Answer:
12.60
Step-by-step explanation:
Start off by the total which is 18. To find the percentage off you will have to make the 30% into a decimal, so simply move it two spaces to the left. Now you have .30. Multiply the decimal, .30, and the full amount, 18. Your equation now should be (.30)*(18). Your answer will be 5.40. 5.40 is what is off of the shirt from the original price, so subtract 5.40 from 18. Your equation now is (18-5.40) which is 12.60.
A equation to represent this situation is:

Now we have to find x.
Step 1: Subtract 0.5x from both sides.
<span><span><span><span>0.75x</span>+7.5</span>−<span>0.5x</span></span>=<span><span><span>0.5x</span>+10</span>−<span>0.5x</span></span></span><span><span><span>0.25x</span>+7.5</span>=10
</span>Step 2: Subtract 7.5 from both sides.
0.25x+7.5−7.5=10−7.50.25x=2.5
Step 3: Divide both sides by 0.25.
<span><span><span>0.25x/</span>0.25</span>=<span>2.5/<span>0.25
</span></span></span>Answer:
<span>x=<span>10</span></span>
They will cost the same after 10 rides.
The answer is Z=18. 2x3=6 then 3x6=18
Answer:
- 33 1/3 liters of 30%
- 16 2/3 liters of 45%
Step-by-step explanation:
Let x represent the liters of 45% solution needed. Then the amount of HCl in the mix is ...
0.45x +0.30(50 -x) = 0.35(50)
0.15x = 0.05(50) . . . . . simplify, subtract 0.30(50)
x = (0.05/0.15)(50) = 50/3 = 16 2/3 . . . liters of 45% HCl
33 1/3 liters of 30% and 16 2/3 liters of 45% HCl are needed.
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<em>Comment on the solution</em>
You may notice that the general solution to a mixture problem of this sort is that the fraction of the mix that is the highest contributor is ...
(mix % - low %) / (high % - low %) = (.35 -.30) / (.45 -.30) = .05/.15 = 1/3
A stem and leaf plot (histogram) shows the mode as the longest list of "leaves." It is the easiest to use for finding mode.
A box-and-whisker plot tells you nothing about relative frequencies.
A scatter plot or line graph would require careful re-interpretation to determine the mode. If the amount of data is large and there are many data values with about the same high frequency, these charts may be unhelpful, too.