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Reil [10]
3 years ago
13

* 50 POINTS* hhhaaaaaalllllpppp

Mathematics
2 answers:
Korolek [52]3 years ago
6 0

K = C^2 / AB

C^2 = K * AB

C = √(KAB)

Hope that helps

Black_prince [1.1K]3 years ago
5 0
K = C^2/ AB
KAB = C^2
√KAB = C
C = √KAB

Answered according to the given information

Hope it helps :)
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ABC has the vertices A(1, 4), B(3, 4), and C(1, 1). Find the coordinates of each point of concurrency. 13. circumcenter of ∆ABC
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Triangle ABC is a right angle triangle. The point of circumcentre and the point of orthocentre is the same point, as shown in the diagram below. The coordinate of the point is (2.5, 2.5)

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P(hemoglobin level between 9 and 11) = P(age between 25 and 35 years)..
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Math help on questions #4 and #5
Vanyuwa [196]
<h3>Answer:</h3>

4. -3

5. 3

<h3>Step-by-step explanation:</h3>

4. For x > -2, the value of a is the slope of the line. The line goes down 3 units for each 1 to the right, so the slope is -3/1 = -3. Then a = -3.

___

5. The ordered pair (h, k) is typically used to name the point to which a function is translated. The vertex of the function f(x) = |x| is (0, 0). When it is translated to (h, k), the function becomes ...

... q(x) = |x -h| +k

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7 0
3 years ago
Suppose X has an exponential distribution with mean equal to 23. Determine the following:
e-lub [12.9K]

Answer:

a) P(X > 10) = 0.6473

b) P(X > 20) = 0.4190

c) P(X < 30) = 0.7288

d) x = 68.87

Step-by-step explanation:

Exponential distribution:

The exponential probability distribution, with mean m, is described by the following equation:

f(x) = \mu e^{-\mu x}

In which \mu = \frac{1}{m} is the decay parameter.

The probability that x is lower or equal to a is given by:

P(X \leq x) = \int\limits^a_0 {f(x)} \, dx

Which has the following solution:

P(X \leq x) = 1 - e^{-\mu x}

The probability of finding a value higher than x is:

P(X > x) = 1 - P(X \leq x) = 1 - (1 - e^{-\mu x}) = e^{-\mu x}

Mean equal to 23.

This means that m = 23, \mu = \frac{1}{23} = 0.0435

(a) P(X >10)

P(X > 10) = e^{-0.0435*10} = 0.6473

So

P(X > 10) = 0.6473

(b) P(X >20)

P(X > 20) = e^{-0.0435*20} = 0.4190

So

P(X > 20) = 0.4190

(c) P(X <30)

P(X \leq 30) = 1 - e^{-0.0435*30} = 0.7288

So

P(X < 30) = 0.7288

(d) Find the value of x such that P(X > x) = 0.05

So

P(X > x) = e^{-\mu x}

0.05 = e^{-0.0435x}

\ln{e^{-0.0435x}} = \ln{0.05}

-0.0435x = \ln{0.05}

x = -\frac{\ln{0.05}}{0.0435}

x = 68.87

5 0
3 years ago
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