Math.

\( 24 \) Step 1: Find/Simplify Anything Involving Parentheses that \( 7^2 \) really means \( \underline{7\times 7} \), which is \( 49 \). \(  3 (\underline{7^2} + 5) \div 6 \: -\: 3 \) Is now … \( 3(49 + 5) \div 6\: -\: 3 \) Since there is still an expression in parentheses, we add the two numbers in parentheses.  \( 3(\underline{49 + 5}) \div 6\:, \: 3 \) Which is now … \( 3(54) \div 6 \:- 3\:  \) Step 2: Once Parentheses Are Complete, Move On Since there are no more operations in parentheses and no more exponents, we now move on to multiplication and division, whichever comes first working from left to right.  \( \underline{3(54)} \div 6\:, \:3\) Becomes:  \( 162 \div 6\:-\:3 \) We then perform the division. \( \underline{162 \div 6}\:-\:3 \) And get:  \( 27\: -\: 3 \) Step 3: Continue Following Order of Operations Since we only have one operation left, we can solve it. \( 27, 3 = 24 \)

53.137  Step 1: Line up the numbers by their decimal point. 68.567, 13, 2.43 Would look like:       \(\begin{align} 68.567 \\ 13 \phantom{0}\\ \underline{-\quad 2.43}\phantom{0} \end{align} \) Now all of our place values are lined up. Step 2: Fill in any empty spots with a placeholder zero. \(\begin{align} 68.567 \\ 13.000 \\ \underline{-\quad 2.430}\end{align} \) Step 3: Subtract like normal. \(\begin{align} 68.567 \\ 13.000 \\ \underline{-\quad 02.430} \\53.137 \end{align} \) 68.567, 13, 2.43 is equal to 53.137

17 marbles. Step 1: Interpret the problem First, we must understand what is happening in this situation. To find the number of marbles that Steven has, we need to multiply the number of marbles Max has by 0.2. Step 2: Calculate the number of marbles that Max has Since Max has 2.5 times the number of marbles Jacob has, we can multiply the number of marbles Jacob has by 2.5. with decimals, we pretend as if the decimals are not there. We do not line up numbers by their decimal point. number of marbles of Jacob × number of marbels of Max = total number of marbels Max has \(\begin{align} 34 \\ \underline{\times \quad 2.5} \end{align} \) Multiply Like Normal and fill in any empty spots with a placeholder zero. \(\begin{align} 34\\ \underline{\times \quad 2.5} \\ 170 \\ \underline{+ \quad 680} \\ 850  \end{align} \) Put the Decimal Place in the Right Location Since there is 1 number to the right of the decimal point in the problem, there will be 1 number that is behind the decimal in the product. Therefore, the decimal point goes between 5 and 0. \( 85.0 \)  Since the 0 is after the decimal point, it can also be dropped since it has no value. So, Max has 85 marbles. Step 3: Calculate the number of marbles that Steven has To find the number of marbles Steven has, we can multiply the number of marbles Max has by 0.2. with decimals, we pretend as if the decimals are not there. We do not line up numbers by their decimal point. total number of marbels of Max × number of marbels of Steven = total number of marbels Steven has \(\begin{align} 85 \\ \underline{\times \quad 0.2} \end{align} \) Multiply Like Normal and fill in any empty spots with a placeholder zero. \(\begin{align} 85\\ \underline{\times \quad 0.2} \\ 170 \\ \underline{+ \quad 000} \\ 170 \end{align} \) Put the Decimal Place in the Right Location Since there is one number to the right of the decimal point in the problem, there will be one number that is behind the decimal in the product. Therefore, the decimal point goes between the 7 and the 0. \( 17.0 \) Since the 0 is after the decimal point, it can also be dropped since it has no value. So, Steven has 17 marbles.

