The formula uppercase S = StartFraction n (a Subscript 1 Baseline plus a Subscript n Baseline) Over 2 EndFraction gives the part
ial sum of an arithmetic sequence. What is the formula solved for an?
a Subscript n Baseline = StartFraction 2 uppercase S minus a Subscript 1 Baseline n Over n EndFraction
a Subscript n Baseline = StartFraction 2 uppercase S + a Subscript 1 Baseline n Over n EndFraction
a Subscript n Baseline = 2 uppercase S + a Subscript 1 Baseline n + n
a Subscript n Baseline = 2 uppercase S minus a Subscript 1 Baseline n + n
a Subscript n Baseline = StartFraction 2 uppercase S minus a Subscript 1 Baseline n Over n EndFraction
a Subscript n Baseline = StartFraction 2 uppercase S + a Subscript 1 Baseline n Over n EndFraction
a Subscript n Baseline = 2 uppercase S + a Subscript 1 Baseline n + n
a Subscript n Baseline = 2 uppercase S minus a Subscript 1 Baseline n + n
2 answers:
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Answer:
Step-by-step explanation:
Hello!
For me, the first step to any statistics exercise is to determine what is the variable of interest and it's distribution.
In this example the variable is:
X: height of a college student. (cm)
There is no information about the variable distribution. To estimate the population mean you need a variable with at least a normal distribution since the mean is a parameter of it.
The option you have is to apply the Central Limit Theorem.
The central limit theorem states that if you have a population with probability function f(X;μ,δ²) from which a random sample of size n is selected. Then the distribution of the sample mean tends to the normal distribution with mean μ and variance δ²/n when the sample size tends to infinity.
As a rule, a sample of size greater than or equal to 30 is considered sufficient to apply the theorem and use the approximation.
The sample size in this exercise is n=50 so we can apply the theorem and approximate the distribution of the sample mean to normal:
X[bar]~~N(μ;σ2/n)
Thanks to this approximation you can use an approximation of the standard normal to calculate the confidence interval:
98% CI
1 - α: 0.98
⇒α: 0.02
α/2: 0.01
X[bar] ±
174.5 ±
[172.22; 176.78]
With a confidence level of 98%, you'd expect that the true average height of college students will be contained in the interval [172.22; 176.78].
I hope it helps!
Answer:
<u>CF = 7</u>
Step-by-step explanation:
Given: Δ ABC
Altitudes ⇒ AD = 12 , BE = 14 and CF = ?
The area of the triangle = 0.5 AD * BC ⇒ (1)
OR area = 0.5 BE * AC ⇒ (2)
OR area = 0.5 CF * AB ⇒ (3)
By equating (1) and (3)
∴ 0.5 CF * AB = 0.5 AD * BC
∴ ⇒ (4)
By equating (1) and (2)
0.5 BE * AC = 0.5 AD * BC
∴ (5)
We should know that about the relation between the sides of the triangle:
The sum of two sides will be greater than the third side
So, AB < BC + AC ⇒ divide both sides by BC
∴
By substitution from (5) with AC/BC
∴
∴
∴ ⇒ (6)
By substitution from (6) at (4)
∴ CF > 12 * 7/13
CF > 6.46
But CF is a positive integer
<u>∴ CF = 7</u>