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xz_007 [3.2K]
4 years ago
8

At a college, 7 out of every 10 students worked either a full-time or part-time job in addition to their studies. If 4,900 stude

nts were enrolled at the college, how many students did not have a full-time or part-time job?
Mathematics
1 answer:
user100 [1]4 years ago
3 0

Remark

If 7/10 do have full or part time jobs then 3/10 do not.

That comes from 1 which represents all the students subtract 7/10 to get 3/10

Solution

Those without jobs = 4900 * 3/10 = 4900 *0.3 =  1470 that do not have any kind of job.

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Step-by-step explanation:

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Which ordered pair is a solution to the system of linear equations 1/2x-3/4y=11/60 and 2/5x+1/6y=3/10
natka813 [3]

ANSWER

( \frac{2}{3} , \frac{1}{5} )

EXPLANATION

The first equation is

\frac{1}{2} x -  \frac{3}{4} y =  \frac{11}{60} ...(1)

The second equation is

\frac{2}{5} x  +  \frac{1}{6} y =  \frac{3}{10} ...(2)

We want to eliminate y, so we multiply the first equation by

\frac{4}{5}

\frac{4}{5}  \times \frac{1}{2} x - \frac{4}{5}    \times \frac{3}{4} y =  \frac{11}{60}  \times  \frac{4}{5}

\frac{2}{5} x - \frac{3}{5} y =  \frac{11}{75} ...(3)

We now subtract equation (3) from (2)

(\frac{2}{3} x  -  \frac{2}{3} x )+ ( \frac{1}{6} y -  -  \frac{3}{5}y ) =(  \frac{3}{10}  -  \frac{11}{75} )

\frac{1}{6} y  +  \frac{3}{5}y  =\frac{3}{10}  -  \frac{11}{75}

\frac{23}{30}y =  \frac{23}{150}

Multiply both sides by

\frac{30}{23}

\implies \:  \frac{30}{23} \times  \frac{23}{30}y=  \frac{23}{150}  \times  \frac{30}{23}

\implies \: y =  \frac{1}{5}

Substitute into the first equation to solve for x .

\frac{1}{2} x -  \frac{3}{4}  \times \frac{1}{5} =  \frac{11}{60}

Multiply to obtain

\frac{1}{2} x -  \frac{3}{20} =  \frac{11}{60}

\frac{1}{2} x = \frac{11}{60} + \frac{3}{20}

\frac{1}{2} x = \frac{1}{3}

Multiply both sides by 2.

2 \times \frac{1}{2} x =2 \times  \frac{1}{3}

x = \frac{2}{3}

The solution is

( \frac{2}{3} , \frac{1}{5} )

5 0
3 years ago
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