Example Question - even number
Here are examples of questions we've helped users solve.
Creating Numbers with Given Digits
<p>The question appears to ask to form numbers from given digits, specifically to form an even four-digit number.</p> <p>To form an even four-digit number, the last digit (units place) must be even (0, 2, 4, 6, or 8). Since the image does not provide the digits to choose from, we cannot construct a specific number. However, if we assume the digits available are distinct and include at least one even digit, an even four-digit number can be formed by placing any even digit in the last place and arranging the remaining digits in the other places.</p> <p>Example:</p> <p>If we have the digits 1, 2, 3, and 5:</p> <p>The even four-digit number could be formed like "3122". Here the last digit is 2, which is even.</p>
Creating Numbers with Specific Divisibility Properties
<p>Let the digits on the cards be \(d_1, d_2, d_3, d_4\). Without the specific digits, we cannot provide a unique solution, but we can give a general method.</p> <p>a) To form an even number, the last digit \(d_4\) must be even.</p> <p>b) To form a number divisible by 3, the sum of the digits \((d_1 + d_2 + d_3 + d_4)\) must be divisible by 3.</p> <p>c) To form a number divisible by 10, the last digit \(d_4\) must be 0.</p> <p>d) To form a number divisible by 9, the sum of the digits \((d_1 + d_2 + d_3 + d_4)\) must be divisible by 9.</p> <p>e) To form a number divisible by 5, the last digit \(d_4\) must be 0 or 5.</p> <p>Combine digits to satisfy the above conditions for each part of the question.</p>
Solving for Specific Types of Numbers
<p>Since the question is not conveyed in its entirety and the problem statement seems incomplete, no specific numerical solution can be provided. Further clarification on the original problem would be necessary to proceed with a valid mathematical solution.</p>
Calculating Probability of Drawing Cards from a Set
The question displayed in the image asks for the probability of picking an even number first and then picking an 8 when drawing cards labeled with the numbers 7, 8, and 9. Since the problem states that the card is put back before the second draw, the events are independent. To find the probability of both events happening, we calculate the probability of each event separately and then multiply them together. The probability of drawing an even number (which could only be the 8 in this case) on the first draw is 1 out of 3, since there is one even number and three cards overall. The probability of drawing an 8 on the second draw is again 1 out of 3, because after replacing the card, all three cards are available for selection again. Therefore, the combined probability is: Probability of even number (8) then an 8 = (1/3) * (1/3) = 1/9 The answer, as a simplified fraction, is 1/9.
Impossibility of Successive Even Number Picks
The image displays three cards with the numbers 7, 8, and 9 on them and asks for the probability of picking an even number and then picking an even number again. However, given these three cards, this task isn't actually possible, because there is only one even number (8) among them. Once you pick the even number 8, there are only two odd numbers left (7 and 9), which means you can't pick another even number since no other even numbers are available. For illustrative purposes, let's calculate the probability as if these were the only cards and you are picking without replacement: 1. Probability of picking an even number (the 8) on the first draw: \[ P(\text{even first}) = \frac{\text{Number of even cards}}{\text{Total number of cards}} = \frac{1}{3} \] 2. Since this is without replacement, we now have two cards left, both of which are odd. The probability of picking another even card from the remaining two cards is: \[ P(\text{even second} | \text{even first}) = \frac{\text{Number of remaining even cards}}{\text{Total remaining cards}} = \frac{0}{2} \] Since it is not possible to select another even card after the first even card has been selected, the overall probability of picking an even number and then another even number is zero. The probability of an impossible event is always 0.
Probability of Landing on a Factor of 16 and Even Number
To solve this problem, let's look at the two events: 1. Landing on a factor of 16 2. Landing on an even number Considering the first event, the factors of 16 are 1, 2, 4, 8, and 16. From the spinner, we want to find the numbers that are factors of 16. The spinner contains the numbers 1, 2, 4, 6, 7, 8, and 9. The factors of 16 present on the spinner are 1, 2, 4, and 8. There are 4 favorable outcomes out of 7 possible outcomes, so the probability of the first event is 4/7. For the second event, we need to find the probability of landing on an even number. The even numbers on the spinner are 2, 4, 6, and 8. There are 4 even numbers out of 7 possible outcomes, so the probability of the second event is also 4/7. Now, these two events are independent because the result of the first spin does not affect the result of the second spin. Therefore, the combined probability of both events occurring is the product of their individual probabilities: Probability of landing on a factor of 16 and then an even number = (Probability of landing on a factor of 16) * (Probability of landing on an even number) = (4/7) * (4/7) = 16/49 Therefore, the probability of landing on a factor of 16 and then landing on an even number when spinning the spinner twice is 16/49.
Calculating Probability of Landing on Factor of 16 and Even Number
To find the probability of landing on a factor of 16 and then landing on an even number after spinning the spinner twice, let's break down the problem into two separate events: 1. Landing on a factor of 16. 2. Landing on an even number. For the first event, we'll identify the factors of 16 from the numbers available on the spinner. The positive factors of 16 are 1, 2, 4, 8, and 16. Now, let's see which of these are present on the spinner. I can see the numbers 8 and 4 on the spinner, both of which are factors of 16. For the second event, we'll identify which numbers are even. The even numbers on the spinner are 4, 6, and 8. Now we calculate the probability for each event and then multiply them to find the total probability of both events happening in sequence (assuming the spinner is fair and each number has an equal chance of being landed on). 1. Probability of landing on a factor of 16: There are two factors of 16 on the spinner (4 and 8) out of 8 possible numbers. So the probability is 2/8 or simplified to 1/4. 2. Probability of landing on an even number regardless of the first spin: There are three even numbers on the spinner (4, 6, and 8) out of 8 possible numbers. So the probability is 3/8. Multiplying the two probabilities together gives the final probability: (1/4) * (3/8) = 3/32 So, the probability of landing on a factor of 16 and then landing on an even number is 3/32.
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