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koban [17]
3 years ago
10

Terry runs a snowplowing business. Income from snowplowing is given by the function f(x) = 35.5x + 6, where f(x) is the income i

n dollars and x is the snowfall in inches received during a winter. If during the years 2006 to 2011, Terry’s town received 53, 42, 55, 63, 58, and 47 inches of snowfall, what was his income (in dollars) during those years? {1,444.4, 1,598.5, 1,825.5, 1,954.5, 2,043, 2,112.5} {1,335.5, 1,742, 1,777.5, 1,835, 2,104.5, 2,224.5} {1,497, 1,674.5, 1,887.5, 1,958.5, 2,065, 2,242.5} {1,254, 1,749.5, 1,837.5, 1,897.5, 2,110.4, 2,310.5} {,1353.5, 1,652.5, 1,753.5, 1,857, 2,034, 2,256.5}
Mathematics
1 answer:
eduard3 years ago
5 0
It's C they are just out of order for the years.

f(x)=35.5x+6
For each year your plug in the snow total for x and solve
2006 x=53 so 35.5(53)+6=1887.5
2007 x=42 so 35.5(42)+6=1497
2008 x=55 so 35.5(55)+6=1958.5
2009 x=63 so 35.5(63)+6=2242.5
2010 x=58 so 35.5(58)+6=2065
2011 x=47 so 35.5(47)+6=1674.5

Hope that helps.
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A random sample of 16 one-kilogram sugar packets is obtained and the actual weights of the packets are measured. The sample mean
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The 99% two-sided confidence interval for the average sugar packet weight is between 0.882 kg and 1.224 kg.

Step-by-step explanation:

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The first step to solve this problem is finding how many degrees of freedom, we have. This is the sample size subtracted by 1. So

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Now, we have to find a value of T, which is found looking at the t table, with 35 degrees of freedom(y-axis) and a confidence level of 1 - \frac{1 - 0.99}{2} = 0.905([tex]t_{995}). So we have T = 2.9467

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The upper end of the interval is the sample mean added to M. So it is 1.053 + 0.171 = 1.224 kg.

The 99% two-sided confidence interval for the average sugar packet weight is between 0.882 kg and 1.224 kg.

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