(A) The mean is 82
(B) The mean absolute deviation is approximately 9.71
(D) The IQR is 11
<u>Explanation:</u>
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(A) 89, 87, 54, 78, 87, 99, 80
Mean of the data =
Therefore, mean of the data is 82
(B) The mean absolute deviation, MAD is
data - 54, 78, 80, 87, 87, 89, 99
Mean of the data, x' is 82
|x - x'| : 54 - 82 = 28
78 - 82 = 4
80 - 82 = 2
87 - 82 = 5
87 - 82 = 5
89 - 82 = 7
99 - 82 = 17
Total of |x - x'| = 68
MAD = |x - x'| / n
MAD = 68/7
MAD = 9.71
(D) IQR is the interquartile range
Data = 54, 78, 80, 87, 87, 89, 99
IQR = median of lower quartile range - median of upper quartile range
IQR = 89 - 78
IQR = 11
Answer:
B .3.5 points of coffee did he drink
hope it helps
Answer:
x=6 y=2
Step-by-step explanation:
y=2
so 6x+9(2)=-24
6x+18=-24
6x=42
x=6
Answer:
a. Mean = 20
Sd = 4
b. Probability of X = 20 = 0.1960
Step-by-step explanation:
we have
n = 25
p = 80% = 0.8
mean = np
= 0.8 * 25
= 20
standard deviation = √np(1-p)
= √25*0.8(1-0.8)
=√4
= 2
probability that exactly 20 favours ban
it follows a binomial distribution
= 25C20 × 0.8²⁰ × 0.2⁵
= 53130 × 0.01153 × 0.00032
= 0.1960
Probability of X = 20 = 0.1960
Answer:
123m squared
Step-by-step explanation:
First let's get the area of the square by doing 10*10 to get 100, now let's find the area of the semicircle. We can pretend it is a full circle and then cut it in half. We do the radius which is 5 times, pi which is 3.14 to get 15.7. We then square that to get 246.49. Since we want a semicircle we cut that in half to get 123.245. We add this to 100 to get 123.245. Since we have to round, the answer is: The area of the figure is about 123 m2.