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Darya [45]
3 years ago
5

Luke paid $12.69 for 3 greeting cards. Each card cost the same amount,

Mathematics
2 answers:
german3 years ago
7 0

Answer:

Here, c represents the cost of one greeting cards,

Since, each card costs same,

Thus, the cost of 3 greeting cards = 3 × the cost of one greeting cards

= 3 × c

= 3c

According to the question,

The cost of 3 greeting cards = $ 12.69

⇒ 3c = 12.69

Which is the required equation that represents the given situation,

After solving this,

We get,

c = 4.32,

Thus, the cost of one card is $ 4.32.

Which is the required solution.

n200080 [17]3 years ago
3 0
3c = 12.69
where c = # of greeting cars
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2 years ago
Consider a uniform distribution from aequals4 to bequals29. ​(a) Find the probability that x lies between 7 and 27. ​(b) Find th
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Answer:

a) 80% probability that x lies between 7 and 27.

b) 28% probability that x lies between 6 and 13.

c) 44% probability that x lies between 9 and 20.

d) 28% probability that x lies between 11 and 18.

Step-by-step explanation:

An uniform probability is a case of probability in which each outcome is equally as likely.

For this situation, we have a lower limit of the distribution that we call a and an upper limit that we call b.

The probability that we find a value x between c and d, in which d is larger than c, is given by the following formula.

P(c \leq x \leq d) = \frac{d - c}{b - a}

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So c = 7, d = 27

P(7 \leq x \leq 27) = \frac{27 - 7}{29 - 4} = 0.8

80% probability that x lies between 7 and 27.

​(b) Find the probability that x lies between 6 and 13. ​

So c = 6, d = 13

P(6 \leq x \leq 13) = \frac{13 - 6}{29 - 4} = 0.28

28% probability that x lies between 6 and 13.

(c) Find the probability that x lies between 9 and 20.

​So c = 9, d = 20

P(9 \leq x \leq 20) = \frac{20 - 9}{29 - 4} = 0.44

44% probability that x lies between 9 and 20.

(d) Find the probability that x lies between 11 and 18.

So c = 11, d = 18

P(11 \leq x \leq 18) = \frac{18 - 11}{29 - 4} = 0.28

28% probability that x lies between 11 and 18.

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