Respuesta:
Kansas, Misuri = 8
Paradise, Michigan = 3
Explicación paso a paso:
El rango intercuartílico (IQR) = Q3 - Q1
Q3 = 3/4 (n + 1) th término
n = número de muestras
Q1 = 1/4 (n + 1) th término
Kansas Misuri:
23,25,28,28,32,33,35
Datos pedidos: 23, 25, 28, 28, 32, 33, 35
n = 7
Q1 = 1/4 * 8 = 2do trimestre = 25
Q3 = 3/4 * 8 = 6. ° tem = 33
Q1 = 33 - 25 = 8
Paradise, Michigan:
16,24,26,26,26,27,28
Datos pedidos: 16,24,26,26,26,27,28
Q1 = 1/4 * 8 = 2do = 24
Q3 =. 3/4 * 8 = 6to = 27
Q3 - Q1 = 27 - 24 = 3
Step-by-step explanation:
Let the height above which the ball is released be H
This problem can be tackled using geometric progression.
The nth term of a Geometric progression is given by the above, where n is the term index, a is the first term and the sum for such a progression up to the Nth term is
To find the total distance travel one has to sum over up to n=3. But there is little subtle point here. For the first bounce ( n=1 ), the ball has only travel H and not 2H. For subsequent bounces ( n=2,3,4,5...... ), the distance travel is 2×(3/4)n×H
a=2H..........r=3/4
However we have to subtract H because up to the first bounce, the ball only travel H instead of 2H
Therefore the total distance travel up to the Nth bounce is
For N=3 one obtains
D=3.625H
Answer:
The Basic Identities are :



So for this question :




4 625/1,000 to 4 25/40 to 4 5/8
The box and whisker plot is attached.
We first order the data from least to greatest:
6, 7, 11, 13, 14, 15, 15, 19, 21
The median is the middle value, or 14.
The lower quartile is the median of the lower half (split by the median). This is between 7 and 11: (7+11)/2 = 18/2 = 9
The upper quartile is the median of the upper half (split by the median). This is between 15 and 19: (15+19)/2 = 34/2 = 17
The highest value is 21.
The lowest value is 6.
We draw the middle line of the box at 14, the median. We draw the left side of the box at the lower quartile, 9. We draw the right side of the box at the upper quartile, 17. From the right side of the box, we draw a whisker to the highest value, 21. From the left side of the box, we draw a whisker to the lowest value, 6.