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ruslelena [56]
3 years ago
14

Does anybody know how to solve this? (Click for full picture if it doesn’t show up)

Mathematics
2 answers:
Semenov [28]3 years ago
5 0

Step-by-step explanation:

angle A = 38

now come with me inside the triangle ABC

use two property of isoceles traingle and sum of angles of triangle

you get x= 71°

zavuch27 [327]3 years ago
5 0

Answer:

x = 71°

Step-by-step explanation:

Notice that angle D is the inscribed angle of arc BC. By definition of inscribed angles, the measure of an arc's inscribed angle is half the measure of the arc. Also we see that angle A is the central angle of arc BC. By definition, the central angle measure is equal to the arc measure.

Here, ∠D = 19, which means arc BC = 19 * 2 = 38°. Then, that means ∠A = 38.

Notice that ABC forms a triangle and remember that all the angles of a triangle add up to 180°. Since ΔABC is isosceles, we know that the base angles are congruent, so let's say they're both x. Since ∠A = 38°, ∠B = ∠C = x, we can add them up to get 180:

∠A + ∠B + ∠C = 180

38 + x + x = 180

38 + 2x = 180

2x = 142

x = 142/2 = 71

Thus, x = 71°.

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An angle measures 15 degrees more than its complement. Find the measures of the two angles
ale4655 [162]

x = one angle

y = other angle

y = 15+x

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x+y = 90

substitute for y (15+x)

x + 15+x = 90

combine like terms

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3 0
3 years ago
In a certain assembly plant, three machines B1, B2, and B3, make 30%, 20%, and 50%, respectively. It is known from past experien
diamong [38]

Answer:

The probability that a randomly selected non-defective product is produced by machine B1 is 11.38%.

Step-by-step explanation:

Using Bayes' Theorem

P(A|B) = \frac{P(B|A)P(A)}{P(B)} = \frac{P(B|A)P(A)}{P(B|A)P(A) + P(B|a)P(a)}

where

P(B|A) is probability of event B given event A

P(B|a) is probability of event B not given event A  

and P(A), P(B), and P(a) are the probabilities of events A,B, and event A not happening respectively.

For this problem,

Let P(B1) = Probability of machine B1 = 0.3

P(B2) = Probability of machine B2 = 0.2

P(B3) = Probability of machine B3 = 0.5

Let P(D) = Probability of a defective product

P(N) = Probability of a Non-defective product

P(D|B1) be probability of a defective product produced by machine 1 = 0.3 x 0.01 = 0.003

P(D|B2) be probability of a defective product produced by machine 2 = 0.2 x 0.03 = 0.006

P(D|B3) be probability of a defective product produced by machine 3 = 0.5 x 0.02 = 0.010

Likewise,

P(N|B1) be probability of a non-defective product produced by machine 1 = 1 - P(D|B1) = 1 - 0.003 = 0.997

P(N|B2) be probability of a non-defective product produced by machine 2  = 1 - P(D|B2) = 1 - 0.006 = 0.994

P(N|B3) be probability of a non-defective product produced by machine 3 = 1 - P(D|B3) = 1 - 0.010 = 0.990

For the probability of a finished product produced by machine B1 given it's non-defective; represented by P(B1|N)

P(B1|N) =\frac{P(N|B1)P(B1)}{P(N|B1)P(B1) + P(N|B2)P(B2) + (P(N|B3)P(B3)} = \frac{(0.297)(0.3)}{(0.297)(0.3) + (0.994)(0.2) + (0.990)(0.5)} = 0.1138

Hence the probability that a non-defective product is produced by machine B1 is 11.38%.

4 0
3 years ago
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