P^ = 9/300 = 0.03
H0 = p < 5%
H1 = p > 5%
standard deviation of sample distribution = sqrt[p(1 - p) / n] = sqrt[0.05(1 - 0.05)/300] = sqrt(0.0001583) = 0.01258
test statistics, z = (p^ - p) / standard deviation = (0.03 - 0.05) / 0.01258 = -1.589
P(-1.589) = 1 - P(1.589) = 1 - 0.94402 = 0.05598
Since, p = 0.05598 < significant level of 0.1, we reject the H0.
i.e. There is no sufficient evidence to suggest that the proportion of defective batteries is less than 5%.
The p-value of the test is 0.05598
Answer:
The answer is option (C), T = 100 • 0.1 • 10
Step-by-step explanation:
Total interest (T) earned after a 10 years can be expressed as;
Total interest (T)=Principal amount×interest rate×Number of years
where;
Total interest=T
Principal amount=$100
Interest rate=10%=10/100=0.1
Replacing;
T=100×0.1×10
The equation for calculating the total interest amount after 10 years can be written as;
T = 100 • 0.1 • 10
It’s a blank screen .. what’s the question ?
Answer: 123
They have to add up to 3 4 5 6 or 7 so it has to be that
