Answer:
1i) nth term = 12(-b/4)^(n-1)
ii) nth term = 3(-1/9)^(n-1)
2i) Sixth term= (-3b^5)/256
ii) Eight term = -1/1594323
Question:
The complete question as found on brainly (question-10746111):
Write down the nth term of each of the following G.Ps whose first two terms are given as follows. Also find the term stated besides each G.P.
i) 12,-3b,......sixth term
ii) 3,-1/3,..........;8th term?
Step-by-step explanation:
1) To solve terms involving Geometric progressions (GPs), we would first state the variables that are to be incorporated in the formula.
i) 12,-3b,......;sixth term
1st term = a= 12
r = common ratio = 2nd term /1st term
r = -3b/12 = -b/4
Then we would find the nth term
See attachment for details
ii) 3,-1/3,..........;8th term
1st term = a= 3
r = common ratio = 2nd term /1st term
r = (-⅓)/3 = (-⅓)(⅓) = -1/9
Then we would find the nth term
See attachment for details
2) we are to determine the sixth term in (i) and eight term in (ii)
See attachment for details
Sixth term= (-3b^5)/256
Eight term = -1/1594323
Answer:
0.60% probability that the sample mean would differ from the true mean by more than 104 dollars if a sample of 476 persons is randomly selected
Step-by-step explanation:
To solve this question, we have to understand the normal probability distribution and the central limit theorem.
Normal probability distribution:
Problems of normally distributed samples are solved using the z-score formula.
In a set with mean and standard deviation
, the zscore of a measure X is given by:
The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.
Central limit theorem:
The Central Limit Theorem estabilishes that, for a random variable X, with mean and standard deviation
, a large sample size can be approximated to a normal distribution with mean \mu and standard deviation, which is also called standard error
In this problem, we have that:
The standard deviation is the square root of the variance. So
What is the probability that the sample mean would differ from the true mean by more than 104 dollars if a sample of 476 persons is randomly selected
Either it differs by 104 or less dollars, or it differs by more than 104 dollars. The sum of the probabilities of these events is 100. I am going to find the probability that it differs by 104 or less dollars first.
Probability that it differs by 104 or less dollars first.
pvalue of Z when X = 16127 + 104 = 16231 subtracted by the pvalue of Z when X = 16127 - 104 = 16023. So
X = 16231
By the Central Limit Theorem
has a pvalue of 0.9970
X = 16023
has a pvalue of 0.0030
0.9970 - 0.0030 = 0.9940
99.40% probability that it differs by 104 or less.
What is the probability that the sample mean would differ from the true mean by more than 104 dollars if a sample of 476 persons is randomly selected
p + 99.40 = 100
p = 0.60
0.60% probability that the sample mean would differ from the true mean by more than 104 dollars if a sample of 476 persons is randomly selected