How to find nth term of a sequence in math?
1 answer:
Salutations!
To find the nth term, you will be needing a formula.
a stands for the first term (first number in the sequence)
d stands for the difference between two terms. (always remember to subtract from the 2nd term to the 1st term)
and the n is what we are going to find
so in your sequence: 5, 10, 15, 20
5 is a (as it is the first term)
d would be 5 ( 10 -5 =5)
your nth term is
Hope this helps!
Have a great day!
To find the nth term, you will be needing a formula.
a stands for the first term (first number in the sequence)
d stands for the difference between two terms. (always remember to subtract from the 2nd term to the 1st term)
and the n is what we are going to find
so in your sequence: 5, 10, 15, 20
5 is a (as it is the first term)
d would be 5 ( 10 -5 =5)
your nth term is
Hope this helps!
Have a great day!
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Answer:
0.1587 = 15.87% probability that the average fracture strength of 100 randomly selected pieces of this glass exceeds 14.2.
Step-by-step explanation:
To solve this question, we need to understand the normal probability distribution and the central limit theorem.
Normal Probability Distribution:
Problems of normal distributions can be solved using the z-score formula.
In a set with mean and standard deviation
, the z-score 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 p-value, we get the probability that the value of the measure is greater than X.
Central Limit Theorem
The Central Limit Theorem establishes that, for a normally distributed random variable X, with mean and standard deviation
, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean
and standard deviation
.
For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.
The fracture strength of tempered glass averages 14 (measured in thousands of pounds per square inch) and has standard deviation 2.
This means that
Sample of 100:
This means that
What is the probability that the average fracture strength of 100 randomly selected pieces of this glass exceeds 14.2?
This is 1 subtracted by the p-value of Z when X = 14.2. So
By the Central Limit Theorem
has a p-value of 0.8413.
1 - 0.8413 = 0.1587
0.1587 = 15.87% probability that the average fracture strength of 100 randomly selected pieces of this glass exceeds 14.2.