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Sagot :
The store should charge a price between ($0, $0.94] to sell at least 625 pads each week. The store should charge a price between ($0, $1.75) to sell more than 300 pads each week. The store should charge a price between ($2.00, $2.50) to sell less than 200 pads each week. The store should charge a price between [$1.31, $2.50) to sell at most 475 pads each week.
a)
In order to solve part a of the problem, we need to set our equation as an inequality. Part a asks us to find the price the store should charge to sell at least 625 pads each week. The word at least gives us a clue on what symbol to use. In this case we should use the [tex]\leq[/tex] symbol, like this:
[tex]625\leq 1000-400p[/tex]
and now we can solve the inequality, we can start by switching places between the 625 and the 400p so we get:
[tex]400p\leq 1000-625[/tex]
[tex]400p\leq 375[/tex]
and we divide both sides into 400, so we get:
[tex]p\leq \frac{375}{400}[/tex]
[tex]p\leq $0.94[/tex]
So this can be rewritten in interval notation like this:
($0, $0.94] The price they should charge can be located between those two values.
b)
In order to solve part b of the problem, we need to set our equation as an inequality. Part b asks us to find the price the store should charge to sell at more than 300 pads each week. The word more than gives us a clue on what symbol to use. In this case we should use the < symbol, like this:
[tex]300< 1000-400p[/tex]
and now we can solve the equation, we can start by switching places between the 300 and the 400p so we get:
[tex]400p< 1000-300[/tex]
[tex]400p< 700[/tex]
and we divide both sides into 400, so we get:
[tex]p< \frac{700}{400}[/tex]
[tex]p< $1.75[/tex]
So this can be rewritten in interval notation like this:
($0, $1.75) The price they should charge can be located between those two values.
c)
In order to solve part c of the problem, we need to set our equation as an inequality. Part c asks us to find the price the store should charge to sell less than 200 pads each week. The word less than gives us a clue on what symbol to use. In this case we should use the > symbol, like this:
[tex]200> 1000-400p[/tex]
and now we can solve the inequality, we can start by switching places between the 200 and the 400p so we get:
[tex]400p> 1000-200[/tex]
[tex]400p> 800[/tex]
and we divide both sides into 400, so we get:
[tex]p> \frac{800}{400}[/tex]
[tex]p> $2.00[/tex]
at this point we need to know what the maximum price the store can charge is. We can do so by setting the equation equal to zero and solve for p, so we get:
0<1000-400p
so we can now move the 400p to the other side of the equation, so we get:
400p<1000
and divide both sides into 400 so we get:
[tex]p<\frac{1000}{400}[/tex]
which simplifies to:
p<$2.50
if the store charges more than $2.50 they will not sell any item at all.
So this can be rewritten in interval notation like this:
($2.00, $2.50) The price they should charge can be located between those two values to sell less than 200 pads each week.
d)
In order to solve part d of the problem, we need to set our equation as an inequality. Part d asks us to find the price the store should charge to sell at most 475 pads each week. The word at most gives us a clue on what symbol to use. In this case we should use the [tex]\geq[/tex] symbol, like this:
[tex]475\geq 1000-400p[/tex]
and now we can solve the inequality, we can start by switching places between the 475 and the 400p so we get:
[tex]400p\geq 1000-475[/tex]
[tex]400p\geq 525[/tex]
and we divide both sides into 400, so we get:
[tex]p\geq \frac{525}{400}[/tex]
[tex]p\geq $1.31[/tex]
So this can be rewritten in interval notation like this:
[$1.31, $2.50) The price they should charge can be located between those two values to sell 475 pads at most each week.
You can check the following link for further information.
https://brainly.com/question/13728818?referrer=searchResults
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