Find the minimum, first quartile, median, third quartile, and maximum of each data set, 12, 10, 11, 7, 9, 10, 5
Sliva [168]
To find the median:
Least to greatest order:
5, 7, 9, 10, 10, 11, 12
10 is the number in the middle. Therefore the median is 10
Answer:
(-2, -6)
Step-by-step explanation:
g(x) = (x+2)²-6
= x²+4x+4-6
= x²+4x-3
the vertex :
x = -4/2 = -2
y = (-2+2)²-6 =0-6 = -6
point of vertex : (-2, -6)
Answer:
Required equation is y - 36 = (24 - 36)/(2 - 1) (x - 1)
y - 36 = -12(x - 1)
y - 36 = -12x + 12
12x + y = 12 + 36
12x + y = 48
Step-by-step explanation:
Answer:
95% confidence interval for the mean μ is (6,14)
The Population mean μ lies between ( 6, 14 )
Step-by-step explanation:
<u><em>Explanation</em></u>:-
Given random sample 'n' = 1200
95% confidence interval for the mean μ is determined by

Level of significance = 95% 0r 0.05
Z₀.₀₅ = 1.96
= 10 ± 4
Mean of the small sample = 10
95% of confidence intervals are
( 10 ±4 )
( 10 -4 , 10+4)
( 6 , 14 )
95% confidence interval for the mean μ lies between ( 6, 14 )
He can use it to make sure he lined his place values up properly