While driving past stores, Tatum counted the number of cars in the parking lots.

Mathematics

2 answers:

poizon [28]3 years ago

8 0

Answer:

median=9

Step-by-step explanation:

i used a median calculator i would recommend https://www.calculatorsoup.com/calculators/statistics/mean-median-mode.php

Vlad [161]3 years ago

7 0

Answer:

Step-by-step explanation:

1, 3, 5, 9, 15, 19, 19

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Examples of benchmarks

NeTakaya

Examples of benchmarks are: 1/2, 1/4,1, 0.
These are just a few examples of benchmarks. I hope this helps. Remember benchmarks are numbers that you can use on a number line as a guideline. Mark as brainliest! :)

7 0

2 years ago

Look at the picture

Sonbull [250]

$\large\displaystyle\text{$\begin{gathered}\sf 9|x-8| < 36 \end{gathered}$}$

$\large\displaystyle\text{$\begin{gathered}\sf Divide \ both \ sides \ by \ 9. \end{gathered}$}$

$\large\displaystyle\text{$\begin{gathered}\sf \frac{9(|x-8|)}{9} < \frac{36}{9} \end{gathered}$}$

$\large\displaystyle\text{$\begin{gathered}\sf |x-8| < 4 \end{gathered}$}$

$\large\displaystyle\text{$\begin{gathered}\sf Solve \ Absolute \ Value. \end{gathered}$}$

$\large\displaystyle\text{$\begin{gathered}\sf |x-8| < 4 \end{gathered}$}$

$\large\displaystyle\text{$\begin{gathered}\sf We \ know \ x-8 < 4 \ and \ x-8 > -4 \end{gathered}$}$

$\large\displaystyle\text{$\begin{gathered}\sf x-8 < 4 \ (Condition \ 1) \end{gathered}$}\\\large\displaystyle\text{$\begin{gathered}\sf x-8+8 < 4+8 \ (Add \ 8 \ to \ both \ sides) \end{gathered}$}\\\large\displaystyle\text{$\begin{gathered}\sf x < 12 \end{gathered}$}$

$\large\displaystyle\text{$\begin{gathered}\sf x-8 > -4 \ (Condition \ 2) \end{gathered}$}\\\large\displaystyle\text{$\begin{gathered}\sf x-8+8 > -4+8 \ (Add \ 8 \ to \ both \ \ sides) \end{gathered}$}\\\large\displaystyle\text{$\begin{gathered}\sf x > 4 \end{gathered}$}$

$\underline{\boldsymbol{\sf{Answer}}}$

$\boxed{\large\displaystyle\text{$\begin{gathered}\sf x < 12 \ and \ x > 4 \end{gathered}$} }$

$\large\displaystyle\text{$\begin{gathered}\sf Therefore,\bf{\underline{the \ correct \ option}} \ \end{gathered}$}\large\displaystyle\text{$\begin{gathered}\sf is \ \bf{\underline{"A"}}. \end{gathered}$}$