We have to define an interval about the mean that contains 75% of the values. This means half of the values will lie above the mean and half of the values lie below the mean.
So, 37.5% of the values will lie above the mean and 37.5% of the values lie below the mean.
In a Z-table, mean is located at the center of the data. So the position of the mean is at 50% of the data. So the position of point 37.5% above the mean will be located at 50 + 37.5 = 87.5% of the overall data
Similarly position of the point 37.5% below the mean will be located at
50 - 37.5% = 12.5% of the overall data
From the z table, we can find the z value for both these points. 12.5% converted to z score is -1.15 and 87.5% converted to z score is 1.15.
Using these z scores, we can find the values which contain 75% of the values about the mean.
z score of -1.15 means 1.15 standard deviations below the mean. So this value comes out to be:
150 - 1.15(25) = 121.25
z score of 1.15 means 1.15 standard deviations above the mean. So this value comes out to be:
150 + 1.15(25) = 178.75
So, the interval from 121.25 to 178.75 contains the 75% of the data values.
Take note that the vine was 3.6 ft and now it is 2.1 times as long.
So since is it 2.1 times as long it will be multiplied by 3.6.
3.6(2.1)= 7.56
So it is 7.56 ft long, rounded it is 7.6 ft
The answer or value to this question will be D
Answer:
Most likely (B)
Step-by-step explanation:
Points of ABCD:
A (3,1)
B (3,4)
C (5,5)
D (5,2)
The algebraic rule for reflecting across the y axis:
(x,y) ---> (-x, y)
Points of ABCD after being reflected: (shown by figure 2)
A (-3, 1)
B (-3, 4)
C (-5, 5)
D (-5, 2)
Then, the figure got translated two units to the left, resulting in figure F in the picture, and A’B’C’D’ in the question.
Points of ABCD after being translated by (x-2, y) : (shown by figure F)
A (-5, 1)
B (-5, 4)
C (-7, 5)
D (-7, 2)
This should be the coordinates of A’B’C’D’.
