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Alekssandra [29.7K]
3 years ago
15

the sum of the first nine terms of an arithmetic series is 162, and the sum of the first 12 terms is 288. Determine the first fi

ve terms of the series.
Mathematics
1 answer:
Natali5045456 [20]3 years ago
8 0

Answer:

\boxed{\pink{\tt \leadsto Sum \ of \ first \ five \ terms \ is \ 50 . }}

Step-by-step explanation:

Given that , the sum of the first nine terms of an arithmetic series is 162 and the sum of the first 12 terms is 288.

\boxed{\red{\bf \bigg\lgroup For \ answer \ refer \ to \ attachment \bigg\rgroup  }}

<h3><u>Related</u><u> </u><u>Infor</u><u>mation</u><u> </u><u>:</u><u>-</u><u> </u></h3>

• The sum of n terms of an AP is

\boxed{\orange{\sf S_n = \dfrac{n}{2}[2a+(n-1)d] }}

• nth term of an AP is given by ,

\boxed{\blue{\sf T_n = a+(n-1)d }}

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Visit: brainly.com/question/21979359

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GIVING BRAINIEST EXPLAIN ANSWER
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Step-by-step explanation:

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From left to right, complete the table of values for the function<br> y - 4 = -3(x + 2).
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Step-by-step explanation:

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What’s the sun of 15 and twice a number
Vaselesa [24]
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3 years ago
A random sample of 36 observations has been drawn from a normal distribution with mean 50 and standard deviation 12. Find the pr
Gre4nikov [31]

Answer:

z= \frac{47 -50}{\frac{12}{\sqrt{36}}}=-1.5

z= \frac{53 -50}{\frac{12}{\sqrt{36}}}=1.5

And using a calculator, excel ir the normal standard table we have that:

P(47 \leq \bar X \leq 53) =P(-1.5 \leq Z \leq 1.5)

And we can calculate the probability like this:P(-1.5 \leq Z \leq 1.5) = P(zStep-by-step explanation:

A random sample of 36 observations has been drawn from a normal distribution with mean 50 and standard deviation 12. Find the probability that the sample mean is in the interval 47<=X<53. Is the assumption of normality important. Why?

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".  

Solution to the problem

Let X the random variable that represent the variable of interest of a population, and for this case we know the distribution for X is given by:

X \sim N(50,12)  

Where \mu=50 and \sigma=12

Since the distribution for X is normal then we know that the distribution for the sample mean \bar X is given by:

\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})

We can find the probability required like this:

z= \frac{47 -50}{\frac{12}{\sqrt{36}}}=-1.5

z= \frac{53 -50}{\frac{12}{\sqrt{36}}}=1.5

And using a calculator, excel ir the normal standard table we have that:

P(47 \leq \bar X \leq 53) =P(-1.5 \leq Z \leq 1.5)

And we can calculate the probability like this:

P(-1.5 \leq Z \leq 1.5) = P(z

4 0
3 years ago
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