Find the median, Q1, Q3, the IQR for the following

2 answers:

7 0

First, you would normally need to order the numbers from least to greatest, but since the numbers are already in order we can skip this step.

24,24,28,32,40,42

The median is the data value in the middle of the data set or, in this case because there is an even number of values, the average of the two middle values.

Median: 30

The first quartile value is the median for the numbers before the median.
So find the median of 24, 24, 28, which is just the middle number.

First quartile: 24

Finding the third quartile is exactly the same as the first, except use the values after the median. (32, 40, 42)

Third quartile: 40

The IQR (Inter-quartile Range) is just the the third quartile minus the first quartile.

40 - 24 = 16

So here are your answers

Median: 30
First quartile: 24
Third quartile: 40
Inter-quartile range: 16

Let me know if you need help understanding anything and I'll try to explain as best I can.

Hope this helps!!

ELEN [110]3 years ago

4 0

24,24,28, 32,40,42
median = (28+32)/2 = 30
Q1 = 24
Q3 = 40
IQR = 40 -24 = 16

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Find the tenth term of the geometric sequence, given the first term and common ratio.

Natalija [7]

Answer:

$a_{10}=\dfrac{1}{128}$

Step-by-step explanation:

In the geometric series:

$a_1=4\\ \\r=\dfrac{1}{2}$

The nth term of the geometric sequence can be calculated using formula

$a_n=a_1\cdot r^{n-1}$

In your case, n = 10, then

$a_{10}\\ \\=4\cdot \left(\dfrac{1}{2}\right)^{10-1}\\ \\=4\cdot \left(\dfrac{1}{2}\right)^9\\ \\=2^2\cdot \dfrac{1}{2^9}\\ \\=\dfrac{1}{2^{9-2}}\\ \\=\dfrac{1}{2^7}\\ \\=\dfrac{1}{128}$