Adding two mixed numbers we can add the whole parts and then the fractional parts. \(  1 \large \frac {3}{10} \normalsize + 2 \large \frac{5}{8} \normalsize = (1 + 2) + (\large \frac{3}{10} + \frac {5}{8} )\) Step 1: Add the Whole parts Together \( 1 + 2 = 3 \) Step 2: Find the Least Common Multiple of the denominators Since \( 10 \) and \( 8 \) are the denominators of the fractions, we will find the LCM of \( 10 \) and \( 8 \). Factors of: 10: 10, 20, 30, 40, 50 8: 8, 16, 24, 32, 40 The least common multiple of \( 10 \) and \( 8 \) is \( 40 \).   Step 3: Create equivalent fractions using the LCM as the new denominator Each fraction will now have a denominator of \( 40 \). Let’s start with \( \large  \frac{3}{10} \). Since \( 10 \times 4 = 40 \), we have to multiply the numerator by the same number. \( \large \frac {3}{10}  =  \large  \frac{3\, \times \, 4}{10 \, \times \, 4}  = \large \frac{12}{40} \) Let’s create an equivalent fraction for \( \large \frac{5}{8} \) with 40 as the new denominator. Since \( 8 \times 5 = 40 \), we will multiply the numerator by the same number. \(  \large \frac {5}{8}  =  \large \frac{5 \, \times \, 5}{8 \, \times \, 5} = \large  \frac{25}{40} \)   Step 4: Add the fractions together (simplify or convert to a mixed number if necessary) Since our fractions now have the same denominator, we can add them together. \(  \large \frac {12}{40}  +  \frac{25}{40} = \frac {12 + 25}{40}  =  \frac{37}{40} \)   Step 5: Add the whole parts and the fractional parts. \(  3  +  \large \frac{37}{40} = 3 \large \frac {37}{40} \) Answer: \( 3 \large \frac {37}{40} \)

Step 1: Convert the mixed numbers into improper fractions \( 3\frac{5}{8} = \frac {3\times8 + 5}{8} \) \(  = \frac{29}{8}\)   \( 2\frac{2}{5} = \frac {2\times5 + 2}{5} \) \(  = \frac{12}{5}\)   \( 3\frac{5}{8} \times 2\frac{2}{5} \) now becomes  \( \frac{29}{8} \times \frac{12}{5} \) Step 2: Cross Simplify [If Applicable] and multiply When we look at the fractions to cross-simplify, we find that \( 12 \) and \( 8 \) have a common factor of \( 4 \). We can now reduce the fraction by \( 4 \).  \( \frac{29}{8} \times  \frac{12}{5} = \frac {29\times (12 \div 4)}{(8 \div 4) \times 5} \) \( =\frac{29\times 3}{2\times5}  \) \( = \frac{87}{10}\) Step 3: Simplify [If Possible] and convert into a mixed number Convert the improper fraction into a mixed number. \( \frac{87}{10} = 87 \div 10\) \(  8 \frac {7}{10}\) Answer: \(  8 \frac {7}{10}\)

254,604 Step 1: Determine the population growth in 2020. We need to find what is \( \frac{1}{6} \) of \(155,880\). This is the population growth in \( 2020 \). The keywords “has grown by \( \frac{1}{6} \) ” point to multiplication. \( 155,880 \times \large \frac{1}{6} = \frac{155,880}{1} \times \frac{1}{6} \) \(= \large \frac{155,880 \: \times \: 1}{1 \: \times \: 6}  \) \( = \large \frac{155,880}{6} \) \(=155,880 \div6 \) \(  = 25,980 \) Step 2: Find the total population in 2020. The total population in \( 2020 \) is the population in 2019 plus the population growth in \( 2020 \). \( 155,880 + 25,980 = 181,860 \) Step 3: Determine the population growth in 2021. We need to find what is \( \frac{2}{5} \) of 181,860. Again, we find the keywords “has grown by \( \frac{2}{5} \) ” to be sure that we should use multiplication. \( 181,860 \times \large \frac{2}{5} = \frac{181,860}{1} \times \frac{2}{5} \) \(= \large \frac{181,860 \: \times \: 2}{1 \: \times \: 5}  \) \(= \large \frac{363,720}{5} \) \(=363,720 \div5 \) \(  = 72,744 \) Step 4: Find the total population in 2021. The total population in 2021 is the population in 2020 plus the population growth in 2021. \( 181,860 + 72,744 = 254,604 \) The city’s population in 2021 is 254,604 people.

