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3 years ago

What is the value of the fourth term in a geometric sequence for which a1= 15 and r= 1/3

1 answer:

3 0

$\bf \begin{array}{cll}
term&value\\
\text{\textemdash\textemdash\textemdash}&\text{\textemdash\textemdash\textemdash}\\
a_1&15\\
a_2&15\left( \frac{1}{3} \right)\\\\
a_3&15\left( \frac{1}{3} \right)\left( \frac{1}{3} \right)\\\\
a_4&15\left( \frac{1}{3} \right)\left( \frac{1}{3} \right)\left( \frac{1}{3} \right)
\end{array}\qquad \implies 15\left( \frac{1}{3} \right)^3\implies 15\cdot \cfrac{1^3}{3^3}
\\\\\\
\cfrac{15}{27}\implies \stackrel{simplified}{\cfrac{5}{9}}$

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Step-by-step explanation:

For this case we have the following data:

350, 70, 65, 50, 60

Part a

We can calculate the mean with the following formula:

$\bar X = \frac{\sum_{i=1}^n X_i}{n}$

And replacing we got:

$\bar X= \frac{350+ 70+ 65 +50+ 60}{5} =119$

And for the median first we need to order the dataset on increasing way:

50, 60, 65, 70, 350

Since the sample size is an odd number we can calculate the median as the middle position for the dataset, for this case the 3th position and we got:

Median = 65

Part b

For this case we can see that we have an outlier present in the data 350, and for this case if we want to give a measure of central tendency is better use the median since this meaure is not affected by outliers. So Lauren should use the median.

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