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maxonik [38]
3 years ago
13

If m∠XWY = 20°, m∠XWZ = 40°, and XY = 16, what is the value of YZ?

Mathematics
1 answer:
Jobisdone [24]3 years ago
7 0
YZ = 16. The triangles are congruent. This would be AAS so YZ = 16 because of CPCTC.
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Vladimir79 [104]

Answer:

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Step-by-step explanation:

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3 years ago
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Which pair of angles are alternate exterior angles ?
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8 and 1 are alternate exterior angles.

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Tammy has 4 times as many books as Willy . Willy has 180 fewer books than Tammy . How many books does Willy have ?​
enyata [817]

Answer:

60 books

Step-by-step explanation:

let the number of books Willy has be x , then

Tammy has 4x books

Willy has 180 fewer books , that is Tammy has 180 more than Willy , so

4x = x + 180 ( subtract x from both sides )

3x = 180 ( divide both sides by 3 )

x = 60

Thus

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which is 180 more than Willy

7 0
1 year ago
Kellie is given the following information:
Murljashka [212]

Answer:

She used inductive reasoning. (False)

She used the law of detachment.  (True)

Her conclusion is valid.  (True)

The statements can be represented as "if p, then q and if q, then r."  (False)

Her conclusion is true. (True)

Step-by-step explanation:

p = Two lines are perpendicular

q = They intersect at Right angles.

Given: A and B are perpendicular

Conclusion: A and B intersect at right angle.

According to the law of detachment, There are two premises (statements that are accepted as true) and a conclusion. They must follow the pattern as shown below.

Statement 1: If p, then q.

Statement 2: p

Conclusion: q

In our case the pattern is followed. The truth of the premises logically guarantees the truth of the conclusion. So her conclusion is true and valid.

3 0
3 years ago
The probability that a professor arrives on time is 0.8 and the probability that a student arrives on time is 0.6. Assuming thes
saul85 [17]

Answer:

a)0.08  , b)0.4  , C) i)0.84  , ii)0.56

Step-by-step explanation:

Given data

P(A) =  professor arrives on time

P(A) = 0.8

P(B) =  Student aarive on time

P(B) = 0.6

According to the question A & B are Independent  

P(A∩B) = P(A) . P(B)

Therefore  

{A}' & {B}' is also independent

{A}' = 1-0.8 = 0.2

{B}' = 1-0.6 = 0.4

part a)

Probability of both student and the professor are late

P(A'∩B') = P(A') . P(B')  (only for independent cases)

= 0.2 x 0.4

= 0.08

Part b)

The probability that the student is late given that the professor is on time

P(\frac{B'}{A}) = \frac{P(B'\cap A)}{P(A)} = \frac{0.4\times 0.8}{0.8} = 0.4

Part c)

Assume the events are not independent

Given Data

P(\frac{{A}'}{{B}'}) = 0.4

=\frac{P({A}'\cap {B}')}{P({B}')} = 0.4

P({A}'\cap {B}') = 0.4 x P({B}')

= 0.4 x 0.4 = 0.16

P({A}'\cap {B}') = 0.16

i)

The probability that at least one of them is on time

P(A\cup B) = 1- P({A}'\cap {B}')  

=  1 - 0.16 = 0.84

ii)The probability that they are both on time

P(A\cap  B) = 1 - P({A}'\cup {B}') = 1 - [P({A}')+P({B}') - P({A}'\cap {B}')]

= 1 - [0.2+0.4-0.16] = 1-0.44 = 0.56

6 0
3 years ago
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