Is there a picture? to go with it
A.
For this case, let us set the variables:
s = 0 (final destination on the ground)
t = unknown
vo = 0
so = 8000 ft
Using the equation, we calculate for t:
0 = -16 t^2 + 0t + 8000
t = 22.36 s
B.
For this case:
s = unknown
t = 22.36 s
vo = 600 miles/hr = 880 ft/s
so = 0
Using the equation, we calculate for s:
s = -16*(22.36)^2 + 880*22.36 + 0
s = 11, 677.29 ft = 2.21 miles
Well, +10 and -10 can do that job.
Answer:
Kindly check explanation
Step-by-step explanation:
Calculate the 25th, 50th, and 75th percentiles.
b. Calculate the interquartile range.
c. Construct a boxplot. Are there any outliers?
Reordered data:
25, 40, 52, 55, 60, 70, 70, 72, 74, 75, 75, 80, 80, 85, 85, 86, 87, 90, 92, 99
25 th percentile :
1/4 (n + 1)th, tetm
n = 20
1/4(21) = 5.25th term
(5 + 6) th term / 2
= 65
50th percentile :
1/2 (n + 1)th, tetm
n = 20
1/2(21) = 10.5th term
(10 + 11) th term / 2
= 75
75th percentile:
3/4 (n + 1)th, tetm
n = 20
3/4(21) = 15.75th term
(15 + 16) th term / 2
= 85.5
Interquartile range :
Q3 - Q1
75th percentile - 25th percentile
85.5 - 65
= 20.5
25 is an Outlier
1.
There are 4 possible suits, hearts (H), clubs (C) spades (S) and diamonds (D)
The probability of each suit is 13/52 or 1/4
In the first draw, there will be 4 branches, each leading to one suit and each will have a probability of 1/4.
In the second draw, each of the branches after the first draw will have 4 more branches coming out of them, each will again have a probability of 1/4.
2.
A fair die means that there will be 6 initial branches, with the probability of 1/6 for each.
The toss of a coin means that each branch will have 2 branches in front of it, and each of these branches will lead to either heads or tails, with a 1/2 probability of each.
