few_shots: 
Example 1:
Problem/activity:
Task
Materials
One 10-frame per pair of students (see example below)
[IMG]
Counters, 10 per pair of students
Decks of number cards 0-5 (with 4 of each number) or two 0-5 dice
Setup
This activity is done in pairs. The students sit side by side with 10-frame and deck of cards or dice. If they have cards, the cards are shuffled and placed face down in front of them.
Action
Player 1 flips a card or rolls the die. They place that many counters on the 10-frame. Player 2 flips a card or rolls the die. They place that many more counters on the 10-frame next to the counters already on the ten frame. The students count the total together.
A recording sheet can be added when the students have been introduced to equations and number sentences. For those students who can handle higher numbers, a deck of 0-10 cards and two 10-frames can be used.
This game can also be played alone.


Concept/skill: 
Represent addition and subtraction with objects, fingers, mental images, drawings (Drawings need not show details, but should show the mathematics in the problem. (This applies wherever drawings are mentioned in the Standards.)), sounds (e.g., claps), acting out situations, verbal explanations, expressions, or equations.

Answer: 
Yes. This task involves representing addition with objects such as number cards and dice.

Example 2:
Problem/activity:
Task
 Arrange the fractions in order from least to greatest. Explain your answer with a picture.
 $\frac{1}{5}, \frac{1}{7}, \frac{1}{3} $
 $\frac{2}{5}, \frac{2}{7}, \frac{2}{3} $
 $\frac{5}{6}, \frac{3}{6}, \frac{1}{6} $
 $\frac{5}{12}, \frac{8}{12}, \frac{4}{12} $

Concept/skill: 
Explain equivalence of fractions in special cases, and compare fractions by reasoning about their size.
 Grade 3 expectations in this domain are limited to fractions with denominators 2, 3, 4, 6, and 8.

Answer: 
Yes. By arranging the fractions in this problem, students learn to compare fractions by reasoning about their size.

Example 3:
Problem/activity:
Problem 1
Solve using the standard algorithm. If you get stuck, use an area model and/or the partial products algorithm to help. Then assess the reasonableness of your answer.
a.
$$5,803\times46$$
b.
$$63\times2,116$$
Problem 2
Dunkin' introduced a croissant doughnut in 2014. In the first three months of its debut, they sold 7,870 dozen of them per day. How many croissant doughnuts were sold in the first three months?


Concept/skill: 
Fluently multiply multi-digit whole numbers using the standard algorithm.

Answer: 
Yes. This set of problems involves multi-digit whole numbers because the numbers in each problem are four and two digit integers, and the mathematical calculations involved in both problems focus on multiplication. The first problem explicitly asks the student to solve its subproblems using the standard algorithm.

Now, does the problem or activity shown below enable students to learn the full intent of the following concept or skill?

shot_info: im_task_001477_pos_0

K

im_task_000157_pos_0

3

fl_problem_000674_pos_0

5
standard_description: Use variables to represent numbers and write expressions when solving a real-world or mathematical problem; understand that a variable can represent an unknown number, or, depending on the purpose at hand, any number in a specified set.
problem_activity: Activity
The mathematical purpose of this activity is to introduce students to curves representing normal distributions. Students compare normal curves with different means and standard deviations to determine how those statistics affect the curve. In particular, ensure students understand that normal curves are entirely defined by the mean and standard deviation, that the mean defines the center of the distribution, and the standard deviation affects the spread of the distribution.
Student Facing
These curves represent
normal distributions
with different means and standard deviations. What do you notice?
mean = 10. standard deviation = 1
[IMG]
mean = 10. standard deviation = 2
[IMG]
mean = 8. standard deviation = 2
[IMG]
mean = 12. standard deviation = 1
[IMG]
mean = 10. standard deviation = 0.8
[IMG]
Student Response
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for free access to Student Response.
Student Facing
Are you ready for more?
The equation for the curve for a normal distribution is
\(f(x) = \frac{1}{\sigma \sqrt{2\pi}} \boldcdot e^{\frac{-(x-\mu)^2}{2\sigma^2}}\)
where
\(\sigma\)
is the standard deviation,
\(\mu\)
is the mean, and
\(e\)
is a particular number close to 2.718.
Notice the part of the function that is
\((x - \mu)\)
. Using your understanding of transformations of functions, how does changing
\(\mu\)
affect the graph of the function?
Student Response
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click here to register or sign in
for free access to Extension Student Response.
Activity Synthesis
The important things for students to notice are:
The curves are entirely defined by the mean and standard deviation.
The curve is symmetric around the mean.
The standard deviation affects the width and height of the curve. A larger standard deviation is wider and shorter. A smaller standard deviation is thinner and taller.
Here are some questions for discussion:
“Why do you think the normal curve for a larger standard deviation is wider and shorter than one with a smaller standard deviation?” (We would expect a larger standard deviation to mean that data is more spread out, so it makes sense that is wider. Since the area must remain 1 under the curve, it has to get shorter to get wider.)
“What is the shape of the normal curve?” (Bell-shaped and symmetric)
“Where is the line of symmetry for each curve?” (
\(x = 10\)
,
\(x = 8\)
,
\(x = 12\)
, and
\(x = 10\)
, respectively)
Tell students,
“Although it is difficult to see in these graphs, the
\(x\)
-axis is a horizontal asymptote for normal curves. As you follow the curve to the right (or left), the curve gets closer and closer to the
\(x\)
-axis without ever actually reaching it.”
“Normal curves are bell-shaped and are most reliably used to model bell-shaped or approximately bell-shaped distributions.”
“The area under the normal curve (between the curve and the
\(x\)
-axis) is always 1. When the standard deviation increases, the curve spreads out to the right and left. To maintain the same area, though, the height of the peak must decrease.”


aligns: no