Step 1: Understand the Fraction Pieces The numerator (top number) is \( 15 \) and the denominator (bottom number) is \( 8 \). The fraction bar shows division. We can write this division problem as: \( 15 \) divided by \( 8 \) or \( 15\div 8 \). Because \( 15 \) can also be written as \( 15.0 \), this division problem can be set up as: \( \require{enclose}  \begin {array}{r} 8 \enclose {longdiv}{15.000} \end{array}\) Step 2: Complete the division Eight does not go into 1 so we move to the next number inside the division bar. Eight goes into \( 15 \)one time so we put a 1 over the \( 5 \). \( \require{enclose}  \begin {array}{r} 1 \phantom{.000}\\ \ 8 \enclose {longdiv}{15.000} \end{array}\) \( 8 \times 1 = 8\)  so we subtract \( 8 \) from \( 15 \) and get \( 7 \). \( \require{enclose}  \begin {array}{r} 1. \phantom{.000}\\ \ 8 \enclose {longdiv}{15.000} \\ \underline{-8}\phantom{.000} \\ 70\phantom {.000} \end{array}\) We drop down the 0 and we also keep the decimal point where it is. Then we ask: How many times does \( 8 \) go into \( 70 \)? Because \( 8\times 8 = 64 \), \( 8 \) goes into \( 70 \), eight times. \( \require{enclose}  \begin {array}{r} 1.8 \phantom{.00}\\ \ 8 \enclose {longdiv}{15.000} \\ \underline{-8}\phantom{.000} \\ 70\phantom {.000} \\ \underline{-64} \phantom{.00} \\ 6 \phantom {.00} \end{array}\) We still have a remainder so we bring down the next zero and keep dividing. Because \( 8\times 7 = 56 \), \( 8 \) goes into \( 60 \) seven times. \( \require{enclose}  \begin {array}{r} 1.87 \phantom{0}\\ \ 8 \enclose {longdiv}{15.000} \\ \underline{-8}\phantom{.000} \\ 70\phantom {.000} \\ \underline{-64} \phantom{.00} \\ 60 \phantom {.00} \\ \underline{-56} \phantom {.00} \\ 4 \phantom{.00}\end{array}\) Because \(  8 \times 5 = 40\), \( 8 \) goes into \( 40 \) five times. \( \require{enclose}  \begin {array}{r} 1.875 \\ \ 8 \enclose {longdiv}{15.000} \\ \underline{-8}\phantom{.000} \\ 70\phantom {.000} \\ \underline{-64} \phantom{.00} \\ 60 \phantom {.00} \\ \underline{56} \phantom {-56} \\ 40 \phantom{.00} \\ \underline{-40} \phantom {.00} \\ 0 \phantom {.00}\end{array}\) Forty minus \( 40 \) is zero so our answer is \( 1.875 \).  \(  \frac{15}{8}\) is \( 1.875 \) when written as a decimal.

Step 1: Read the Decimal Out Loud A few reminders: a. We read everything to the left of the decimal by itself. b. The decimal says the word “and”. c. We then read everything to the right of the decimal. d. We end it by saying the place value of the last number. Therefore, 0.932 is said as: Nine Hundred Thirty-two Thousandths Step 2: Write the Fraction You Read Out Loud Zero came before the decimal, so there is no whole number. We also said nine hundred thirty-two thousandths out loud. Nine hundred thirty-two would be our numerator and one thousand would be our denominator. \( \frac {932}{1000}\) Let’s check and see if the fraction can be simplified. 932 and 1000 have a factor of 4 in common so we can simplify the fraction by dividing both the numerator and denominator by a common factor, 4. \( \frac {932}{1000} \div \frac{4}{4}  =  \frac {932 \:\div\: 4}{1000\: \div\: 4} = \frac {233}{250}\) Let’s check and see if the fraction can be further simplified. Because the only factor that 233 and 250 have in common is 1, the fraction is completely simplified. 0.932 is \( \frac {233}{250} \) when written as a fraction.

