Averages for 5th Grade: Determining the average without calculation

To find the average of the numbers $33, 38, 49, 45, 40$ without calculating directly, we can roughly estimate it by identifying a central value.

Step 1: Arrange the numbers for clarity: $33, 38, 40, 45, 49$.

Step 2: Observe that these numbers range from $33$ to $49$, with $40$ being the middle value.

The numbers $38, 40, 45$ are tightly centered around $40$, suggesting that the average would be closer to $40$.

Step 3: Evaluate the closest given choices. The choice of $41$ aligns closely with our estimation, just slightly above $40$ due to the higher numbers $45$ and $49$. This suggests a slight increase around the middle.

Therefore, the approximate average of the numbers is $41$.

To solve this problem, we'll visually estimate a balance of the numbers provided:

1. First, identifying the extremities: - The lowest number is $106$ - The highest number is $199$.
2. The middle range involves $126$, $133$, and $176$, which cluster around $148$.
3. Although $199$ and $106$ stretch the average, $133$ and $126$ counteract $176$'s high point.
4. Thus, balancing these extremities by approximating allows $148$ to emerge as a logical middle value.
5. Therefore, the estimated average that best represents the centrality of these numbers is $148$.

Accordingly, we choose 148 as the best estimate for the average, which is represented by Choice 3: $148$.

To solve this problem, let's first quickly glance at the numbers given: $10, 80, 96, 84, 95$.

Since calculating the precise average is not required, we can estimate by observing:

• One number is relatively small: $10$.
• The other numbers, $80, 96, 84, 95$, are all fairly close to each other and significantly larger than $10$, indicating that the average will lean towards the higher values.

Examining the choices, which are $11, 9, 73, 97$:

• $73$ is a choice that logically fits between the smallest and largest numbers in our list.
• $11, 9$, and $97$ appear to be far from typical average values given the numbers provided.

As $73$ is nearer to the typical values seen in the majority of the list, \textbf{73} would be the most reasonable choice for the average.

Therefore, without full computation, the most sensible choice for the average of these numbers is $\textbf{73}$.

To solve this problem, we'll employ reasonable estimation and logic:

• Step 1: Identify the range of numbers, which is from 7 to 20.
• Step 2: Consider that the average (mean) is typically near the middle of the list when the numbers are evenly distributed.
• Step 3: Identify the number configuration: $7, 8, 11, 14, 20$. Notice $11$ is close to the middle, but $20$ as an outlier pulls the average higher.

Reviewing the numbers:

• 11 and 14 are central, flanked by 8 and 7 on the lower side, and 20 on the upper side.
• Hence, guess an average slightly above the median (11).

Since we're not calculating, a reasonable average looking at the numbers and distribution is $12$.

The solution to the problem is thus $12$.

To solve this problem, let us examine the range of the numbers given:

The smallest number in the set is $10$, and the largest is $14$. Hence, any average of these numbers must fall within this range, i.e., between 10 and 14.

Considering the choices:

• $12$ is within the range $10$ to $14$.
• $9$ is outside of this range.
• $13$ is within the range $10$ to $14$.
• $11$ is within the range $10$ to $14$.

Thus, it is clear that the number that cannot be the average of $(10, 11, 12, 13, 14)$ is $9$, as it does not lie within the required range.

To solve this problem, we'll follow these steps:

• Step 1: Calculate the sum of the numbers.
• Step 2: Find the average by dividing this sum by the number of numbers.
• Step 3: Compare this calculated average to the given options to identify which cannot be the average.

Now, let's work through each step:
Step 1: Calculate the sum of the numbers: $8 + 7 + 5 + 10 = 30$.
Step 2: Find the average by dividing the sum by the total number of numbers: $\frac{30}{4} = 7.5$.
Step 3: Compare the calculated average to the options provided:

• Option 1: $6$ - not equal to $7.5$.
• Option 2: $8$ - not equal to $7.5$, but close enough to an integer comparison standard.
• Option 3: $12$ - clearly not possible since $7.5 < 12$.
• Option 4: $9$ - not equal to $7.5$.

Of all these options, the choice that cannot possibly be the average based on the numbers given is $12$. Therefore, the answer is $12$.