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3 years ago

i have to submit this in an hour plz do this asap,the document is attached plz download it and answer it

Mathematics

1 answer:

insens350 [35]3 years ago

8 0

Answer:

1. a. Q° = 95°

b. a = 20.3; b = 19.3

2. p = 36; q = 14.4

Step-by-step explanation:

Question 1:

Recall: The Exterior Angle Theorem of a Triangle states that the size of the exterior angle of a triangle equals the sum of the two sizes of the two opposite angles of the triangle.

a. Based on the theorem,

Q° = 41° + 54° = 95°

✔️122° = 4a + 2a (exterior angle theorem)

122 = 6a

Divide both sides by 6

122/6 = a

a = 20.3

✔️122° + 3b = 180° (angles on a straight line)

3b = 180 - 122

3b = 58

b = 58/3

b = 19.3

Question 2:

Recall: Opposite angles of a parallelogram are equal while adjacent/consecutive angles are supplementary.

✔️2p + 3p = 180° (adjacent angles)

5p = 180

p = 180/5

p = 36

✔️2p = 5q (opposite angles)

Plug in the value of p

2(36) = 5q

72 = 5q

72/5 = q

q = 14.4

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Your company is considering offering 900 employees the opportunity to transfer to its new headquarters in Ottawa and, as personn

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Answer:

a) $P(100M) = \frac{100C100*1100C800}{1200C900}$

b) $P(1M) = \frac{100C1*1199C899}{1200C900}$

Step-by-step explanation:

Extract information from question:

Manager = M = 100

Factory Workers = FW = 200

Miscellaneous Staff = MS = 900

Total Workers = 1200

Moving Workers = 900

'Combinations' are a way to calculate the total outcomes of an event where order of the outcomes does not matter. To calculate combinations the formula below is used:

$nCr=\frac{n!}{r!(n-r)!}$

where 'n' is the total number of items

'r' is the number of items being chosen at a time

The '!' is the factorial function which is the product of all integers equal to and less than the given integer.

Combinations can be calculated manually using the formula above or by using the nCr function on a scientific calculator.

Part a)

Combinations of all 100 managers being selected

$C = nCr=\frac{n!}{r!(n-r)!}$

$C(100M) = 100C100$

Combinations of 800 of factory workers and miscellaneous staff being selected out of 1100 remaining staff

$C = nCr=\frac{n!}{r!(n-r)!}$

$C(800FW/MS) = 1100C800$

Total possible unspecified combinations of 900 staff being selected out of 1200 total staff

$C = nCr=\frac{n!}{r!(n-r)!}$

$C(900M/FW/MS) = 1200C900$

Probability of all 100 managers being selected:

$P(100M) = \frac{100C100*1100C800}{1200C900}$

Part b)

Combinations of 1 out of 100 managers being selected

$C = nCr=\frac{n!}{r!(n-r)!}$

$C(1M) = 100C1$

Combinations of 899 of managers, factory workers and miscellaneous staff being selected out of 1199 remaining staff

$C = nCr=\frac{n!}{r!(n-r)!}$

$C(899M/FW/MS) = 1199C899$

Total possible unspecified combinations of 900 staff being selected out of 1200 total staff

$C = nCr=\frac{n!}{r!(n-r)!}$

$C(900M/FW/MS) = 1200C900$

Probability of being the 1 manager being selected:

$P(1M) = \frac{100C1*1199C899}{1200C900}$

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3 years ago

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ANEK [815]

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