Step 1: Interpret the Problem In this problem, we are given a monetary value that incurs a fee percentage and a tip percentage to find the final cost amount. We will do this in two parts. First, we will determine the fee percentage amount and add that to the initial monetary value. When calculating a percentage amount, we multiply the percentage by the given total. \( Fee \: Percentage \times Initial \: Total = Fee \: Percentage \: Amount \)   Step 2: Convert the fee percent to a decimal We do not use actual percentages when solving math problems, so we have to change \( 7\% \) into a decimal. To change a percent into a decimal, we divide the percent by 100, which is the same as moving the decimal two spots to the left. Since there is no decimal in 7, we assume it is to the right of the ones place, so: \( 7 \% = 7. \%\) Now we can slide the decimal point two spaces left. Note that there will be a space between the decimal and the 7 that is filled by a 0. \( 7. \% \to 0.07  \) \( 7\% \) as a decimal is \(0.07\)   Step 3: Determine the fee percentage amount and find the first total Now that our percent has been turned into a decimal, we can multiply: \( 0.07 \times 122.19 = 8.5533  \) *Note: We will not be rounding here; we round at the end of the problem. To determine our first total, we add the percentage fee amount to our initial amount. \( 122.19 + 8.5533 = 130.7433  \)   Step 4: Convert the tip percent to a decimal We do not use actual percentages when solving math problems, so we have to change \( 15\% \) into a decimal. To change a percent into a decimal, we divide the percent by 100, which is the same as moving the decimal two spots to the left. Since there is no decimal in 15, we assume it is to the right of the ones place, so: \( 15 \% = 15. \% \) Now we can slide the decimal point two spaces left. \( 15. \% \to 0.15 \) \( 15\% \) as a decimal is \(0.15\).   Step 5: Determine the tip percentage amount and find the final total that we are finding the tip based on the bill after the service fee, so we will use the total amount calculated in Step 3 here. Now that our percent has been turned into a decimal, we can multiply it with our first total: \( 0.15 \times 130.7433 = 19.611495 \) *Note: We will not be rounding here; we round at the end of the problem. To determine our final total, we add the tip percentage amount to our first total. \( 130.7433 + 19.611495 = 150.354795 \) When a math problem has a final answer representing a monetary quantity, the final answer must be rounded to the nearest hundredth, unless otherwise directed. \( 150.354795 = $ 150.35 \) The final total of Oscar’s order was \($150.35\).

Step 1: Interpret the Problem Calculating the patient’s original heart rate before the drop requires some manipulation of the percentages. We need to consider that the original heart rate represents \( 100\% \). If the original heart rate represents \( 100\% \) we must calculate how much of the original heart rate is remaining after a percent drop. Given the heart rate dropped by \( 30\% \), we know, \( 100\%, 30\% = 70 \%  \) Therefore, the remaining heart rate percent is \( 70\% \).   Step 2: Turn The Percent Into a Decimal We never use percentages in our actual math problems, so we first must turn our percentages into a decimal. To turn a percentage into a decimal, we divide by 100, which is the same as moving the decimal point two spaces to the left. \( 70\% = 70. \% \) \( 70. \%  \div 100 = 0.70 \) \( 70\% \)as a decimal is \( 0.70 \). zeros at the end of a number after a decimal does not add any value, so we can just write it as .70. \( 0.70 = 0.7 \)   Step 3: Determine the Original Heart Rate We know \( 63 = 70 \%\), and we are looking for the value that represents \( 100\% \). This means whatever \( 100\% \) is, will be more than \( 63 \). Once you calculate the percent paid and turn that into a decimal, we can use the rule below to calculate the original price of an item by using division. \( Current \: Heart \: Rate \div Decimal = Original \: Heart \: Rate  \) \( 63 \div 0.7 = 90  \) The patient’s heart rate before the drop was \(90\) beats per minute.

\( 79.4 \% \) Step 1: Calculate Part (Non-\(100 \%\)) In reading the prompt, the first step is to identify the amounts needed to be added to calculate the portion of the budget that has already been spent, which represents the part (non-100%). \( 75 + 65 = 140 \) Making 140 the part (non-\(100 \%\)).   Step 2: Divide the Part (Non-\(100 \%\)) by the Whole (\(100 \%\)) To find the percentage, we have to divide the part of the total available amount by the total amount available. In this case, the non-\(100 \%\) part that we have, \(140\), will be divided by the total amount available, which represents \(100 \%\), which is \(680\) \(140 \div 680 = 0.2058823529 \) We can round this off to the third-place value. Check your answer choices to see if numbers were rounded to a different place value. So let’s look at the third-place value: \(0.20\underline{5}8823529\) We look to the right and see an 8. Since an 8 is 5 or more, that means the underlined digit goes up by one so the \(5\) becomes a \(6 \). \( 0.206 \) Every other number to the right of the 5 would turn to 0, and since they would have no value, we can get rid of them.   Step 3: Multiply by \(100\) Now that we have our decimal, we have to turn it into a percent. we turn a decimal into a percent by multiplying it by 100, which is the same as moving the decimal point two spots to the right. \( 0.206 \times 100 = 20.6 \%\) This tells us how much of the budget has already been spent. Finally, we have to find the percentage of what she has left.   Step 4: Subtract Budget Spent from Total Budget The event coordinator spent \(20.6 \%\)  of her budget. In order to find the percentage that is left we will subtract the percentage she spent from \( 100 \%\). \( 100 \%, 20.6 \% = 79.4 \%\) The event coordinator has \(79.4 \%\) of her budget left.

