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Brrunno [24]
3 years ago
7

Two customers went to a post office to buy postcards and large envelopes. Each postcard

Mathematics
1 answer:
astraxan [27]3 years ago
4 0

The right answer is Option A.

Step-by-step explanation:

Let,

x be the postcards

y be the large envelops

According to given statement;

14x+5y=12   Eqn 1

10x+15y=24.80   Eqn 2

Multiplying Eqn 1 by 3;

3(14x+5y=12)\\42x+15y=36\ \ \ Eqn \ 3

Subtracting Eqn 2 from Eqn 3;

(42x+15y)-(10x+15y)=36-24.80\\42x+15y-10x-15y=11.20\\32x=11.20\\

Dividing both sides by 32

\frac{32x}{32}=\frac{11.20}{32}\\x=0.35

Putting x=0.35 in Eqn 1;

14(0.35)+5y=12\\4.90+5y=12\\5y=12-4.90\\5y=7.10

Dividing both sides by 5

\frac{5y}{5}=\frac{7.10}{5}\\y=1.42

Therefore, one large envelope costs $1.42

The right answer is Option A.

Keywords: linear equations, subtraction

Learn more about linear equations at:

  • brainly.com/question/9738996
  • brainly.com/question/10015690

#LearnwithBrainly

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Doubles mean you have to roll the same number simultaneously so let’s say we want to calculate the probability for double ones: then it’s 1/6 on the first dice for a one, and 1/6 on the second dice to land on a one as well.

I personally like to imagine a box like this:
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If you have one dice then it’s just a random segment on one of the lines. If you want the specific result from two dice then you want two specific segments which is also the 1 specific tile out of 36 (6 width times 6 height). So you multiply.

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And 1/36 chance to roll double twos, threes, fours, fives, and sixes. But we don’t count the double fours because any four will do. So:

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So for the probability of either doubles or containing a four is the probability of doubles of either number plus the probability of either dice being a four:

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