help! hurry!! NO SPAM!!!! I will mark brainliest if you get it correct and show your work.Identify the 15th term of the arithmet
2 answers:
Answer:
a₁₅ = 30
Step-by-step explanation:
The nth term of an arithmetic sequence is
= a₁ + (n - 1)d
where a₁ is the first term and d the common difference
Given a₅ = 10 and a₁₀ = 20 , then
a₁ + 4d = 10 → (1)
a₁ + 9d = 20 → (2)
Subtract (1) from (2) term by term to eliminate a₁
5d = 10 ( divide both sides by 5 )
d = 2
Substitute d = 2 into (1) and solve for a₁
a₁ + 4(2) = 10
a₁ + 8 = 10 ( subtract 8 from both sides )
a₁ = 2
Then
a₁₅ = 2 + (14 × 2) = 2 + 28 = 30
You might be interested in
Using the normal distribution, it is found that:
a)
b) There is a 0.5859 = 58.59% probability that the college graduate has between $14,200 and $33,950 in student loan debt.
c) Low: $23,519.65, High: $26,580.35.
<h3>Normal Probability Distribution</h3>The z-score of a measure X of a normally distributed variable with mean and standard deviation
is given by:
- The z-score measures how many standard deviations the measure is above or below the mean.
- Looking at the z-score table, the p-value associated with this z-score is found, which is the percentile of X.
The mean and the standard deviation are given, respectively, by:
.
Hence the distribution of X can be described as follows:
The probability that the college graduate has between $14,200 and $33,950 in student loan debt is the <u>p-value of Z when X = 33950 subtracted by the p-value of Z when X = 14200</u>, hence:
X = 33950:
Z = 0.73
Z = 0.73 has a p-value of 0.7673.
X = 14200:
Z = -0.91
Z = -0.91 has a p-value of 0.1814.
0.7673 - 0.1814 = 0.5859.
There is a 0.5859 = 58.59% probability that the college graduate has between $14,200 and $33,950 in student loan debt.
Considering the symmetry of the normal distribution, the middle 10% is between the 45th percentile(X when Z = -0.127) and the 55th percentile(X when Z = 0.127), hence:
X - 25150 = -0.127(12050)
X = $23,519.65.
X - 25150 = 0.127(12050)
X = $26,580.35.
More can be learned about the normal distribution at brainly.com/question/4079902
#SPJ1