When solving equations, remember to use: Inverse Operations, operations that undo one another. So if subtraction is present, we use addition, etc. What we do to one side we MUST do to the other Our goal is to ISOLATE the variable which means to have JUST one of the variables. Step 1: Distribute In this problem, we are first going to remove the parentheses from the equation. To do this we must multiply the term in front of each parenthesis by each term inside its respective parenthesis. \( 2(3x, 1), 2 (x + 5) = 12 \) \( 6x, 2, 2x, 10 = 12 \) Step 2: Combine Any Like Terms Now we can rearrange the equation so that we can combine all whole numbers and all terms with the same variable. Also, remember to include the sign IN FRONT of each term. \( 6x, 2, 2x, 10 = 12 \) If we rearrange, it becomes: \( 6x, 2x, 2, 10 = 12 \) This simplifies to: \( 4x, 12 =12 \) Step 3: Solve the Equation When we solve equations, we can start by applying the inverse of the constant term on both sides of the equation. The constant is the term with no variable attached to it. We see that \( 12 \) is being subtracted from \( 4x \), the \( -12  \) is the constant term because there is no \( x \) attached, so we have to undo subtraction using addition. Therefore we are going to add 12 to both sides. \( 4x, 12 + 12 = 12 + 12 \) Which then becomes: \( 4x = 24 \) Now we can solve: Since \( 4 \) is being multiplied by \( x \), we have to undo the multiplication using division. Therefore we are going to divide both sides by \( 4 \). \( \large \frac{4x}{4} = \frac{24}{4}\) This simplifies to: \( 1x = 6 \) Since \( 1x \) and \( x \) are the same thing, our final answer is: \(  x = 6\)

Step 1: Manipulate the Equations so that a term can cancel out When we have a system of equations in which both equations are written in the form \( ax + by = c \), we can solve the system using the method called ELIMINATION. To use elimination we need to make sure that one of the pairs of variables cancels up to zero. Looking at the system we notice that the two \( x \) terms are \(2x\) and \( 4x \)and the two \( y \)-terms are \( 5y \) and \( 3y \). Neither of the pairs cancels up to zero so there is some work to do. We need to decide what to multiply the equations by so that we can make one pair cancel up to zero. If we look at the \( x \)-terms we can multiply \( 2x \) by \( -2 \) and it will become \( -4x \) which if added to the other term, 4x, will cancel up to zero. This means that we will multiply all terms in the top equation by \( -2  \): \( -2\times (2x + 5y) = 19\times -2  \) \(  -4x\:, 10y = -38 \) So our system becomes, \( -4x, 10y = -38 \) \(  4x + 3y = 17 \) Step 2: Add the Equations and solve Once we have one pair of the variable terms cancel up to zero we can add the equations to each other. We add \(x\)-term to \(x  \)-term and \(  y\)-term to \( y \)-term vertically. \( \begin{array}{ccc}\require{enclose}-4x &, &10y &= &-38\\4x& + &3y& = &17 \\ \hline x\: &-& 7y &= &-21 \end{array} \) So our \( x \)-terms zero out (cancel out) giving us an equation that only has y so we can solve for \( y \). \( -7y = -21 \) Since \( -7 \) is being multiplied by \( y \), we have to undo the multiplication by using division. Therefore we are going to divide both sides by \( -7 \). \( large \frac {-7y}{-7} = \frac{-21}{-7} \) This simplifies to: \( 1y = 3 \) Since \( 1y\) and \( y \) are the same thing, our answer is: \( y = 3 \) Step 3: Find the value of the other term Now that we have the value of the \( y\)-term we can substitute it into either of the original equations to find the value of the \( x \)-term. \( 2x + 5y =19 \) \(2x + 5(3) = 19  \) \(   2x + 15 = 19 \) \( 2x + 15 -15 = 19 -15 \) \( 2x = 4 \) Since \(  2\) is being multiplied by \( x \), we have to undo the multiplication by using division. Therefore, we are going to divide both sides by \( 2 \). \(  \large \frac{2x}{2} = \frac {4}{2}\) This simplifies to: \( 1x =2 \) Since \( 1x  \) and \( x \) are the same thing, our answer is: \( x = 2 \) So for this system, our solution is \( x = 2\) and \( y = 3 \).

