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IRISSAK [1]
3 years ago
11

Given △ACP ≅ △LNX, find each missing measure.

Mathematics
1 answer:
shtirl [24]3 years ago
5 0

9514 1404 393

Answer:

  XL = 13 cm, m∠L = 29°

  AC = 21 cm, m∠C = 29°

  PC = 13 cm, m∠Y = ??  (there is no angle Y in the figure)

Step-by-step explanation:

The figure on the right shows the triangles are isosceles, with XL = XN. That means also PA = PC. Those side lengths are all the same: 13 cm.

  PA = PC = XL = XN = 13 cm

In the same fashion, ...

  LN = AC = 21 cm

__

Of course, the angles have a similar relationship:

  ∠A = ∠C = ∠L = ∠N = 29°

The two base angles in the triangle add to 58°, so the third angle is ...

  180° -58° = 122°

  ∠P =∠X = 122°

_____

<em>Additional comment</em>

As is often the case in over-specified problems, the numbers shown are not consistent The attached figures show the length LN is actually incorrect for the angle and side specified for triangle ACP, or the angles are incorrect for the side lengths specified. What this means is that the answers you get will depend on how you work the problem. Above, we have taken the given numbers and relationships at face value, even though we know they are incorrect.

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Answer:

Step-by-step explanation:

For a fair die, there are six likely options; 1, 2, 3, 4, 5, and 6

the probability of a even number is 3/6 = 0.5

Since the results of the die roll is independent and each trial is mutually exclusive, the distribution to explain the probability of occurrence will follow a binomial distribution such that n is the number of trials

x = number of successful throws

therefore for a Binomial distribution where

P(X =x) = nCx . P^x . (1-P)^ (n-x)

since p = 0.5, and n = 12, the distribution follows

P(X = x) = 12Cx . 0.5^x . (1 - 0.5)^(12- x)

= 12Cx . 0.5^x . 0.5)^(12- x)

where x = (0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12)

since we are interested in the probability of the number of times an even number occurs

it can occur either as P(X = 0), P(X =1), P(X =2), P(X =3), P(X =4), P(X =5), P(X =6), P(X =7), P(X =8), P(X =9), P(X =10), P(X =11), and P(X =12)

For no even number in 12 rolls,

P(X = 0) = 12C0 . 0.5^0 . 0.5^(12- 0) = 0.000244

For one even number in 12 rolls,

P(X = 1) = 12C1 . 0.5^1 . 0.5^(12- 1) = 0.002930

For two even number in 12 rolls,

P(X = 2) = 12C2 . 0.5^2 . 0.5^(12- 2) = 0.016113  

For three even number in 12 rolls,

P(X = 3) = 12C3 . 0.5^3 . 0.5^(12- 3) = 0.053711  

For four even number in 12 rolls,

P(X = 4) = 12C4 . 0.5^4 . 0.5^(12- 4) = 0.120850

For five even number in 12 rolls,

P(X = 5) = 12C5 . 0.5^5 . 0.5^(12- 5) = 0.193359

For six even number in 12 rolls,

P(X = 6) = 12C6 . 0.5^6 . 0.5^(12- 6) = 0.225586

For seven even number in 12 rolls,

P(X = 7) = 12C7 . 0.5^7 . 0.5^(12- 7) = 0.193359

For eight even number in 12 rolls,

P(X = 8) = 12C8 . 0.5^8 . 0.5^(12- 8) = 0.120850

For nine even number in 12 rolls,

P(X = 9) = 12C9 . 0.5^9 . 0.5^(12- 9) = 0.053711

For ten even number in 12 rolls,

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For twelve even number in 12 rolls,

P(X = 12) = 12C12 . 0.5^12 . 0.5^(12- 12) = 0.000244

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the probability value stands

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