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Amiraneli [1.4K]
2 years ago
10

Mr. Smith had 33 dozen cans. He sold 340 of them at 15$ each. He sold the remaining cans at a discount of 20% each. How much mon

ey did he collect from selling the cans.
Mathematics
1 answer:
iren2701 [21]2 years ago
3 0

Answer:

Mr. Smith collects $5748 from selling the can.

Step-by-step explanation:

Given total number of cans = 33\times12=396

Selling price of one can = $15.

\therefore \textrm{Selling price of 340 cans} = \textrm {selling price of one can} \times \textrm {number of cans}=15\times340=\$5100

Now remaining number of cans = total number of cans - sold number of cans =396-340=54

For remaining number of cans he gives a discount of 20% each,

∴ Selling price of one can = 15-\frac{15\times20}{100}=15-3=12

∴ Selling price of 54 cans = \textrm{selling price of one can} \times \textrm{number of cans}=12\times54=\$648

So total money he earns =  \$5100+\$648= \$5748.

Thus Mr. Smith earns $5748 by selling the cans.

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By definition, the arc length is given by:
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 arc = (π / 5) * (2.8)
 Rewriting we have:
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8 0
2 years ago
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eimsori [14]
Use the FOIL method (First, Outside, Inside, Last)

6r(-8r) = -48r²
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-48r² - 18r + 8r + 3

Combine like terms:


-48r² - 18r + 8r + 3

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2 years ago
Problem 5 (4+4+4=12) We roll two fair 6-sided dice. Each one of the 36 possible outcomes is assumed to be equally likely. 1) Fin
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Answer:

1

p(b) =  \frac{1}{6}

2

p(k) =  \frac{1}{3}

3

P(a) =  \frac{1}{3}

Step-by-step explanation:

Generally when two fair 6-sided dice is rolled the doubles are

(1 1) , ( 2 2) , (3 3) , (4 4) , ( 5 5 ), (6 6)

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The total outcome of the rolling the two fair 6-sided dice is

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Generally the probability that doubles (i.e., having an equal number on the two dice) were rolled is mathematically evaluated as

p(b) =  \frac{N}{n}

p(b) =  \frac{6}{36}

p(b) =  \frac{1}{6}

Generally when two fair 6-sided dice is rolled the outcome whose sum is 4 or less is

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Looking at this outcome we see that there are two doubles present

So

The conditional probability that doubles were rolled is mathematically represented as

p(k) =  \frac{2}{6}

p(k) =  \frac{1}{3}

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The conditional probability that at least one die is a 1 is mathematically represented as

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2 years ago
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We will use the addition rule of probability of two events to solve the question. According to the rule given two events A and B;

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Required

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