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Likurg_2 [28]
2 years ago
14

Is g = 2.5 a solution for 11.8 = 9.3 + g Yes or No please show work

Mathematics
1 answer:
Phoenix [80]2 years ago
8 0

Answer:

Yes

Step-by-step explanation:

9.3

+2.5

---------

11.8

9+2=11

.3+.5=.8

11+.8=11.8

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Can someone help me with this real quick?​
valentina_108 [34]

Answer:

11. A) 6x = 13

Step-by-step explanation:

11. Subtract 10 from both sides to make it equal 6x=13

12. For #12... None of the answers seem right? If you distribute and subtract, it should be

-4x + 44 = 66 then

-4x = 22...

Sorry if this didn’t help much, I know that 11 is right though ;)

6 0
3 years ago
Can someone help me on this?
dezoksy [38]

Answer:

C

Step-by-step explanation:

divide 22.5 by 10

=2.5

so

f=2 1/2c

8 0
2 years ago
A business owner collected data on the people working for her company. She determined the average number of minutes each employe
Naddika [18.5K]

Answer:

The comparison using median and IQR is best because one of the graphs is not symmetrical.<em> </em>

Step-by-step explanation:

The following information is missing

<em>A box plot titled Number of Minutes Women Spend on Breaks. The number line goes from 30 to 70. The whiskers range from 30 to 54, and the box ranges from 34 to 50. A line divides the box at 48.  </em>

<em>A box plot titled Number of Minutes Men Spend on Breaks. The number line goes from 30 to 70. The whiskers range from 30 to 68, and the box ranges from 36 to 60. A line divides the box at 48.  </em>

<em>The business owner uses the median and IQR to determine the center and variability of the data sets. Which best describes the comparison? </em>

<em>The comparison would be more accurate using the mean and MAD because one of the graphs is symmetric. </em>

<em>The comparison would be more accurate using the mean and MAD because the median of both data sets is the same. </em>

<em>The comparison using median and IQR is best because one of the graphs is not symmetrical. </em>

<em>The comparison using median and IQR is best because the median is greater than the IQR for both data sets.</em>

<em />

Mean and MAD are useful for comparison when both data sets are symmetrical.

In the women box plot the Q1 is at 34, the median is at 48, and the Q3 is at 50, so it is not symmetrical (the difference between the median and the Q1, and the Q3 and the median is not the same)

8 0
3 years ago
Problem page a long distance runner starts at the beginning of a trail and runs at a rate of 6 miles per hour. one hour later, a
alukav5142 [94]
The thing you have to figure out about this is the distance for each person and the time it takes for the biker to meet the runner.  The rates we are told.  The formula is distance = rate times time.  Let's do that for the runner first.  His rate is 6 so the formula so far is d = 6t.  Now let's work on the time.  If the biker left an hour later than the runner, then the runner has been running an hour more than the biker.  Therefore, the runner's time is t + 1.  Hold off on the distance part til we do for the biker what we just did for the runner.  The biker's rate is 14, and we already decided that his time is t.  His equation is d = 14t.  Now at the exact moment the biker meets the runner their distances are the same.  So if the equation for the runner is d = 6t + 6 and the equation for the biker is d = 14t and their distances are the same, by the transitive property, their rates and times are the same as well, meaning we set them equal to each other and solve for t.  6t + 6 = 14t.  6 = 8t and t = 3/4.  This means that it took 45 minutes for the biker to meet the runner.
5 0
2 years ago
What is interest earned on 3500 invested at 6% compounded quarterly or 12 years
enyata [817]
\bf \qquad \textit{Compound Interest Earned Amount}&#10;\\\\&#10;A=P\left(1+\frac{r}{n}\right)^{nt}&#10;\quad &#10;\begin{cases}&#10;A=\textit{accumulated amount}\\&#10;P=\textit{original amount deposited}\to &\$3500\\&#10;r=rate\to 6\%\to \frac{6}{100}\to &0.06\\&#10;n=&#10;\begin{array}{llll}&#10;\textit{times it compounds per year}\\&#10;\textit{quarterly, thus four}&#10;\end{array}\to &4\\&#10;t=years\to &12&#10;\end{cases}&#10;\\\\\\&#10;A=3500\left(1+\frac{0.06}{4}\right)^{4\cdot 12}\implies A=3500(1.015)^{48}

so, the interest earned then will be  A - 3500.
6 0
2 years ago
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