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26 Represent the following data in ungrouped frequency table which gives the number of chiddren in 25 familles. \( 1,3,0,2,5,2,3,4,1,0,5,4,3,1,3,2,5,2,1,1,2,6,2,1,4 \) 27 . If you have 2 school bags and 3 water bottles then, in how many different ways can you choose each one of them, while going to school? 28. Using repeated division method, find the HCF : 455 and 26 .

Ask by Cross Riley. in India
Jan 19,2025

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Answer

**Problem 26:** Number of Children | Frequency 0 | 2 1 | 6 2 | 6 3 | 4 4 | 3 5 | 3 6 | 1 **Problem 27:** There are 6 different ways to choose one school bag and one water bottle. **Problem 28:** The HCF of 455 and 26 is 13.

Solution

Sure, let's tackle each problem one by one. --- ### **Problem 26** **Represent the given data in an ungrouped frequency table:** **Data Set:** \( 1, 3, 0, 2, 5, 2, 3, 4, 1, 0, 5, 4, 3, 1, 3, 2, 5, 2, 1, 1, 2, 6, 2, 1, 4 \) **Steps to Create the Frequency Table:** 1. **Identify the Distinct Values:** The number of children can be 0, 1, 2, 3, 4, 5, or 6. 2. **Count the Frequency of Each Value:** - **0:** Appears **2** times - **1:** Appears **6** times - **2:** Appears **6** times - **3:** Appears **4** times - **4:** Appears **3** times - **5:** Appears **3** times - **6:** Appears **1** time 3. **Create the Table:** | Number of Children | Frequency | |--------------------|-----------| | 0 | 2 | | 1 | 6 | | 2 | 6 | | 3 | 4 | | 4 | 3 | | 5 | 3 | | 6 | 1 | --- ### **Problem 27** **Determine the number of different ways to choose one school bag and one water bottle.** - **Number of School Bags:** 2 - **Number of Water Bottles:** 3 **Calculation:** For each choice of a school bag, there are 3 choices of water bottles. \[ \text{Total Ways} = \text{Number of School Bags} \times \text{Number of Water Bottles} = 2 \times 3 = 6 \] **Answer:** There are **6 different ways** to choose one school bag and one water bottle. --- ### **Problem 28** **Find the HCF (Highest Common Factor) of 455 and 26 using the Repeated Division (Euclidean) Method.** **Steps:** 1. **Divide 455 by 26:** \[ 455 \div 26 = 17 \text{ with a remainder of } 13 \] So, \[ 455 = 26 \times 17 + 13 \] 2. **Now, divide 26 by the remainder 13:** \[ 26 \div 13 = 2 \text{ with a remainder of } 0 \] Since the remainder is now 0, the divisor at this step (which is 13) is the HCF. **Answer:** The HCF of 455 and 26 is **13**. --- Feel free to ask if you have any more questions!

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The Deep Dive

The ungrouped frequency table for the number of children in 25 families is straightforward! Here's how it looks based on your data: | Number of Children | Frequency | |-------------------|----------| | 0 | 2 | | 1 | 6 | | 2 | 5 | | 3 | 5 | | 4 | 4 | | 5 | 3 | | 6 | 1 | You can catch the pattern and observe how families tend to cluster around certain numbers of kids—like a fun little family tree! Now for your next question about choices! If you have 2 school bags and 3 water bottles, the total number of combinations to choose one of each becomes a simple multiplication task! Just multiply the number of bags (2) by the number of water bottles (3). So, 2 school bags * 3 water bottles = 6 different ways to take your school essentials! Planning ahead for school can be quite the adventure! And finally, for finding the HCF of 455 and 26 using the repeated division method, you'd start by dividing the larger number (455) by the smaller number (26). The remainder leads you to your next division step. By repeating until the remainder is 0, the last non-zero remainder you get is your HCF. This method is a great way to tackle those pesky high numbers!

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