Item 4 Find the median, first quartile, third quartile, and interquartile range of the data. 132,127,106,140,158,135,129,138 med
Ket [755]
Answer:
133.5, 127.5, 139.5, 12
Step-by-step explanation:
order data:
106, 127, 129, 132, 135, 138, 140, 158
Median:
The middle number: (8+1)/2 = 4.5 between the 4&5 numbers
= (135-132)/2= 1.5
1.5 + 132 = 133.5
lower quartile (1st quartile):
(8+1)/4 = 2.25 between the 2&3 numbers
(129+127)/4=0.5
0.5+127 = 127.5
upper quartile(3rd quartile):
(8+1)/4 x3 = 6.75 between the 6&7 numbers
(140-138)/4 x3 = 1.5
1.5 + 138 =139.5
Interquartile range:
139.5-127.5= 12
hope this helps
Step-by-step explanation:
2(X-1)=16
2X-2=16
2X=16+2
X=18÷2
X=9
I don't understand (3×1)-(5=86)-10
I think it is wrong or sth I don't know!
Answer:
-2
Step-by-step explanation:
since the parabola opens upwards, the minimum is at the vertex of the parabola and the maximum is infinity, because the lines of a parabola goes on forever
you look for the smallest x value for the parabola and since you can see that the vertex of the parabola lies on x=-2, the minimum value of the function is -2.
Hope that helps :)
Answer:
The third answer
Step-by-step explanation:
