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Anna007 [38]
3 years ago
15

At a collage bookstore Clara purchased a math textbook and a novel that cost a total of $54 not including tax if the price of th

e math textbook m is $8 more than 3 times the price of the novel n which system of linear equations could be used to determine the price of each book
Mathematics
1 answer:
diamong [38]3 years ago
3 0

Answer:

m+n=54 and m=3n+8 is the system of equations that could be used to determine the price of each book.

Step-by-step explanation:

Given,

Total cost of maths book and novel = $54

Let,

Cost of maths book = m

Cost of novel = n

According to given statement;

m+n=54      Eqn 1

the price of the math textbook, m, is $8 more than 3 times the price of the novel

m = 3n+8     Eqn 2

m+n=54 and m=3n+8 is the system of equations that could be used to determine the price of each book.

Step-by-step explanation:

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Answer:

y = -2x + 8

Step-by-step explanation:

y = mx + b

-8 = -2(8) + b

-8 = -16 + b

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3 years ago
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. A box in a certain supply room contains four 40-W lightbulbs, five 60-W bulbs, and six 75-W bulbs. Suppose that three bulbs ar
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Answer:

a) 59.34%

b) 44.82%

c) 26.37%

d) 4.19%

Step-by-step explanation:

(a)

There are in total <em>4+5+6 = 15 bulbs</em>. If we want to select 3 randomly there are  K ways of doing this, where K is the<em> combination of 15 elements taken 3 at a time </em>

K=\binom{15}{3}=\frac{15!}{3!(15-3)!}=\frac{15!}{3!12!}=\frac{15.14.13}{6}=455

As there are 9 non 75-W bulbs, by the fundamental rule of counting, there are 6*5*9 = 270 ways of selecting 3 bulbs with exactly two 75-W bulbs.

So, the probability of selecting exactly 2 bulbs of 75 W is

\frac{270}{455}=0.5934=59.34\%

(b)

The probability of selecting three 40-W bulbs is

\frac{4*3*2}{455}=0.0527=5.27\%

The probability of selecting three 60-W bulbs is

\frac{5*4*3}{455}=0.1318=13.18\%

The probability of selecting three 75-W bulbs is

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The probability that it is necessary to examine at least six bulbs until a 75-W bulb is found, <em>supposing there is no replacement</em>, is the same as the probability of taking 5 bulbs one after another without replacement and none of them is 75-W.

As there are 15 bulbs and 9 of them are not 75-W, the probability a non 75-W bulb is \frac{9}{15}=0.6

Since there are no replacement, the probability of taking a second non 75-W bulb is now \frac{8}{14}=0.5714

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\frac{9}{15},\frac{8}{14},\frac{7}{13},\frac{6}{12},\frac{5}{11}

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0.6, 0.5714, 0.5384, 0.5, 0.4545

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