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aksik [14]
3 years ago
5

Maradee works at Cox's Jewelers for $7.50 per hour with time-and-a-half for all hours over 40 in a week. Thus, her weekly salary

is given by
S(h) = 11.25h + 300
where h is the number of overtime hours and S(h) is her weekly salary. (See Example 5.)
(a) Find S(8.5). (Round your answer to two decimal places.)
$

(b) One week, Maradee worked 47 hours. Find her salary for that week.
$

(c) Maradee's Valentine week salary was $398.61. How many hours overtime did she work? (Round your answer to one decimal place.)
hr

Mathematics
1 answer:
BigorU [14]3 years ago
3 0

Answer:

a) 395.725

b) 378.75

c) 8.8

Step-by-step explanation:

a) S(8.5) = 11.25(8.5) + 300 = 395.625

b) if she worked 47 hrs, that means there were 7 overtime, h=7 and we have to find S(7).

S(7) = 11.25(7) + 300 = 378.75

c) her salary was 398.61, that means S(h) = 398.61 and we have to find h.

398.61 = 11.25(h) + 300

98.61 = 11.25(h)

h = 98.61 / 11.25

h= 8.8

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Answer and explanation:

To find : List the sample space and tell whether you think the events are equally likely ?

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a) Toss 2 coins; record the order of heads and tails.

Let H is getting head and t is getting tail.

When two coins are tossed the sample space is {HH,HT,TH,TT}.

Total number of outcome = 4

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Let B denote boy and G denote girl.

If there are 3 children then the sample space is

{GGG,GGB,GBG,BGG,BBG,GBB,BGB,BBB}

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Number of boys      Favorable outcome    Probability

           0                      GGG                        \frac{1}{8}

           1                    GGB,GBG,BGG          \frac{3}{8}

           2                   GBB,BGB,BBG           \frac{3}{8}

           3                       BBB                         \frac{1}{8}

Since the probabilities are not equal the events are not equally likely.

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Getting a head in a trial is dependent on the previous toss.

Similarly getting 3 consecutive tails also dependent on previous toss.

Hence, the probabilities cannot be equal and events cannot be equally likely.

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

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

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

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

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           2                                  3                    \frac{3}{36}

           3                                  5                    \frac{5}{36}

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           5                                  9                    \frac{9}{36}

           6                                  11                    \frac{11}{36}

Since the probabilities are not the same the events are not equally likely.

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