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gogolik [260]
3 years ago
6

Stephanie started selling earrings online at a handmade shop. She makes a profit of $8.50 on each pair of earrings. If she wants

to make a profit of no less than $765 this month, how many pairs of earrings does she need to sell? Write and solve an inequality to model this solution.
Mathematics
1 answer:
Vesnalui [34]3 years ago
3 0

Answer:

90 pairs

Step-by-step explanation:

We can use clues from the problem in order to set up our inequality.

First, we know that a pair of earrings, p, costs $8.50. So, we can say that 8.5p= total profit.

That's a function we could use, but we need an inequality. Well, it says that Stephanie wants to make no less, or <u>at least</u> $765. This indicates that we need the 'greater than or equal to' symbol. Just replace the equal sign:

8.5p\geq 765

Now we can isolate p to find how many earrings she needs to sell to make a profit of at least $765.

8.5p\geq 765\\\\p\geq \frac{765}{8.5}\\\\p\geq 90

Stephanie needs to sell at least 90 pairs of earrings.

Hope this helps!

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With a little care, you can find the median and the quartiles from the histogram. what are these numbers? you can also find the
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Complete Question

The histogram from this question is shown on the first uploaded image

Answer:

The first quartile is  1st Q = 19^{th} girl (subject ) = 1 \serving\ of\ fruit\ per\ day

The median is   Median  = 38 ^{th} \ girl (subject) =  2 \serving\ of\ fruit\ per\ day

The third quartile  is 3rd Q =  56^{th} \  girl (subject) = 4 \serving\ of\ fruit\ per\ day  

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Step-by-step explanation:

From the question we are told that

   The total  number of girls is n =  74

Generally the median is mathematically represented as

       Median  =  \frac{\frac{n}{2}  +[ \frac{n}{2} + 1]  }{2}

So

          Median  =  \frac{\frac{74}{2}  +[ \frac{74}{2} + 1]  }{2}

=>         Median  =  \frac{37  +38 }{2}

=>         Median  =37.5 \  girl    

From the histogram 37.5 \approx 38^{th} \ girl fall under 2 fruits per day

Generally the first quartile is mathematically represented as

      1st \ Q =  \frac{\frac{n}{4} + [\frac{n}{4} + 1]  }{2}

=>    1st \ Q =  \frac{\frac{74}{4} + [\frac{74}{4} + 1]  }{2}

=>     1st \ Q = 19 \  girl

From the histogram 19^{th}girl fall under 1 fruit per day

 Generally the third quartile is mathematically represented as

     3rd\  Q =  \frac{\frac{n}{2}  + n }{2}

=>    3rd\  Q =  \frac{\frac{74}{2}  + 74 }{2}

=>    3rd\  Q =  55.5

From the histogram 55.5 \approx 56^{th} \ girl fall under 4 fruits per day

Generally from the histogram table the frequency  [number of subjects (girls) ]  and the servings of fruit per day can be represented as

                                                                                                               Total

x(servings per day )        0     1     2     3     4     5    6     7      8    

f (number of subjects )   15    11   15     11    8     5    3     3      3    \sum f  =  74

xf                                       0    11    30   33   32  25   18   21     24    \sum xf =  194

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      \= x = \frac{1}{ \sum f}  * [\sum xf]

=>     \= x = \frac{1}{ 74}  *194

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