Answer:
-1, 0, 1, 2, 3
Step-by-step explanation:
Solve two equations:
-5 < 2n-1 => -4 < 2n => n > -2
and
2n-1 ≤ 5 => 2n ≤ 6 => n ≤ 3
so
-2 < n ≤ 3
then enumerate the possible values for n
-1, 0, 1, 2, 3
Answer:
55
Step-by-step explanation:
plug in the numbers to the equation
7(5)+5(4)=55
3÷3×7÷9-2÷3×9÷9
Step-by-step explanation:
3×7=21
3×9=27
21÷27=7÷9
2×9=18
3×9=27
18÷27=2÷3
7÷9-2÷3=5÷6
Answer:
Yes, both np and n(1-p) are ≥ 10
Mean = 0.12 ; Standard deviation = 0.02004
Yes. There is a less than 5% chance of this happening by random variation. 0.034839
Step-by-step explanation:
Given that :
p = 12% = 0.12 ;
Sample size, n = 263
np = 263 * 0.12 = 31.56
n(1 - p) = 263(1 - 0.12) = 263 * 0.88 = 231.44
According to the central limit theorem, distribution of sample proportion approximately follow normal distribution with mean of p = 0.12 and standard deviation sqrt(p*(1 - p)/n) = sqrt (0.12 *0.88)/n = sqrt(0.0004015) = 0.02004
Z = (x - mean) / standard deviation
x = 22 / 263 = 0.08365
Z = (0.08365 - 0.12) / 0.02004
Z = −1.813872
Z = - 1.814
P(Z < −1.814) = 0.034839 (Z probability calculator)
Yes, it is unusual
0.034 < 0.05 (Hence, There is a less than 5% chance of this happening by random variation.
