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sattari [20]
3 years ago
8

A bedroom set lists for $7,000 and carries a trade discount of 20 percent. Freight (FOB shipping point) of $120 is not of the li

st price. Calculate the final price (also include cost of freight) of the bedroom set assuming terms of 2/10, n/30 ROG. The date of the invoice was july 10 with the goods being received August 11. The bill was paid on august 20
Mathematics
1 answer:
kirza4 [7]3 years ago
5 0

Answer:

The total price is $5720

Step-by-step explanation:

According to the given statement a bedroom set lists for $7,000 and carries a trade discount of 20 percent. Freight (FOB shipping point) of $120 is not of the list price.

A bedroom lists for = $7000

Discount = 20% = 0.2

Now multiply 7000 by 0.2 = 7000* 0.2 = 1400

It means the discount value was $1400

Now subtract 1400 from 7000 = 7000 - 1400 = $5600

Now to get the total price add 5600 with 120 we get,

$5600+120 = $5720

Thus the total price is $5720 .....

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Answer:

11 mm, 21 mm, and 16 mm.

Step-by-step explanation:

since 11+21>16, 11+ 16>21>11, You can form a triangle with 11,21,16

4 0
2 years ago
If B = {−13, −9, −7, −3}, choose the set A that will make the following statement false.
Svetach [21]

Answer: c.A is the set of rational numbers

Step-by-step explanation:

B ⊆ A means that every element of B is an element of A

B = { -13 , -9 , -7 , - 3 }

The element of B are negative integers , this mean that the element of A must also be integers therefore :

Option a is correct.

Option b is also correct

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clearly , this does not necessarily define A , so option c is the odd one out

5 0
3 years ago
Pens come in packages of 6. Pencils come in
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A. The Least common multiple between 6 and 8 is 24, so you have to buy 4 packages of pens and 3 packages of pencils.

2. Since you bought 24 pens and 24 pencils, you have 48 pens and pencils in all

8 0
2 years ago
Read 2 more answers
A 5-card hand is dealt from a perfectly shuffled deck. Define the events: A: the hand is a four of a kind (all four cards of one
TiliK225 [7]

In a hand of 5 cards, you want 4 of them to be of the same rank, and the fifth can be any of the remaining 48 cards. So if the rank of the 4-of-a-kind is fixed, there are \binom44\binom{48}1=48 possible hands. To account for any choice of rank, we choose 1 of the 13 possible ranks and multiply this count by \binom{13}1=13. So there are 624 possible hands containing a 4-of-a-kind. Hence A occurs with probability

\dfrac{\binom{13}1\binom44\binom{48}1}{\binom{52}5}=\dfrac{624}{2,598,960}\approx0.00024

There are 4 aces in the deck. If exactly 1 occurs in the hand, the remaining 4 cards can be any of the remaining 48 non-ace cards, contributing \binom41\binom{48}4=778,320 possible hands. Exactly 2 aces are drawn in \binom42\binom{48}3=103,776 hands. And so on. This gives a total of

\displaystyle\sum_{a=1}^4\binom4a\binom{48}{5-a}=886,656

possible hands containing at least 1 ace, and hence B occurs with probability

\dfrac{\sum\limits_{a=1}^4\binom4a\binom{48}{5-a}}{\binom{52}5}=\dfrac{18,472}{54,145}\approx0.3412

The product of these probability is approximately 0.000082.

A and B are independent if the probability of both events occurring simultaneously is the same as the above probability, i.e. P(A\cap B)=P(A)P(B). This happens if

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The above "sub-events" are mutually exclusive and share no overlap. There are 48 possible non-aces to choose from, so the first sub-event consists of 48 possible hands. There are 12 non-ace 4-of-a-kinds and 4 choices of ace for the fifth card, so the second sub-event has a total of 12*4 = 48 possible hands. So A\cap B consists of 96 possible hands, which occurs with probability

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3 years ago
A city has 3 new houses for every 9 old houses. If there are 21 new houses in the city, how many old houses are there?
kipiarov [429]
To find your answer you would divide 21 by 3 which would be 7, once you've got 7 you would multiply it by 9 which would give you the amount of old houses that there is which would be 63.
5 0
2 years ago
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