few_shots: 
Example:
Problem/activity:
Task
Cruz and Erica were both getting ready for soccer.
Cruz ran 1 lap around the school.
Erica ran 3 laps around the playground.
Erica said,
I ran more laps, so I ran farther.
Cruz said,
4 laps around the school is 1 mile, but it takes 12 laps around the playground to go 1 mile. My laps are much longer, so I ran farther.
Who is right? Draw a picture to help you explain your answer.


Description: 
Explain why a fraction a/b is equivalent to a fraction (n × a)/(n × b) by using visual fraction models, with attention to how the number and size of the parts differ even though the two fractions themselves are the same size. Use this principle to recognize and generate equivalent fractions.
Grade 4 expectations in this domain are limited to fractions with denominators 2, 3, 4, 5, 6, 8, 10, 12, and 100.

Answer: 
Yes. This task asks students to compare two fractions that are equal. The students need to explain why the two fractions are equal by drawing a picture, which involves using visual fraction models.

Now, assess whether the following problem or activity aligns with the provided description.

shot_info: im_task_000212_pos_0

4
standard_description: Use the mean and standard deviation of a data set to fit it to a normal distribution and to estimate population percentages. Recognize that there are data sets for which such a procedure is not appropriate. Use calculators, spreadsheets, and tables to estimate areas under the normal curve.
problem_activity: Task
Below is an isosceles triangle $ABC$ with $|AB| = |AC|$:
[IMG]
Three students propose different arguments for why $m(\angle B) = m(\angle C)$.
Ravi says
If I draw the bisector of $\angle A$ then this is a line of symmetry for $\triangle ABC$ and so $m(\angle B) = m(\angle C)$.
Brittney says
If $M$ is the midpoint of $\overline{BC}$ then $\triangle ABM$ is congruent to $\triangle ACM$ and so $\angle B$ and $\angle C$ are congruent.
Courtney says
If $P$ is a point on $\overline{BC}$ such that $\overleftrightarrow{AP}$ is perpendicular to $\overline{BC}$ then $\triangle ABP$ is congruent to $\triangle ACP$ and so $\angle B$ and $\angle C$ are congruent.
Fill in the details in each argument to show why $m(\angle B) = m(\angle C)$. Can you find another different argument showing that $m(\angle B) = m(\angle C)$?


aligns: no