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USPshnik [31]
3 years ago
12

The areas of the squares adjacent to two sides of a right triangle are 32 units^2 2 squared and 32 units^2 Find the length, xxx,

of the third side of the triangle.
Mathematics
1 answer:
NikAS [45]3 years ago
6 0

Answer:

8 units

Step-by-step explanation:

If the area of the squares are 32 units^2, we have that:

Area = Side * Side

Side^2 = 32

Side = sqrt(32) = 5.657 units

So as the squares are adjacent to two sides of the triangle, we have that the triangle has two sides of 5.657 units

As it is a right triangle, we can use the Pythagoras' theorem:

c^2 = a^2 + b^2

c^2 = 5.657^2 + 5.657^2

c^2 = 64

c = 8 units

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How do you write 125% as a fraction mixed number or whole number
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2 years ago
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Suppose in the University of Manitoba, 40% of the students live in apartments. If 600 students are randomly selected, then the a
just olya [345]

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:

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And we are interested on the following probability:

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

So then the correct answer would be:

B) .9996

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