac{200-240}{12}\leq \frac{X-\mu}{\sigma}\leq \frac{400-240}{12})=P(-3.33\leq Z \leq 13.33)

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So then the correct answer would be:

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Step-by-step explanation:

The exact way to solve this problem is using the binomial distribution, assuming that our random variable of interest is "number of students living in apartments" represented by X and X \sim Bin(n=600, p =0.4)

And we want this probability:

P(200 \leq X \leq 400) [tex]In order to find this probability we can use the foloowing excel code:"=BINOM.DIST(400,600,0.4,TRUE)-BINOM.DIST(199,600,0.4,TRUE)"And we got:[tex] P(200 \leq X \leq 400) =0.999675[tex]But for this case the problem says that we need to approximate, so then we can use the normal approximation to the normal distribution. We need to check the conditions in order to use the normal approximation.
[tex]np=600*0.4=240\geq 10

n(1-p)=600*(1-0.4)=360 \geq 10

So we see that we satisfy the conditions and then we can apply the approximation.

If we appply the approximation the new mean and standard deviation are:

E(X)=np=600*0.4=240

\sigma=\sqrt{np(1-p)}=\sqrt{600*0.4(1-0.4)}=12

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P(200\leq X\leq 400)=P(\frac{200-240}{12}\leq \frac{X-\mu}{\sigma}\leq \frac{400-240}{12})=P(-3.33\leq Z \leq 13.33)

=P(Z

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B) .9996

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Answer:
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