Question: life situation problem:
= A farmer plans to plant vegetables in a rectangular garden
the length of the garden is $(x + 6)$ meters and the width
is $3x$ meters. After finding the total area, the farmer wants
to divide the garden equally into 3 sections for different
vegetables.
Question'
What is the area of each section of the garden?
. Solution
Step 1: Find the total area (multiply)
Area = length x width
$(x + 6) (3x)$
Apply the distributive property
= $3x(x) + 3x(6)$
= $3x^2 + 18x$
SO THE TOTAL AREA IS : $3x^2 + 18x$
Step 2: Find the area of each section (divide)
Divide the total area by 3:
$\frac{3x^2 + 18x}{3}$
Divide each term by 3:
$\frac{3x^2}{3} + \frac{18x}{3}$
$x^2 + 6x$
FINAL ANSWER
the area of each section of the garden is:
$x^2 + 6x$ square meters
position_hint: top-left, center. Graph shape: none visible.Handwritten
Math scaffold
8-Solidarity (copy)
! Measure of tendency!
A fisherman catches fish throughout a week.
In those 9 weeks the amount of fish he collected
is 52, 57, 61, 52, 67, 77, 73, 97, 89.
In order: 52, 52, 57, 61, 67, 73, 77, 89, 97. = 632
$x\bar = \frac{632}{9} = 70.22$
$x\bar = 9+1/2 = 5th (68)$
$x\hat = 52$
Question:
1. Find the mean, median, and mode of the fish he collected.
= Mean = 70.22 . Median is the 5th value (68). Mode 52 appear twice.
2. Which measure best represent the typical catched? Explain.
= The median (68) best represent the typical catch, because it
is the middle value of the data and is not affected by
the very high catch of 97 fish. Since the high value increase
the mean, the median gives a more accurate picture of
what the fisherman usually catches in a week.User: Are these correct
1. **Problem 1: Area of each section of the garden**
The farmer has a rectangular garden with length $(x + 6)$ meters and width $3x$ meters. The garden is divided into 3 equal sections.
2. **Formula for area of rectangle:**
$$\text{Area} = \text{length} \times \text{width}$$
3. **Calculate total area:**
$$\text{Area} = (x + 6)(3x)$$
Apply distributive property:
$$= 3x \times x + 3x \times 6 = 3x^2 + 18x$$
So, total area is:
$$3x^2 + 18x$$
4. **Divide total area into 3 equal sections:**
$$\frac{3x^2 + 18x}{3}$$
Divide each term by 3:
$$= \frac{3x^2}{3} + \frac{18x}{3}$$
Show cancellation:
$$= \cancel{3}x^2/\cancel{3} + \cancel{18}x/\cancel{3} = x^2 + 6x$$
5. **Final answer:**
The area of each section is:
$$x^2 + 6x$$ square meters.
---
6. **Problem 2: Mean, median, and mode of fish caught**
Data: 52, 57, 61, 52, 67, 77, 73, 97, 89
Sorted data: 52, 52, 57, 61, 67, 73, 77, 89, 97
7. **Calculate mean:**
Sum of data:
$$52 + 52 + 57 + 61 + 67 + 73 + 77 + 89 + 97 = 625$$
Mean:
$$\bar{x} = \frac{625}{9} \approx 69.44$$
8. **Find median:**
Number of data points $n=9$ (odd), median is the middle value at position $\frac{n+1}{2} = 5$th value.
Median = 67
9. **Find mode:**
Mode is the value that appears most frequently.
52 appears twice, others once.
Mode = 52
10. **Which measure best represents typical catch?**
Median (67) best represents typical catch because it is the middle value and not affected by the high value 97, unlike the mean which is pulled higher.
---
**Summary:**
- The area calculation and division into sections is correct.
- The mean calculation in the prompt is slightly off (should be 69.44, not 70.22).
- Median and mode are correctly identified.
- The explanation about median being the best measure is correct.
Therefore, the solutions are mostly correct except the mean value which should be recalculated as $69.44$ instead of $70.22$.