Answer:
Results:
0.1432
0.0045
0.0905
0.0483
Step-by-step explanation:
Step a:
P(A or 10, A or 10) = 20/52 * 19/51 = 5/13 / 19/51 = 95/663 = 0.1432
Step b:
P(A A) = 4/52 * 3/51 = 1/13 * 1/17 = 1/221 = 0.0045
Step c:
P(10 10) = 16/52 * 15/51 = 4/13 * 5/17 = 20/221 = 0.0905
Step d:
P(A 10 or 10 A) = 2 * 4/52 * 16/51 = 2/13 * 16/51 = 32/663 = 0.0483
As well we get the probability by subtracting a, b and c:
P(blackjack): 0.1432 - 0.0905 - 0.0045 = 0.0482
The amount of days would be the independent variable and the ripeness of the fruit would be the dependent variable
Answer:
Answer:
a).
The amount spent on school materials for each term of all ST201students
b).
a).
It is not a random sample. This looks like a convenience sampling and there is sampling bias. This sample is not representative of the entire population. Since it is not a random sample it is not appropriate to generalize the results to all students.
b).
The sample size is 80 which is greater than 30. It is large enough to assume normal distribution according to central limit theorem.
c).
mean: $617
z critical value at 95%: 1.96
standard error = σ/sqrt(n) =500/sqrt(80) = 55.9017
lower limit= mean-1.96*se = 617-1.96*55.9017=507.43
upper limit= mean+1.96*se = 617+1.96*55.9017=726.57
d).
The amount spent on school materials for each term for the 80 ST201students is $617. We are 95% confident that amount spent on school materials for each term of all ST201students falls in the interval ($507.43, $726.57).
Step-by-step explanation:
Answer:
c
Step-by-step explanation:
