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Kazeer [188]
2 years ago
15

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

Mathematics
2 answers:
yanalaym [24]2 years ago
5 0

Answer:

Step-by-step explanation

hope this helps

zmey [24]2 years ago
5 0

Answer:

a.

Step-by-step explanation:

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2 years ago
The heights of a random sample of 50 college students showed a mean of 174.5 centimeters and a standard deviation of 6.9 centime
Minchanka [31]

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

Z_{1-\alpha /2}= Z_{1-0.01}= Z_{0.99} =2.334

X[bar] ± Z_{1-\alpha /2} * \frac{S}{\sqrt{n} }

174.5 ± 2.334* \frac{6.9}{\sqrt{50} }

[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!

4 0
2 years ago
Express 0.009238 to 3 significant figures​
adell [148]

Answer: = 9.238 * 10 ^ 3

6 0
2 years ago
Triangle $ABC$ has altitudes $\overline{AD},$ $\overline{BE},$ and $\overline{CF}.$ If $AD = 12,$ $BE = 14,$ and $CF$ is a posit
Shalnov [3]

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

∴ CF = \frac{AD*BC}{AB} =\frac{12BC}{AB} ⇒ (4)

By equating (1) and (2)

0.5 BE * AC = 0.5 AD * BC

∴ \frac{AC}{BC} = \frac{AD}{BE} =\frac{12}{14} =\frac{6}{7} (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

∴ \frac{AB}{BC} < 1+\frac{AC}{BC}

By substitution from (5) with AC/BC

∴ \frac{AB}{BC}

∴ \frac{AB}{BC}

∴ \frac{BC}{AB} > \frac{7}{13} ⇒ (6)

By substitution from (6) at (4)

∴ CF > 12 * 7/13

CF > 6.46

But CF is a positive integer

<u>∴ CF = 7</u>

4 0
2 years ago
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