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iogann1982 [59]
3 years ago
7

A seafood-sales manager collected data on the maximum daily temperature, T, and the daily revenue from salmon sales, R, using sa

les receipts for 30 days selected at random. Using the data, the manager conducted a regression analysis and found the least-squares regression line to be Rˆ=126+2.37T. A hypothesis test was conducted to investigate whether there is a linear relationship between maximum daily temperature and the daily revenue from salmon sales. The standard error for the slope of the regression line is SEb1=0.65. Assuming the conditions for inference have been met, which of the following is closest to the value of the test statistic for the hypothesis test?
a. t=0.274

b. t=0.65

c. t=1.54

d. t=3.65

e. t=193.85
Mathematics
1 answer:
konstantin123 [22]3 years ago
4 0

Answer: d

Step-by-step explanation:

t= b1- B0/ SE(b1) = 2.37-0/0.65= 3.646 -> 3.65

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The given matrix is the augmented matrix for a linear system. Use technology to perform the row operations needed to transform t
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Answer:

x_{1} = \frac{176}{127} + \frac{71}{127}x_{4}\\\\ x_{2} = \frac{284}{127} + \frac{131}{254}x_{4}\\\\x_{3} = \frac{845}{127} + \frac{663}{254}x_{4}\\

Step-by-step explanation:

As the given Augmented matrix is

\left[\begin{array}{ccccc}9&-2&0&-4&:8\\0&7&-1&-1&:9\\8&12&-6&5&:-2\end{array}\right]

Step 1 :

r_{1}↔r_{1} - r_{2}

\left[\begin{array}{ccccc}1&-14&6&-9&:10\\0&7&-1&-1&:9\\8&12&-6&5&:-2\end{array}\right]

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r_{3}↔r_{3} - 8r_{1}

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r_{2}↔\frac{r_{2}}{7}

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Step 5 :

r_{3}↔\frac{r_{3}. 7}{254}

\left[\begin{array}{ccccc}1&0&4&-11&:-8\\0&1&-\frac{1}{7} &-\frac{1}{7} &:\frac{9}{7} \\0&0&1&-\frac{663}{254} &:-\frac{1690}{254} \end{array}\right]

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