Answer: ?????
Step-by-step explanation: Are you finding who would have the most after a certain number of years have passed?
Answer:
Rate = 13.173% per year
Step-by-step explanation:
**Not 100% sure about this answer, but I think it's right.**
Calculation Steps:
Solving for rate r as a decimal
r = n[(A/P)^1/^nt - 1]
r = 1 × [(668.00/359.80)^1/^(1)^(5) - 1]
r = 0.131731
Then convert r to R as a percentage
R = r * 100
R = 0.131731 * 100
R = 13.173%/year
Answer: There is a 90% chance that the true proportion of teenagers who drive their own car to school will lie in (0.5907, 0.9093).
Step-by-step explanation:
Interpretation of a% confidence interval : A person can be a% confident that the true population parameter lies in it.
Here, A 90% confidence interval to estimate the true proportion of teenagers who drive their own car to school is found to be (0.5907, 0.9093).
i.e. A person can be 90% confident that the true proportion of teenagers who drive their own car to school lies in (0.5907, 0.9093).
Hence, correct interpretation is : There is a 90% chance that the true proportion of teenagers who drive their own car to school will lie in (0.5907, 0.9093).
Answer:
0.04
Step-by-step explanation:
becausse thats the tenth power
Answer: a. 0.05
b. 0.40
c. 0.85
Step-by-step explanation:
Let F= Event that a certain motorist must stop at the first signal.
S = Event that a certain motorist must stop at the second signal.
As per given,
P(F) = 0.45 , P(S) = 0.5 and P(F or S) = 0.9
a. Using general probability formula:
P(F and S) =P(F) + P(S)- P(F or S)
= 0.45+0.5-0.9
= 0.05
∴ the probability that he must stop at both signals = 0.05
b. Required probability = P(F but (not s)) = P(F) - P(F and S)
= 0.45-0.05= 0.40
∴ the probability that he must stop at the first signal but not at the second one =0.40
c. Required probability = P(exactly one)= P(F or S) - P(F and S)
= 0.9-0.05
= 0.85
∴ the probability that he must stop at exactly one signal = 0.85
