Answer:
P(A&B) = 0.4
Explanation:
Because it is a random process and there are no special constraints the probability for everybody is the same, the probability of choosing a particular site is 1/7, the person originally seated in chair number seven has 5/7 chance of not seating in chair number six and seven, the same goes for the person originally seated in chair number six; Because we want the probability of the two events happening, we want the probability of the intersection of the two events, and because the selection of a chair change the probability for the others (Dependents events) the probability P(A&B) = P(A) * P(B/A) where P(A) is 5/7 and the probability of choosing the right chair after the event A is 4/7, therefore, P(A&B) = 4/7*5/7 = 0.4.
If the events were independent the probability would be 0.51.
The answer is b because 6(4x+3y) would be 6*4=24x then 6*3=18y
The correct option is B or,
t=d(2+3.5)
5.5 hours a day times however many days he or she decides to do so, equals the total number of hours :)
Answer:
False
Step-by-step explanation:
H0: µ ≤ 42
HA: µ > 42
S represents the sample standard deviation, then we must use the t-student test.
t-statistic formula:
t= (xbar-m)/(S/(sqrt(n)))
xbar: sample mean
m: hypothesized value
S: sample standard deviation
n: number of observations
t=(45-42)/(1.2/sqrt(15))
t-statistic= 9.682
The critical value from the t-student distribution with, 15-1 degrees of freedom and 1% significance level , is 2.6245
Because the t-statistic is greater than the critical value, we must reject the null hypothesis.
