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kirill115 [55]
3 years ago
13

Write a paragraph proof. Given: ∠T and ∠V are right angles. Prove: ∆TUW ∆VWU

Mathematics
2 answers:
Veseljchak [2.6K]3 years ago
8 0

Step-by-step explanation:

in ∆TUW & ∆VWU

∠UTW = ∠UVW (data / given)

WU = WU (common sides) ∠TWU = ∠VUW (TW//UV alternate∠ )

so ∆TUW & ∆VWU are congruent

Oduvanchick [21]3 years ago
5 0

Answer:

∆TUW≅∆VWU by AAS.

Step-by-step explanation:

Given information: ∠T and ∠V are right angles, TW║UV.

Prove: ∆TUW≅∆VWU

Proof:

If a transversal line intersect two parallel lines, then alternate interior angles are congruent.

In ∆TUW and ∆VWU,

\angle T\cong \angle V                 (Right angles)

\angle TWU\cong \angle VUW                 (Alternate interior angles)

UW\cong UW                     (Reflection property)

By AAS postulate, ∆TUW and ∆VWU are congruent.

\triangle TUW\cong \triangle VWU

Hence proved.

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The market and Stock J have the following probability distributions:
denis-greek [22]

Answer:

1) E(M) = 14*0.3 + 10*0.4 + 19*0.3 = 13.9 \%

2) E(J)= 22*0.3 + 4*0.4 + 12*0.3 = 11.8 \%

3) E(M^2) = 14^2*0.3 + 10^2*0.4 + 19^2*0.3 = 207.1

And the variance would be given by:

Var (M)= E(M^2) -[E(M)]^2 = 207.1 -(13.9^2)= 13.89

And the deviation would be:

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And the variance would be given by:

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And the deviation would be:

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Step-by-step explanation:

For this case we have the following distributions given:

Probability  M   J

0.3           14%  22%

0.4           10%    4%

0.3           19%    12%

Part 1

The expected value is given by this formula:

E(X)=\sum_{i=1}^n X_i P(X_i)

And replacing we got:

E(M) = 14*0.3 + 10*0.4 + 19*0.3 = 13.9 \%

Part 2

E(J)= 22*0.3 + 4*0.4 + 12*0.3 = 11.8 \%

Part 3

We can calculate the second moment first with the following formula:

E(M^2) = 14^2*0.3 + 10^2*0.4 + 19^2*0.3 = 207.1

And the variance would be given by:

Var (M)= E(M^2) -[E(M)]^2 = 207.1 -(13.9^2)= 13.89

And the deviation would be:

Sd(M) = \sqrt{13.89}= 3.73

Part 4

We can calculate the second moment first with the following formula:

E(J^2) = 22^2*0.3 + 4^2*0.4 + 12^2*0.3 =194.8

And the variance would be given by:

Var (J)= E(J^2) -[E(J)]^2 = 194.8 -(11.8^2)= 55.56

And the deviation would be:

Sd(M) = \sqrt{55.56}= 7.45

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