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Alchen [17]
2 years ago
5

A worker carries jugs of liquid soap from a production line to a packing area, carrying 4 jugs per trip. If the jugs are packed

into cartons that hold 7 jugs each, how many jugs are needed to fill the last partially filled carton after the worker has made 17 trips?
A. 1
B. 2
C. 4
D. 5
E. 6
Mathematics
1 answer:
Allisa [31]2 years ago
7 0

Answer:

B. 2

Step-by-step explanation:

We have been given that a worker carries jugs of liquid soap from a production line to a packing area, carrying 4 jugs per trip.

Since worker has made 17 trips, so total number of hugs of liquid soap would be 17 times number of jugs carried per trip.

17\times 4=68.

Since each carton holds 7 jugs, so number of cartoons would be \frac{68}{7}.

\frac{68}{7}\Rightarrow \frac{63+5}{7}=9\frac{5}{7}

We can see that there are 9 completely filled cartoons and there are 5 jugs remaining, so we need 7-5=2 jugs.

Therefore, we need 2 jugs to fill the 10th cartoon and option B is the correct choice.

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Since we are given the common ratio (3/2), all we need to find to define the geometric sequence, is its multiplicative factor (a) that corresponds to the first term of the sequence - remember that all consecutive terms will be generated by multiplying this first value repeatedly by the common ratio (3/2) as shown below:  

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Since we are given the information that f(5)=81 we can use this to find the value of the first term:

f(5)=\,a\,*\,(\frac{3}{2} )^4\\81=\,a\,*(\frac{81}{16} )\\a\,*\,81=\,81\.*\,16\\a\,=\,16

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Step-by-step explanation:

We know that the area of parallelogram is given by:-

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