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3 years ago

♥||Please help as soon as you can!||♥

Mathematics

2 answers:

nika2105 [10]3 years ago

6 0

$60 to $45 would be 25% off

You would do 60 x 25 = 1,500

Remove the two zeros which would be 15

60 - 15 = 45

ankoles [38]3 years ago

3 0

Answer:

the price was reduced by 25%

Step-by-step explanation:

60 - 45 = 15

15 is what percent of 60?

15/60 = .25 x 100 = 25%

also can notice that 15 is 1/4 of 60, so it is equal to 25%.

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

Tresset [83]

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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