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beks73 [17]
3 years ago
10

Elliot has been running a lawn care business since 2000. He cuts grass, trims, and weed whacks yards for his customers throughou

t the season. Each year, he has increased his fee by the same amount. The table shows what Elliot charged each customer for two given years of his business:
Year Lawn Care Fee
2000 $750
2010 $1350


A. What is the rate of change and initial value for Elliot’s business? How do you know?
B. Write an equation in slope-intercept form to represent the fees that Elliot charges each year.
Mathematics
1 answer:
pashok25 [27]3 years ago
8 0
A.) Rate of change = (1350 - 750) / (2010 - 2000) = 600/10 = $60
The initial value is the amount he charged in his first year of business which is $750.

B.) Required equation is given by y - 750 = 60x
y = 60x + 750 where x represent the number of years after 2000.
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M_d=M_1-M_2=2577.75-3809.42=-1231.67

The estimated standard error of the difference between means is computed using the formula:

s_{M_d}=\sqrt{\dfrac{\sigma_1^2}{n_1}+\dfrac{\sigma_2^2}{n_2}}=\sqrt{\dfrac{1916.29^2}{67}+\dfrac{2379.47^2}{33}}\\\\\\s_{M_d}=\sqrt{54808.468+171572.045}=\sqrt{226380.513}=475.795

The t-value for a 95% confidence interval is t=1.96.

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MOE=t\cdot s_M=1.96 \cdot 475.795=932.54

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LL=M_d-t \cdot s_{M_d} = -1231.67-932.54=-2164.21\\\\UL=M_d+t \cdot s_{M_d} = -1231.67+932.54=-299.13

The 95% confidence interval for the difference between means is (-2164.21, -299.13).

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