A triangle has vertices at B(−3, 0), C(2, −1), D(−1, 2). Which transformation would produce an image with vertices B″(−2, 1), C″
(3, 2), D″(0, −1)?
(x, y) → (x, −y), (x, y) → (x + 1, y + 1)
(x, y) → (−x, y), (x, y) → (x + 1, y + 1)
(x, y) → (x, −y), (x, y) → (x + 2, y + 2)
(x, y) → (−x, y), (x, y) → (x + 2, y + 2)
(x, y) → (x, −y), (x, y) → (x + 1, y + 1)
(x, y) → (−x, y), (x, y) → (x + 1, y + 1)
(x, y) → (x, −y), (x, y) → (x + 2, y + 2)
(x, y) → (−x, y), (x, y) → (x + 2, y + 2)
2 answers:
Answer:
A. (x, y) → (x,-y), (x, y) → (x + 1, y + 1)
Step-by-step explanation:
Plot points B, C and D on the coordinate plane (blue points in attached diagram).
1 transformation is reflection across the x-axis with the rule
(x,y)→(x,-y)
and it maps these points to
- B(-3,0)→B'(-3,0)
- C(2,-1)→C'(2,1)
- D(-1,2)→D'(-1,-2)
These image points are marked in green in attached diagram.
2 transformation is translation 1 unit to the right and 1 unit up with the rule
(x,y)→(x+1,y+1)
and it maps previous image points to
- B'(-3,0)→B''(-2,1)
- C'(2,1)→C''(3,2)
- D'(-1,-2)→D''(0,-1)
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Answer:
a) The 98% confidence interval for the mean weight is between 10.00409 grams and 10.00471 grams
b) 49 measurements are needed.
Step-by-step explanation:
Question a:
We have that to find our level, that is the subtraction of 1 by the confidence interval divided by 2. So:
Now, we have to find z in the Ztable as such z has a pvalue of .
That is z with a pvalue of , so Z = 2.327.
Now, find the margin of error M as such
In which is the standard deviation of the population and n is the size of the sample.
The lower end of the interval is the sample mean subtracted by M. So it is 10.0044 - 0.00031 = 10.00409 grams
The upper end of the interval is the sample mean added to M. So it is 10 + 0.00031 = 10.00471 grams
The 98% confidence interval for the mean weight is between 10.00409 grams and 10.00471 grams.
(b) How many measurements must be averaged to get a margin of error of +/- 0.0001 with 98% confidence?
We have to find n for which M = 0.0001. So
Rounding up
49 measurements are needed.
Part 1:
The table showing the decreasing debt is shown as follows:
![\begin{tabular} {|c|c|c|c|} Starting Balance&Interest Accrued&Payment&Closing Balance\\[1ex] \$1,853.42&\$22.39&\$400&\$1,475.81\\\$1,475.81&\$17.83&\$400&\$1,093.64\\\$1,093.64&\$13.21&\$200&\$906.85\\\$906.85&\$10.96&\$200&\$717.81\\\$717.81&\$8.67&\$200&\$526.48\\\$526.48&\$6.36&\$200&\$332.84\\\$332.84&\$4.02&\$200&\$136.86\\\$136.86&\$1.65&\$138.51&- \end{tabular}](https://tex.z-dn.net/?f=%5Cbegin%7Btabular%7D%0A%7B%7Cc%7Cc%7Cc%7Cc%7C%7D%0AStarting%20Balance%26Interest%20Accrued%26Payment%26Closing%20Balance%5C%5C%5B1ex%5D%0A%5C%241%2C853.42%26%5C%2422.39%26%5C%24400%26%5C%241%2C475.81%5C%5C%5C%241%2C475.81%26%5C%2417.83%26%5C%24400%26%5C%241%2C093.64%5C%5C%5C%241%2C093.64%26%5C%2413.21%26%5C%24200%26%5C%24906.85%5C%5C%5C%24906.85%26%5C%2410.96%26%5C%24200%26%5C%24717.81%5C%5C%5C%24717.81%26%5C%248.67%26%5C%24200%26%5C%24526.48%5C%5C%5C%24526.48%26%5C%246.36%26%5C%24200%26%5C%24332.84%5C%5C%5C%24332.84%26%5C%244.02%26%5C%24200%26%5C%24136.86%5C%5C%5C%24136.86%26%5C%241.65%26%5C%24138.51%26-%0A%5Cend%7Btabular%7D)
The column for interest accrued is obtained by the folmular
where P is the month's starting balance.
Part 2:
From the table, the last payment is $138.51
Part 3:
The total amount paid by the time the credit card is payed of is the summation of the "payment" column of the table.
Thus, the total amount is given by:
Part 4:
The debt ratio is given by
The table showing the decreasing debt is shown as follows:
The column for interest accrued is obtained by the folmular
where P is the month's starting balance.
Part 2:
From the table, the last payment is $138.51
Part 3:
The total amount paid by the time the credit card is payed of is the summation of the "payment" column of the table.
Thus, the total amount is given by:
Part 4:
The debt ratio is given by