Answer:
(48.106 ; 53.494)
Step-by-step explanation:
Given the data:
X : 52 48 49 52 53
Sample mean = ΣX / n
n = 5
Sample mean, xbar = 254 / 5 = 50.8
Standard deviation, s = 2.17 (using calculator)
The standard error (SE) : s/√n =2.17/√5 = 0.970
The degree of freedom, df = n-1
df = 5 - 1 = 4
Tscore(0.05, 4) = 2.776
Confidence interval :
Xbar ± Tscore*standard error
50.8 ± (2.776 * 0.970)
50.8 ± 2.694
Lower boundary = 50.8 - 2.694 = 48.106
Upper boundary = 50.8 + 2.694 = 53.494
(48.106 ; 53.494)
All you would have to do is look at the 2 in the place of the thousands which equals 2000, then look at the 2 that is in the place of the tens, which equals 20, so (this might get complicated) 2000/20 is... (If you need more help just tell me)
Answer:
4m(4m-3)
Step-by-step explanation:
16m^2 − 12m
16 m^2 = 2*2*2*2*m*m
12m = 2*2*3*m
The greatest factor for 16m^2 and 12m is 2*2*m or 4m
Factor out 4m
16 m^2 = 2*2*2*2*m*m = 4m(2*2m)=4m(4m)
12m = 2*2*3*m = 4m(3)
Factoring out 4m
16m^2 − 12m
4m(4m-3)
8x + 6=4x + 38
4x = 32
x = 8
<B = 4x + 38 = 4(8) + 38 = 70
