Answer:
8800 yards
Step-by-step explanation:
The first step to solve this problem is to convert the first mixed number to an improper fraction.

÷ 3 +

Dividing is equivalent to multiplying with the reciprocal value,, so next you will need to change the signs.

x

+

Multiply the fractions

+

Finally,, add the two fractions together to get your final answer

This means that the correct answer to your question is

.
Let me know if you have any further questions
:)
Hi There!
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Full Question:
The box plots show the high temperatures in June and August for Denver in degrees Fahrenheit.
Which can you tell about the mean temperatures for these two months?
There is not enough information to determine the mean temperatures.
The mean temperature for August is higher than June's mean temperature.
The mean temperature for June is equal to the mean temperature for August.
The high interquartile range for August pulls the mean temperature above June's mean temperature.
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Interquartile Range Formula: Q3 - Q1
Interquartile Range for August: 10
Interquartile Range for June: 8
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Median = Mean
June: 82
August: 82
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Answer: The mean temperature for June is equal to the mean temperature for August.
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Hope This Helps :)
You can divide it and see what you get for it
well, let's say that the first solution is 7 liters and has "x" percent of acid, how much acid total in it? well, 7x, <u>keeping in mind that "x" is a decimal form</u>.
likewise, the second solution is 3 liters and has 15% acid, so how much acid is there in that one? (15/100) * 3 = 0.45, so
![\begin{array}{lcccl} &\stackrel{solution}{quantity}&\stackrel{\textit{\% of }}{amount}&\stackrel{\textit{liters of }}{amount}\\ \cline{2-4}&\\ \textit{1st solution}&7&x&7x\\ \textit{2nd solution}&3&0.15&0.45\\ \cline{2-4}&\\ mixture&10&0.29&2.9 \end{array}~\hfill \implies 7x~~ + ~~0.45~~ = ~~2.9 \\\\[-0.35em] ~\dotfill\\\\ 7x=2.45\implies x=\cfrac{2.45}{7}\implies x=0.35~\hfill \stackrel{\textit{converting to \%}}{0.35\cdot 100}\implies \stackrel{\%}{35}](https://tex.z-dn.net/?f=%5Cbegin%7Barray%7D%7Blcccl%7D%20%26%5Cstackrel%7Bsolution%7D%7Bquantity%7D%26%5Cstackrel%7B%5Ctextit%7B%5C%25%20of%20%7D%7D%7Bamount%7D%26%5Cstackrel%7B%5Ctextit%7Bliters%20of%20%7D%7D%7Bamount%7D%5C%5C%20%5Ccline%7B2-4%7D%26%5C%5C%20%5Ctextit%7B1st%20solution%7D%267%26x%267x%5C%5C%20%5Ctextit%7B2nd%20solution%7D%263%260.15%260.45%5C%5C%20%5Ccline%7B2-4%7D%26%5C%5C%20mixture%2610%260.29%262.9%20%5Cend%7Barray%7D~%5Chfill%20%5Cimplies%207x~~%20%2B%20~~0.45~~%20%3D%20~~2.9%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill%5C%5C%5C%5C%207x%3D2.45%5Cimplies%20x%3D%5Ccfrac%7B2.45%7D%7B7%7D%5Cimplies%20x%3D0.35~%5Chfill%20%5Cstackrel%7B%5Ctextit%7Bconverting%20to%20%5C%25%7D%7D%7B0.35%5Ccdot%20100%7D%5Cimplies%20%5Cstackrel%7B%5C%25%7D%7B35%7D)