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mrs_skeptik [129]
3 years ago
13

Can someone solve the hypotenuse for me? Thank you!

Mathematics
1 answer:
dmitriy555 [2]3 years ago
3 0

Answer:

Side a = 88.67626

Side b = 57

Side c = 67.92995

Angle ∠A = 90° = 1.5708 rad = π/2

Angle ∠B = 40° = 0.69813 rad = 2/9π

Angle ∠C = 50° = 0.87266 rad = 5/18π

Area = 1,936.00371

Perimeter p = 213.60621

Semiperimeter s = 106.80311

Height ha = 43.66453  

Height hb = 67.92995

Height hc = 57

Median ma = 44.33813  

Median mb = 73.66633

Median mc = 66.35224

Inradius r = 18.12685  

Circumradius R = 44.33813  

Vertex coordinates: A[0, 0] B[67.92995, 0] C[0, 57]

Centroid: [22.64332, 19]

Inscribed Circle Center: [18.12685, 18.12685]

Circumscribed Circle Center: [33.96498, 28.5]

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Dahasolnce [82]
19.18 / 2.8 = 6.85

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SAT scores are normed so that, in any year, the mean of the verbal or math test should be 500 and the standard deviation 100. as
vovangra [49]

Answer:

a) P(X>625)=P(\frac{X-\mu}{\sigma}>\frac{625-\mu}{\sigma})=P(Z>\frac{625-500}{100})=P(Z>1.25)

P(Z>1.25)=1-P(Z

b) P(400

P(-1

P(-1

c) z=-0.842

And if we solve for a we got

a=500 -0.842*100=415.8

So the value of height that separates the bottom 20% of data from the top 80% is 415.8.  

Step-by-step explanation:

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".  

Part a

Let X the random variable that represent the SAT scores of a population, and for this case we know the distribution for X is given by:

X \sim N(500,100)  

Where \mu=500 and \sigma=100

We are interested on this probability

P(X>625)

And the best way to solve this problem is using the normal standard distribution and the z score given by:

z=\frac{x-\mu}{\sigma}

If we apply this formula to our probability we got this:

P(X>625)=P(\frac{X-\mu}{\sigma}>\frac{625-\mu}{\sigma})=P(Z>\frac{625-500}{100})=P(Z>1.25)

And we can find this probability using the complement rule and with the normal standard table or excel:

P(Z>1.25)=1-P(Z

Part b

We are interested on this probability

P(400

And the best way to solve this problem is using the normal standard distribution and the z score given by:

z=\frac{x-\mu}{\sigma}

If we apply this formula to our probability we got this:

P(400

And we can find this probability with this difference:

P(-1

And in order to find these probabilities we can find tables for the normal standard distribution, excel or a calculator.  

P(-1

Part c

For this part we want to find a value a, such that we satisfy this condition:

P(X>a)=0.8   (a)

P(X   (b)

Both conditions are equivalent on this case. We can use the z score again in order to find the value a.  

As we can see on the figure attached the z value that satisfy the condition with 0.2 of the area on the left and 0.8 of the area on the right it's z=-0.842. On this case P(Z<-0.842)=0.2 and P(Z>-0.842)=0.8

If we use condition (b) from previous we have this:

P(X  

P(z

But we know which value of z satisfy the previous equation so then we can do this:

z=-0.842

And if we solve for a we got

a=500 -0.842*100=415.8

So the value of height that separates the bottom 20% of data from the top 80% is 415.8.  

8 0
3 years ago
I need help with this ​
Korvikt [17]

Answer:

D.

Step-by-step explanation:

According to converse of the Pythagorean Theorem, that if the square of the third side (longest side) of a triangle equals the sum of the other two shorter sides, therefore, the triangle formed must be a right triangle.

From the side lengths given, the following satisfies the Pythagorean triple:

5² + 12² = 13².

Therefore, the procedure to use to confirm the converse of the Pythagorean Theorem would be to draw the two shortest side, 5 cm and 12 cm, so that a right angle will be between them. The side measuring 13 cm should therefore fit in to form a right triangle.

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