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2 years ago

Find the median, first quartile, third quartile, and interquartile range of the data.

Mathematics

1 answer:

vovangra [49]2 years ago

8 0

Answer:

Median: 88, First Quartile: 84, Third Quartile: 92, IQR (interquartile range): 8

Step-by-step explanation:

1. Sort the numbers in numerical order (like 1, 2, 3, etc.)

2. Find the number that is in the middle

3. Find all the quartiles by finding the median of the lower and upper halves of the list

4. Subtract Q1 (quartile 1) and Q3 (quartile 3) like this: Q3 - Q1 to get IQR

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Help ASAP show work please thanksss!!!!

Llana [10]

Answer:

$\displaystyle log_\frac{1}{2}(64)=-6$

Step-by-step explanation:

Properties of Logarithms

We'll recall below the basic properties of logarithms:

$log_b(1) = 0$

Logarithm of the base:

$log_b(b) = 1$

Product rule:

$log_b(xy) = log_b(x) + log_b(y)$

Division rule:

$\displaystyle log_b(\frac{x}{y}) = log_b(x) - log_b(y)$

Power rule:

$log_b(x^n) = n\cdot log_b(x)$

Change of base:

$\displaystyle log_b(x) = \frac{ log_a(x)}{log_a(b)}$

Simplifying logarithms often requires the application of one or more of the above properties.

Simplify

$\displaystyle log_\frac{1}{2}(64)$

Factoring $64=2^6$ .

$\displaystyle log_\frac{1}{2}(64)=\displaystyle log_\frac{1}{2}(2^6)$

Applying the power rule:

$\displaystyle log_\frac{1}{2}(64)=6\cdot log_\frac{1}{2}(2)$

Since

$\displaystyle 2=(1/2)^{-1}$

$\displaystyle log_\frac{1}{2}(64)=6\cdot log_\frac{1}{2}((1/2)^{-1})$

Applying the power rule:

$\displaystyle log_\frac{1}{2}(64)=-6\cdot log_\frac{1}{2}(\frac{1}{2})$

Applying the logarithm of the base:

$\mathbf{\displaystyle log_\frac{1}{2}(64)=-6}$

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Answer:

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Step-by-step explanation:

make a box and multiply each of the parts

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