Example Question - even numbers
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Solving Set Theory Problems with Roster Notation
<p>\text{There are five problems to solve in this question.}</p> <p>\text{1. Convert the set-builder notation } \{x | x \text{ is a natural number less than 5}\} \text{ to roster form. }</p> <p>\text{Solution: } \{1, 2, 3, 4\}</p> <p>\text{2. Express the set } G = \{x | x \text{ is an even number between 1 and 20}\} \text{ in roster form. }</p> <p>\text{Solution: } \{2, 4, 6, 8, 10, 12, 14, 16, 18, 20\}</p> <p>\text{3. Express the set } L = \{x | x \text{ is a letter in the word "MATHEMATICS"}\} \text{ in roster form. }</p> <p>\text{Solution: } \{'M', 'A', 'T', 'H', 'E', 'I', 'C', 'S'\}</p> <p>\text{4. Convert the set-builder notation } \{x | x \text{ is an odd integer greater than 5 and less than 15}\} \text{ to roster form. }</p> <p>\text{Solution: } \{7, 9, 11, 13\}</p> <p>\text{5. If } A = \{x | x \text{ is a prime number less than 10}\}, \text{ what is } A? </p> <p>\text{Solution: } \{2, 3, 5, 7\}</p>
Identifying Even-Digit Numbers
<p>The image does not contain a clear mathematical question that can be answered. It appears to be asking for numbers that have a certain property related to their digits, possibly being even, but without the full context or a specific question, no mathematical solution can be provided.</p>
Finding Probability of Picking Even Numbers from a Set of Cards
This problem is about finding the probability of picking an even number from a set of cards labeled with numbers from 1 to 7, and then, without replacing the first card, picking another even number. Let's determine the probability step by step. 1. The probability of picking an even number (2, 4, 6) on the first draw: There are 3 even numbers out of 7 total numbers, so the probability is \( \frac{3}{7} \). 2. The probability of picking another even number on the second draw: After one even card is removed, there are now 2 even numbers remaining, and only 6 cards in total to choose from. So the probability for the second draw is \( \frac{2}{6} \) which simplifies to \( \frac{1}{3} \). Now we multiply the probabilities of both events happening in sequence, which are independent in this context: \( \frac{3}{7} \times \frac{1}{3} = \frac{3}{21} \), which simplifies to \( \frac{1}{7} \). So, the probability of drawing an even number and then another even number without replacement is \( \frac{1}{7} \).
Impossible Event: Picking Even Numbers
The image shows the numbers 7, 8, and 9, and we are asked to calculate the probability of picking an even number and then picking another even number. Firstly, let's understand the scenario. We have three numbers: 7, 8, and 9. Out of these, only one number is even: 8. Since we are not replacing the first card before picking the second card, the probabilities will be different for each draw. The probability of picking the even number (which is 8) on the first draw is 1 out of 3, since there is only one even number out of three possibilities. This can be written as: \[ P(\text{first draw even}) = \frac{1}{3} \] After picking an even number on the first draw, there are now only two cards left, but since we've already picked the only even card, there are no even cards left. This means that the probability of picking an even number now is 0, as there are no even numbers to choose from. Therefore, the probability of picking an even number second time after already having picked an even number first is: \[ P(\text{second draw even} | \text{first draw even}) = 0 \] To find the combined probability of both events happening in sequence (first picking an even number, then picking another even number), we multiply the two individual probabilities together: \[ P(\text{first even and second even}) = P(\text{first draw even}) \times P(\text{second draw even} | \text{first draw even}) \] \[ P(\text{first even and second even}) = \frac{1}{3} \times 0 = 0 \] Thus, the probability of picking an even number and then picking another even number is 0, or simply impossible in this scenario.
Probability of Drawing Even and Prime Numbers from a Box of Cards
Dựa vào hình ảnh bạn cung cấp, đề bài là: "Một hộp đựng 20 tấm thẻ cùng loại được đánh số thứ tự 1; 2; 3; ...; 19; 20. Rút ngẫu nhiên một tấm thẻ. Tính xác suất của các biến cố sau: a) Số trên thẻ là số chẵn. b) Số trên thẻ là số nguyên tố." Để giải bài toán này, chúng ta cần phân tích từng phần một: a) Xác suất để rút được một số chẵn từ hộp thẻ: Có tổng cộng 10 số chẵn trong 20 số (2, 4, 6, ..., 20). Vậy xác suất để rút được một số chẵn là số số chẵn chia cho tổng số thẻ: P(số chẵn) = 10 / 20 = 1/2 b) Xác suất để rút được một số nguyên tố từ hộp thẻ: Trong 20 số đầu tiên, các số nguyên tố là 2, 3, 5, 7, 11, 13, 17, 19. Có tổng cộng 8 số nguyên tố. P(số nguyên tố) = 8 / 20 = 2/5 Vậy xác suất rút được số chẵn là 1/2 và xác suất rút được số nguyên tố là 2/5.
Mathematical Properties of Consecutive Numbers
Die folgenden Behauptungen werden einzeln überprüft: a. Die Summe von drei aufeinanderfolgenden Zahlen ist immer durch 3 teilbar. Diese Behauptung ist wahr. Nehmen wir drei aufeinanderfolgende Zahlen, bezeichnet als \( n \), \( n+1 \), und \( n+2 \) (wobei \( n \) eine ganze Zahl ist). Ihre Summe ist \( n + (n+1) + (n+2) = 3n + 3 = 3(n + 1) \). Da dies ein Produkt einer ganzen Zahl (\( n+1 \)) und der Zahl 3 ist, ist die Summe durch 3 teilbar. b. Zerlege die Zahl 48 in drei Summanden. Das Produkt dieser Zahlen ist immer eine gerade Zahl. Diese Behauptung ist ebenfalls wahr. Die Zahl 48 ist gerade, und jede Zerlegung in Summanden wird zumindest eine gerade Zahl beinhalten, weil die Summe dreier ungerader Zahlen immer ungerade wäre, und 48 ist gerade. Da das Produkt einer geraden Zahl mit beliebigen anderen Zahlen ebenfalls gerade ist, wird das Produkt der drei Summanden immer eine gerade Zahl sein, unabhängig davon, wie man die 48 zerlegt. c. Wenn \( x \) gerade ist, dann ist \( x^2 \) immer gerade. Diese Behauptung ist wahr. Wenn \( x \) gerade ist, lässt sich \( x \) als \( 2k \) schreiben, wobei \( k \) eine ganze Zahl ist. Das Quadrat von \( x \) ist dann \( (2k)^2 = 4k^2 = 2(2k^2) \), was wiederum die Form einer geraden Zahl (ein Vielfaches von 2) hat. Daher ist \( x^2 \) immer gerade, wenn \( x \) gerade ist.
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