<span>Let n be the number of taxis in NY. The average distance travelled is 60,000 miles, therefore the middle 95% will have the same average as the population, the reason being the mileage is symmetrically distributed about the mean Therefore the total number of miles in one year for the middle 95% is 60,000 * 0.95 * n
</span><span>The range of miles driven by the middle 95% can be found from the empirical rule that says:
For a normal distribution, approximately 95% of the data points lie within the range plus and minus 2 standard deviations of the population mean. In this case the range is
(60,000-22,000) to (60,000 + 22,000)</span>
X = 112
See attachment file below.
Hope it helped!
Answer:
A. 35
Step-by-step explanation:
The median of a data set is the middle value when the data values are placed in order of size.
<u>Given data set</u>:
- {3, 35, 23, 37, 45, 5, 49, 27, 48}
Place the data in <u>order of size</u>:
- {3, 5, 23, 27, 35, 37, 45, 48, 49}
To find the median, divide the total number of data values (n) by 2.
- If n/2 is a whole number, the median is halfway between the values in this position and the position above.
- If n/2 is not a whole number, round it up to find the position of the median.
As there are 9 data values, the median value is:

Therefore, the median of the given data set is 35.
We know that
if two lines are perpendicular
then
the slopes
m1*m2=-1
step 1
find the slope AB
A (0,2)
B (-3,-3)
m=(y2-y1)/(x2-x1)-----> m=(-3-2)/(-3-0)-----> m=-5/-3----> m1=5/3
step 2
find the slope CD
C (-4,1)
D (0,-2)
m=(y2-y1)/(x2-x1)-----> m=(-2-1)/(0+4)-----> m=--3/4----> m2=-3/4
step 3
multiply mi*m2
(5/3)*(-3/4)-----> -15/12
so
15/12 is not -1
therefore
AB is not perpendicular to CD
Answer:
x = 5 is the right answer.
Step-by-step explanation:
As we know sum of all angles in a triangle is 180°.Therefore in this question we will form an equation.
Sum of all the angles of a triangle = 180
91 + 10x - 4 +8x +3 = 180
90 + 18x = 180
18x + 90 - 90 = 180 - 90
18x = 90
x = 90÷18 = 5
Therefore x = 5 is the right answer.