\(154.23\) Step 1: Your Rounding Rules Underline the place value you are rounding to. Look to the right of the number you have underlined. If you see a 5-9, the underlined digit goes up by 1. If you see a 0-4, the underlined digit stays the same. Anything to the right of the underlined digit turns to 0. Anything to the left of the underlined digit usually stays the same (unless a 9 was rounded up)   Step 2: Check Out Each Answer Choice To Determine Which Was Rounded Correctly and is the largest number Answer choice: \(153.62\) Underline 3 to round the number to the nearest whole number. \( 15\underline{3}.62 \) Since the number to the right of 3 is 6, 3 goes up to 4. Any number to the right of 4 turns to 0 and any number to the left stays the same. \( 154.00 \) 153.62 does round to 154 but must be compared to the other answer choices to see if it is the largest number that rounds to 154.   Answer choice: \(154.52\) Underline 4 to round the number to the nearest whole number. \( 15\underline{4}.52 \) Since the number to the right of 4 is 5, 4 goes up to 5. Any number to the right of 5 turns to 0 and any number to the left stays the same. \( 155.00 \) To the nearest whole number, \( 154.52 \) rounds to \(155\) not \(154\) so, this answer choice is not Answer choice: \(153.61\) Underline 3 to round the number to the nearest whole number. \( 15\underline{3}.61 \) Since the number to the right of 3 is 6, 3 goes up to 4. Any number to the right of 4 turns to 0 and any number to the left stays the same. \( 154.00 \) \(153.61\) does round to \(154\) but must be compared to the other answer choices to see if it is the largest number that rounds to \(154\).   Answer choice: \( 154.23 \) Underline 4 to round the number to the nearest whole number. \( 15\underline{4}.23 \) Since the number to the right of 4 is 2, 4 stays the same. Any number to the right of 4 turns to 0 and any number to the left stays the same. \(154.00\) \(154.23\) does round to \(154\) but must be compared to the other answer choices to see if it is the largest number that rounds to \(154\).   Step 3: Compare the numbers in the answer choices that round to 154 to see which is the largest. Comparing answers, we have \(153.61 < 153.62 < 154.23 \), Therefore, \( 154.23 \).

Step 1: Interpret the question To find the average in this scenario, we need to take the ratio of the total cost of the fruits and vegetables to the total number of pounds of them. Therefore, we have a division between the total cost and the total number of pounds. The total cost of the fruits and vegetables will be the sum of their costs. Total cost: \( 5.85 + 9 \) The total number of pounds of fruits and vegetables will be the sum of their weights. Total number of pounds: \( 3 + 2.7 \) To get the average cost per pound for the fruits and vegetables, we divide the \(2\) totals. \( Average \: cost \: per \: pound = (Total \: cost) \div (Total \: number \: of \: pounds) \)   Step 2: Round our numbers Since the question asks “approximate total cost…” we have to round some of these numbers before we add them together. The numbers $5.85 and 2.7 will need to be rounded. \($ 5.85 \) will be rounded to the nearest ones place. Let’s round  \($ 5.85 \) We underline the place value we are rounding to: \(\underline{5}.85\) When we look to the right of the \(5\) we see an \(8\), which tells us the underlined digit goes up by one. So, \(5\) goes to \(6\). Everything to the right of \(6\) turns to \(0\). \( 6.00 \) Since the zero after the decimal has no value, we can eliminate it. \(5.85\) rounds to \(6\). \(2.7\) will also be rounded to the nearest ones place. Let’s round \(2.7\). We underline the place value we are rounding to: \( \underline{2}.7 \) When we look to the right of the \(2\) we see a \(7\), which tells us the underlined digit goes up by one. So, \(2\) goes to \(3\). Everything to the right of \(3\) turns to \(0\). \(3.0\) Since the zero after the decimal has no value, we can eliminate it. \(2.7\) rounds to \(3\).   Step 3: Solve First, find the total cost of the fruits and vegetables. \( 6+ 9 = 15 \) So, the total cost of the fruits and vegetables is \($15\). Next, find the total weight of the fruits and vegetables. \( 3 + 3 = 6 \) So, the total weight of the fruits and vegetables is \(6 \: pounds\). Finally, we get the average cost per pound of the fruits and vegetables by dividing the total cost by the number of pounds. \( 15 \div 6 = 2.5 \) So, the average cost per pound for fruits and vegetables is \( $2.50 \)