The table represents the average daily price of a two-bedroom beachfront condo each month, with January represented as month 1,
February as month 2, and so on. Month (x) Daily Rental Price (y) 1 $154 2 $205 3 $266 4 $358 5 $403 6 $425 7 $437 8 $430 9 $381 10 $285 11 $211 12 $195 Use the graphing tool to determine the curve of best fit for this data. Write the equation of the curve in the space below.
1 answer:
Answer:
y = - 9.1768x2 + 122.2567x + 14.9091
Step-by-step explanation:
Given the following :
Month (x) Daily Rental Price (y) 1 $154 2 $205 3 $266 4 $358 5 $403 6 $425 7 $437 8 $430 9 $381 10 $285 11 $211 12 $195
Using the online regression equation graphing tool ; The quadratic model obtained in the form,
y = Ax^2 + Bx + C is :
y = - 9.1768x2 + 122.2567x + 14.9091
Attached below is a picture of the quadratic regression curve.
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Answer:
We conclude that the calibration point is set too high.
Step-by-step explanation:
We are given the following in the question:
Population mean, μ = 1000 grams
Sample mean, = 1001.1 grams
Sample size, n = 50
Alpha, α = 0.05
Population standard deviation, σ = 2.8 grams
First, we design the null and the alternate hypothesis
We use One-tailed(right) z test to perform this hypothesis.
Formula:
Putting all the values, we have
Now,
Since,
We reject the null hypothesis and accept the alternate hypothesis. We accept the alternate hypothesis. We conclude that the calibration point is set too high.
Answer:
3x - y -6 = 0
Step-by-step explanation:
We need to find the Equation of the line parallel to the given equation of line . The given equation of the line is ,
<u>Slope </u><u>Intercept</u><u> Form</u><u> </u><u>:</u><u>-</u><u> </u>
where ,
- m is slope
- c is y intercept .
Therefore , the Slope of the line is 3 . Let the parallel line passes through ( 3,3) . We know that the parallel lines have same slope . Therefore the slope of the parallel line will also be 3 .
<u>Using</u><u> point</u><u> slope</u><u> form</u><u> </u><u>:</u><u>-</u><u> </u>