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nikitadnepr [17]
3 years ago
10

ZE is the angle bisector of measure YEX and the perpendicular bisector of GF, GX is the angle bisector of measure YGZ and the pe

rpendicular bisector of EF, FY is the angle bisector of measure ZFX and the perpendicular bisector of EG. Point A is the intersection of EZ, GX, and FY.

Mathematics
1 answer:
den301095 [7]3 years ago
7 0

Answer:

C

Step-by-step explanation:

The center of inscribed circle into triangle is point of intersection of all interior angles of triangle.

The center of circumscribed circle over triabgle is point of intersection of perpendicular bisectors to the sides.

Circumscribed circle always passes through the vertices of the triangle.

Inscribed circle is always tangent to the triangle's sides.

In your case angles' bisectors and perpendicular bisectors intesect at one point, so point A is the center of inscribed circle and the center of corcumsribed circle. Thus, these circles pass through the points X, Y, Z and G, E, F, respectively.

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If someone has 8 fl. oz. of 16% wine, how many standard drinks has that person consumed?
UkoKoshka [18]

Answer:

2.1 standard drinks

(or 2 if you need to round to the nearest whole number)

Step-by-step explanation:

The ABV (alcohol by volume) percentage is a measure of the amount of pure alcohol (as a percentage) of the total volume of liquid in a drink.

So for 16% wine, this means that 16% of the volume of the wine is pure alcohol.

Therefore, if 16% of 8 fl oz is pure alcohol, this means that 1.28 fl oz of the 8 fl oz is pure alcohol (since 0.16 × 8 = 1.28).

One standard drink ≈ 0.6 fl oz pure alcohol

To calculate the number of standard drinks, divide the total amount of pure alcohol found by the amount in one standard drink:

1.28 ÷ 0.6 = 2.1

Therefore there is approximately 2.1 standard drinks in 8 fl oz of 16% wine.

5 0
2 years ago
What are whole numbers, integers, and rational numbers?
professor190 [17]

Answer:

Integers

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The number of failures of a testing instrument from contamination particles on the product is a Poisson random variable with a m
Mazyrski [523]

Answer:

The probability that the instrument does not fail in an 8-hour shift is P(X=0) \approx 0.8659

The probability of at least 1 failure in a 24-hour day is P(X\geq 1 )\approx 0.3508

Step-by-step explanation:

The probability distribution of a Poisson random variable X representing the number of successes occurring in a given time interval or a specified region of space is given by the formula:

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

Let X be the number of failures of a testing instrument.

We know that the mean \mu = 0.018 failures per hour.

(a) To find the probability that the instrument does not fail in an 8-hour shift, you need to:

For an 8-hour shift, the mean is \mu=8\cdot 0.018=0.144

P(X=0)=\frac{e^{-0.144}0.144^0}{0!}\\\\P(X=0) \approx 0.8659

(b) To find the probability of at least 1 failure in a 24-hour day, you need to:

For a 24-hour day, the mean is \mu=24\cdot 0.018=0.432

P(X\geq 1 )=1-P(X=0)\\\\P(X\geq 1 )=1-\frac{e^{-0.432}0.432^0}{0!}\\\\P(X\geq 1 )\approx 0.3508

3 0
3 years ago
1. Approximate the given quantity using a Taylor polynomial with n3.
Jet001 [13]

Answer:

See the explanation for the answer.

Step-by-step explanation:

Given function:

f(x) = x^{1/4}

The n-th order Taylor polynomial for function f with its center at a is:

p_{n}(x) = f(a) + f'(a) (x-a)+\frac{f''(a)}{2!} (x-a)^{2} +...+\frac{f^{(n)}a}{n!} (x-a)^{n}

As n = 3  So,

p_{3}(x) = f(a) + f'(a) (x-a)+\frac{f''(a)}{2!} (x-a)^{2} +...+\frac{f^{(3)}a}{3!} (x-a)^{3}

p_{3}(x) = f(a) + f'(a) (x-a)+\frac{f''(a)}{2!} (x-a)^{2} +...+\frac{f^{(3)}a}{6} (x-a)^{3}

p_{3}(x) = a^{1/4} + \frac{1}{4a^{ 3/4} }  (x-a)+ (\frac{1}{2})(-\frac{3}{16a^{7/4} } ) (x-a)^{2} +  (\frac{1}{6})(\frac{21}{64a^{11/4} } ) (x-a)^{3}

p_{3}(x) = 81^{1/4} + \frac{1}{4(81)^{ 3/4} }  (x-81)+ (\frac{1}{2})(-\frac{3}{16(81)^{7/4} } ) (x-81)^{2} +  (\frac{1}{6})(\frac{21}{64(81)^{11/4} } ) (x-81)^{3}

p_{3} (x) = 3 + 0.0092592593 (x - 81) + 1/2 ( - 0.000085733882) (x - 81)² + 1/6  

                                                                                  (0.0000018522752) (x-81)³

p_{3} (x)  =  0.0092592593 x - 0.000042866941 (x - 81)² + 0.00000030871254

                                                                                                       (x-81)³ + 2.25

Hence approximation at given quantity i.e.

x = 94

Putting x = 94

p_{3} (94)  =  0.0092592593 (94) - 0.000042866941 (94 - 81)² +          

                                                                 0.00000030871254 (94-81)³ + 2.25

         = 0.87037 03742 - 0.000042866941 (13)² + 0.00000030871254(13)³ +    

                                                                                                                       2.25

         = 0.87037 03742 - 0.000042866941 (169) +  

                                                                      0.00000030871254(2197) + 2.25

         = 0.87037 03742 - 0.007244513029 + 0.0006782414503 + 2.25

p_{3} (94)  = 3.113804102621

Compute the absolute error in the approximation assuming the exact value is given by a calculator.

Compute \sqrt[4]{94} as 94^{1/4} using calculator

Exact value:

E_{a}(94) = 3.113737258478

Compute absolute error:

Err = | 3.113804102621 - 3.113737258478 |

Err (94)  = 0.000066844143

If you round off the values then you get error as:

|3.11380 - 3.113737| = 0.000063

Err (94)  = 0.000063

If you round off the values up to 4 decimal places then you get error as:

|3.1138 - 3.1137| = 0.0001

Err (94)  = 0.0001

4 0
3 years ago
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