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jekas [21]
3 years ago
11

On March 30, Century Television received an invoice dated March 28 from ACME Manufacturing for 51 televisions at a cost of $100

each. Century received a 8/5/5 chain discount. Shipping terms were FOB shipping point. ACME prepaid the $91 freight. Terms were 5/10 EOM. When Century received the goods, 3 sets were defective. Century returned these sets to ACME. On April 08, Century sent a $275 partial payment. Century will pay the balance on May 06. What is Century’s final payment on May 06? Assume no taxes.
Mathematics
1 answer:
vekshin13 years ago
3 0
<span>You are given that on March 30, Century Television received an invoice dated March 28 from ACME Manufacturing for 51 televisions at a cost of $100 each. Century received a 8/5/5 chain discount. Shipping terms were FOB shipping point. ACME prepaid the $91 freight. Terms were 5/10 EOM. When Century received the goods, 3 sets were defective. Century returned these sets to ACME. On April 08, Century sent a $275 partial payment. Century will pay the balance on May 06. Century’s final payment on May 06 is $580</span>
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If you are dealt five cards from a shuffled deck of 52 cards find the probability of getting three queens and two kings
dexar [7]

The probability of getting three queens and two kings is \frac{1}{1082900}

<u>Solution:</u>

Given that , you are dealt five cards from a shuffled deck of 52 cards  

We have to find the probability of getting three queens and two kings  

Now, we know that, in a deck of 52 cards, we will have 4 queens and 4 kings.

\text { probability of an event }=\frac{\text { favarable possibilities }}{\text { number of possibilities }}

<em><u>Probability of first queen:</u></em>

\text { Probability for } 1^{\text {st }} \text { queen }=\frac{4}{52}=\frac{1}{13}

<em><u>Probability of second queen:</u></em>

\text { Plobability for } 2^{\text {nd }} \text { queen }=\frac{3}{51}=\frac{1}{17}

Here we used 3 for favourable outcome, since we already drew 1 queen out of 4

And now number of outcomes = 52 – 1 = 51

<em><u>Probability of third queen:</u></em>

Similarly here favorable outcome = 2, since we already drew 2 queen out of 4

And now number of outcomes = 51 – 1 = 50

\text { Probability of } 3^{\text {rd }} \text { queen }=\frac{2}{50}=\frac{1}{25}

<em><u>Probability for first king:</u></em>

Here kings are 4, but overall cards are 49 as 3 queens are drawn

\text { probability for } 1^{\text {st }} \text { king }=\frac{4}{49}

<em><u>Probability for second king:</u></em>

Here, kings are 3 and overall cards are 48 as 3 queens and 1 king are drawn

\text { probability of } 2^{\text {nd }} \text { king }=\frac{3}{48}=\frac{1}{16}

<em><u>And, finally the overall probability to get 3 queens and 2 kings is:</u></em>

=\frac{1}{13} \times \frac{1}{17} \times \frac{1}{25} \times \frac{4}{49} \times \frac{1}{16}=\frac{4}{4331600}=\frac{1}{1082900}

Hence, the probability is \frac{1}{1082900}

7 0
3 years ago
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