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Degger [83]
2 years ago
7

HELP PLS!!!!!!!! ASAP

Mathematics
1 answer:
Naddik [55]2 years ago
7 0

ΔAOP ≅ ΔBOQ by AAS theorem. Therefore, AO ≅ BO and PO ≅ OQ by CPCTC.

<h3>What is the Angle-Angle-Side Congruence Theorem (AAS)?</h3>

The angle-angle-side congruence theorem states that if two consecutive angles and a non-included side of a triangle are congruent to two consecutive angles and a corresponding non-included side of another triangle, both triangles are considered congruent to each other.

<h3>What is the CPCTC Theorem?</h3>

The CPCTC theorem says that if two triangles are proven to be congruent triangles, then all their corresponding parts would also be congruent to each other.

Thus:

Angle A ≅ Angle B [given]

AP ≅ BQ [given]

Angle AOP ≅ angle BOQ [vertical angles theorem]

ΔAOP ≅ ΔBOQ [AAS theorem]

AO ≅ BO and PO ≅ OQ [CPCTC]

Learn more about the AAS theorem on:

brainly.com/question/4460411

#SPJ1

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Brian &amp; Paul share a lottery win of £7800 in the ratio 1 : 4. Brian then shares his part between himself, his wife &amp; the
Elza [17]

Answer:

The wife gets £468 over the son

Step-by-step explanation:

Firstly, we shall calculate the total amount Brian made from from what he and Paul won.

The ratio is 1:4, meaning for every 2 part Brain takes, Paul takes 4

Paul’s share is thus 1/5 * 7800 = 7800/5 = £1560

Now we have another ratio for sharing the amount Brain has at home.

The total ratio here is 1+6+3 = 10

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3 0
4 years ago
Compute: 5 ∙ (2 + 3i)
dimulka [17.4K]

Answer:

It will be equal to 40

Step-by-step explanation:

We have to compute 5\times (2+3!)

Let first find 3 !

For this we have to use factorial concept

So 3! will be equal to 3! = 3×2×1 = 6

Now According to 6+2 = 8

And now after solving bracket we have to multiply with 5

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4 0
4 years ago
A given population proportion is .25. What is the probability of getting each of the following sample proportions
anyanavicka [17]

This question is incomplete, the complete question is;

A given population proportion is .25. What is the probability of getting each of the following sample proportions

a) n = 110 and = p^ ≤ 0.21, prob = ?

b) n = 33 and p^ > 0.24, prob = ?

Round all z values to 2 decimal places. Round all intermediate calculation and answers to 4 decimal places.)

Answer:

a) the probability of getting the sample proportion is 0.1660

b) the probability of getting the sample proportion is 0.5517

Step-by-step explanation:

Given the data in the questions

a)

population proportion = 0.25

q = 1 - p = 1 - 0.25 = 0.75

sample size n = 110

mean = μ = 0.25

S.D = √( p( 1 - p) / n ) = √(0.25( 1 - 0.25) / 110 ) √( 0.1875 / 110 ) = 0.0413

Now, P( p^ ≤ 0.21 )

= P[ (( p^ - μ ) /S.D) < (( 0.21 - μ ) / S.D)

= P[ Z < ( 0.21 - 0.25 ) / 0.0413)

= P[ Z < -0.04 / 0.0413]

= P[ Z < -0.97 ]

from z-score table

P( X ≤ 0.21 ) = 0.1660

Therefore, the probability of getting the sample proportion is 0.1660

b)

population proportion = 0.25

q = 1 - p = 1 - 0.25 = 0.75

sample size n = 33

mean = μ = 0.25

S.D = √( p( 1 - p) / n ) = √(0.25( 1 - 0.25) / 33 ) = √( 0.1875 / 33 ) = 0.0754

Now, P( p^ > 0.24 )  

= P[ (( p^ - μ ) /S.D) > (( 0.24 - μ ) / S.D)

= P[ Z > ( 0.24 - 0.25 ) / 0.0754 )

= P[ Z > -0.01 / 0.0754  ]

= P[ Z > -0.13 ]

= 1 - P[ Z < -0.13 ]

from z-score table

{P[ Z < -0.13 ] = 0.4483}

1 - 0.4483

P( p^ > 0.24 )  = 0.5517

Therefore, the probability of getting the sample proportion is 0.5517

6 0
3 years ago
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