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laila [671]
3 years ago
8

A clothing business finds there is a linear relationship between the number of shirts, n, it can sell and the price, p, it can c

harge per shirt. In particular, historical data shows that 1,000 shirts can be sold at a price of $30, while 3,000 shirts can be sold at a price of $5. Find a linear equation in the form p(n)
Mathematics
1 answer:
gulaghasi [49]3 years ago
3 0

Answer:

p=(-0.0125n) + 42.5

Step-by-step explanation:

Let p= price

n = number of shirts

m = slope of the line (note, the more shirts, the lower the price, so we know it's going to be negative)

b = y intercept

There are two points which are (1000, $30) and (3000, $5)

Our slope m = (p1-p2)/(n1-n2)

Filling in from our points m = (30-5)/(1000-3000)

m = 25/-2000

m = -0.0125

Since we have determined our slope, we can now find our equation

p-p1=m(n-n1)

p-30=(-0.0125)(n-1000)

p-30= (-0.0125n) + 12.5

p=(-0.0125n) + 42.5

Then, we can double check with the other point there:

p=(-0.0125n) + 42.5

5? (-0.0125x 3000) + 42.5

5= 5

Therefore,linear equation in the form p(n) is

p=(-0.0125n) + 42.5

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Find the median, first quartiles, third quartiles, and interquartile range of the set of data. 0, 0, 1, 2, 13, 27, 34, 63
alexgriva [62]

Answer:

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3 years ago
Find a linear approximation of the function f(x)=\sqrt[3]{1+x} at a=0, and use it to approximate the numbers \sqrt[3]{.96} and \
34kurt
f(x)=\sqrt[3]{1+x}=(1+x)^{1/3}\implies f'(x)=\dfrac1{3(1+x)^{2/3}}

The linear approximation to f(x) around x=a is then

L(x)=f(a)+f'(a)(x-a)\approx f(x)

So the approximation centered at a=0 will be

L(x)=f(0)+f'(0)x=(1+0)^{1/3}+\dfrac x{3(1+0)^{2/3}}
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which means we have

\sqrt[3]{0.96}\approx L(0.96)=1+\dfrac{0.96}3=1.32

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Compare to the actual values of

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