Question

1. alice wants to know what she needs on her exam (worth 15% of her final grade) to pass the course with a 75. she earned 82% on her culminating assignment (worth 15% of her final grade). her grades throughout the course, which as an average is worth 70% of her final grade are: 74%, 92%, 85%, 71%, 68%, 72%. what grade does she need?

hint: weighted average:
grade = average term × %worth + culminating × %worth + exam mark × % worth

Explanation:

Step1: Calculate the average of term grades

First, sum up the term grades: \(74 + 92 + 85 + 71 + 68 + 72\). Let's calculate that: \(74+92 = 166\), \(166+85 = 251\), \(251+71 = 322\), \(322+68 = 390\), \(390+72 = 462\). Then divide by the number of grades (6) to get the average: \(\frac{462}{6}=77\). So the average term grade is \(77\%\).

Step2: Calculate the weighted contributions of term and culminating

The term average is worth \(70\%\) (or \(0.7\)) of the final grade, so its contribution is \(77\times0.7\). Let's compute that: \(77\times0.7 = 53.9\).

The culminating assignment is \(82\%\) and worth \(15\%\) (or \(0.15\)) of the final grade, so its contribution is \(82\times0.15\). Calculating that: \(82\times0.15 = 12.3\).

Step3: Set up the equation for the final grade

Let \(x\) be the exam mark (worth \(15\%\) or \(0.15\) of the final grade). The final grade should be \(75\), so we have the equation:

\(53.9 + 12.3 + 0.15x = 75\)

First, sum the known contributions: \(53.9 + 12.3 = 66.2\). So the equation becomes:

\(66.2 + 0.15x = 75\)

Subtract \(66.2\) from both sides: \(0.15x = 75 - 66.2 = 8.8\)

Then solve for \(x\): \(x=\frac{8.8}{0.15}\approx58.67\) (rounded to two decimal places). Since grades are typically in whole numbers, we can check if \(x = 59\) (or the exact value) works. But let's do the exact calculation: \(x=\frac{8.8}{0.15}=\frac{880}{15}=\frac{176}{3}\approx58.67\). So she needs approximately \(58.67\%\) on the exam. If we consider rounding to a whole number, it's about \(59\%\). But let's verify with the exact equation.

Wait, let's re - check the term average calculation. Wait, the term grades are \(74\%,92\%,85\%,71\%,68\%,72\%\). Let's recalculate the sum: \(74 + 92 = 166\), \(166+85 = 251\), \(251+71 = 322\), \(322+68 = 390\), \(390+72 = 462\). \(462\div6 = 77\). That's correct. Then term contribution: \(77\times0.7 = 53.9\), culminating: \(82\times0.15 = 12.3\). Sum: \(53.9+12.3 = 66.2\). Then \(75 - 66.2 = 8.8\). Then \(x=\frac{8.8}{0.15}=\frac{880}{15}=\frac{176}{3}\approx58.67\). So she needs approximately \(58.67\%\) on the exam. If we need to present it as a percentage, we can say she needs about \(59\%\) (or the exact fractional value).

Answer:

Alice needs approximately \(\boldsymbol{58.67\%}\) (or about \(59\%\)) on her exam. If we use exact fraction, it is \(\frac{176}{3}\%\approx58.67\%\)