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Ad libitum [116K]
3 years ago
14

Write a two-column proof.

Mathematics
2 answers:
irga5000 [103]3 years ago
8 0

Proof:-

In ∆XYZ and ∆VWZ

\because\sf\begin{cases}\sf XZ\cong VZ(Given)\\ \sf YZ\cong WZ(Given)\\ \sf

Hence

∆XYZ\cong∆VWZ(Side-Angle-Side)

ira [324]3 years ago
8 0
<h3>Given:-</h3>

\begin{gathered}\\ \sf\longmapsto XZ≅VZ\end{gathered}

\begin{gathered}\\ \sf\longmapsto YZ≅WZ\end{gathered}

<h3>To prove:-</h3>

⇒△ XYZ≅△ VWZ

<h3>Proof:-</h3>

In △ XYZ&△ VWZ we have,

  • XZ=VZ [Given]
  • ∠XZY=∠VZW [Vertically opposite angles]
  • YZ=WZ [Given]

△ XYZ≅△ VWZ

Hence, Proved by side-angle-side theorem.

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In an election, the population consists of the people who voted. Although there is overall data on how the population voted, the
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Answer:

Answer:

P(A) = 0.39

Step-by-step explanation:

We are given;

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P(W|A^c ) = 0.3

We are told that 60% of the respondents said they voted for A. Thus;

P(A|W) = 60% = 0.6

Now, using the principle of drawing lots, we can be able to find the probability of the event that they are willing to participate in the exit poll which is P(W).

Thus;

P(W) = [P(W|A) × P(A)] +[P(W∣A^c) × P(A^c)]

Now, P(A^c) can be expressed as 1 - P(A)

Thus, we now have;

P(W) = [P(W|A) × P(A)] + [P(W∣A^c) × (1 - P(A)]

Plugging in the relevant values gives;

P(W) = 0.7P(A) + 0.3(1 - P(A))

P(W) = 0.7P(A) + 0.3 - 0.3P(A)

P(W) = 0.3 + 0.4P(A)

Now,using Baye's theorem, we can find an expression for P(A|W)

Thus;

P(A|W) = [P(A ∩ W)]/P(W)

This can be further expressed as;

P(A|W) = [P(A) × P(W|A)]/P(W)

Plugging in relevant values, we have;

0.6 = 0.7P(A)/(0.3 + 0.4P(A))

Cross multiply to get;

0.6(0.3 + 0.4P(A)) = 0.7P(A)

0.18 + 0.24P(A) = 0.7P(A)

0.18 = 0.7P(A) - 0.24P(A)

0.46P(A) = 0.18

P(A) = 0.18/0.46

P(A) = 0.39

Step-by-step explanation:

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4 0
3 years ago
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motikmotik

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