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saul85 [17]
1 year ago
11

An invoice is dated January 17 with terms of 8/10 , 6/20 , 4/30 , n/40 . find the three final discount dates and the net payment

date. (type a whole number)
Mathematics
1 answer:
IRISSAK [1]1 year ago
6 0

Invoice Date = January 17

• First terms:

8/10 and n/40

This means the final discount can be availed "10" days from Invoice Date (Jan17)

So, it will be:

Jan 17 + 10 days = January 27

Final Discount Date = Jan 27

The net payment date will be "n/40", which is 40 days from invoice date, thus

Net payment date will be:

Jan 17 - Jan 31 = 14 days

Feb 1 - Feb 26 = 26 days

------------------------------------

40 days

Payment Date = Feb 26

• Second terms:

6/20 and n/40

This means the final discount can be availed "20" days from Invoice Date (Jan17)

So, it will be:

Jan 17 + 20 days = Feb 6

Final Discount Date = Feb 6

The net payment date will be "n/40", which is 40 days from invoice date, thus, same as before,

Payment Date = Feb 26

• Third terms:

4/30 and n/40

This means the final discount can be availed "30" days from Invoice Date (Jan17)

So, it will be:

Jan 17 + 20 days = Feb 16

Final Discount Date = Feb 16

The net payment date will be "n/40", which is 40 days from invoice date, thus, same as before,

Payment Date = Feb 26

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Answer:

P(B1) = (11/15)

P(B2) = (4/15)

P(A) = (11/15)

P(B1|A) = (5/7)

P(B2|A) = (2/7)

Step-by-step explanation:

There are 11 red chips and 4 blue chips in a box. Two chips are selected one after the other at random and without replacement from the box.

B1 is the event that the chip removed from the box at the first step of the experiment is red.

B2 is the event that the chip removed from the box at the first step of the experiment is blue. A is the event that the chip selected from the box at the second step of the experiment is red.

Note that the probability of an event is the number of elements in that event divided by the Total number of elements in the sample space.

P(E) = n(E) ÷ n(S)

P(B1) = probability that the first chip selected is a red chip = (11/15)

P(B2) = probability that the first chip selected is a blue chip = (4/15)

P(A) = probability that the second chip selected is a red chip

P(A) = P(B1 n A) + P(B2 n A) (Since events B1 and B2 are mutually exclusive)

P(B1 n A) = (11/15) × (10/14) = (11/21)

P(B2 n A) = (4/15) × (11/14) = (22/105)

P(A) = (11/21) + (22/105) = (77/105) = (11/15)

P(B1|A) = probability that the first chip selected is a red chip given that the second chip selected is a red chip

The conditional probability, P(X|Y) is given mathematically as

P(X|Y) = P(X n Y) ÷ P(Y)

So, P(B1|A) = P(B1 n A) ÷ P(A)

P(B1 n A) = (11/15) × (10/14) = (11/21)

P(A) = (11/15)

P(B1|A) = (11/21) ÷ (11/15) = (15/21) = (5/7)

P(B2|A) = probability that the first chip selected is a blue chip given that the second chip selected is a red chip

P(B2|A) = P(B2 n A) ÷ P(A)

P(B2 n A) = (4/15) × (11/14) = (22/105)

P(A) = (11/15)

P(B2|A) = (22/105) ÷ (11/15) = (2/7)

Hope this Helps!!!

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d = √ (-9-6)2+(-5-7)2

d = √ ((-15)2+(-12)2)

d = √ (225+144)

d = √ 369

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