Answer:
10
Step-by-step explanation:
√81 = 9 (9+9-4*2)
4 * 2 = 8 (9+9-8)
9 + 9 = 18 (18 - 8)
18 - 8 = 10
5 times
the lcm of 3 and 4 is 12
1 minute = 60 seconds
60 divided by 12 is 5
J/-2 + 7 = -12 multiply both sides with 2
-j + 14 = -24 move -14 to the other side and change sign
-j= -24 - 14
-j = -38 and if you divide -38 with -1 you get + 38
j=38
Answer:
The calculated value of t= 0.1908 does not lie in the critical region t= 1.77 Therefore we accept our null hypothesis that fatigue does not significantly increase errors on an attention task at 0.05 significance level
Step-by-step explanation:
We formulate null and alternate hypotheses are
H0 : u1 < u2 against Ha: u1 ≥ u 2
Where u1 is the group tested after they were awake for 24 hours.
The Significance level alpha is chosen to be ∝ = 0.05
The critical region t ≥ t (0.05, 13) = 1.77
Degrees of freedom is calculated df = υ= n1+n2- 2= 5+10-2= 13
Here the difference between the sample means is x`1- x`2= 35-24= 11
The pooled estimate for the common variance σ² is
Sp² = 1/n1+n2 -2 [ ∑ (x1i - x1`)² + ∑ (x2j - x`2)²]
= 1/13 [ 120²+360²]
Sp = 105.25
The test statistic is
t = (x`1- x` ) /. Sp √1/n1 + 1/n2
t= 11/ 105.25 √1/5+ 1/10
t= 11/57.65
t= 0.1908
The calculated value of t= 0.1908 does not lie in the critical region t= 1.77 Therefore we accept our null hypothesis that fatigue does not significantly increase errors on an attention task at 0.05 significance level
