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题目
题目

KNSFHP 1139.14 AU2025 (11088) Emergency Contact Info and Media Waiver v.2

多项填空题

By entering the last four (4) digits of my OSU ID (Note: OSU ID can be found on your BuckID.  It is the shorter of the two numbers)  [Fill in the blank],  and the Date [Fill in the blank],  I represent and certify that my true age is stated above.  Furthermore,  I acknowledge that I HAVE READ AND UNDERSTAND THE FOREGOING REQUEST AND RELEASE.  

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标准答案
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思路分析
The item presents a completion task with two blanks to fill in and does not include any multiple-choice options to evaluate. To proceed, we focus on what each blank is asking for based on the surrounding text. Blank 1: "By entering the last four (4) digits of my OSU ID ... [Fill in the blank]" - The instruction explicitly specifies that the blank should contain the last four digits of the OSU ID. This is a numeric identifier used to......Login to view full explanation

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Question text The students below each construct a sinusoidal function based on a shared scenario involving water level fluctuations.Match each student’s reasoning to the explanation that best fits their thinking. Scenario:The depth of water , in metres, at a canal dock fluctuates due to the system of canal locks lowering and raising the water for container ships. The water varies sinusoidally from a minimum of 1.5 m to a maximum of 4.5 m. The depth reaches its minimum at 5:00 AM, and one full cycle of the water level is completed every 4 hours. The horizontal axis represents time , in hours, where corresponds to midnight. Student 1:I was asked to find the water depth at 3 AM.The equation I created was Since the minimum occurs at 5:00 AM, I shifted the function by 5.Using this function, I found the depth at 3 AM was 1.5 m.Answer 1 Question 22[select: , All parameters are correct. Therefore, the initial evaluation is correct, and the correct depth at 3 AM is 1.5 m. , All parameters are correct except the phase shift: there should be a phase shift of -5. Therefore, the initial evaluation is incorrect; using the new equation, the correct depth at 3 AM is 4.5 m. , All parameters are correct except the k-value: the k-value should be pi/4. The initial evaluation is still correct; the depth at 3 AM is 1.5 m.] Student 2:I was asked to find the water depth at 6 AM.The equation I created was I chose sine because the water starts rising after its lowest point.Using my equation, I found the depth at 6 AM was 3 m.Answer 2 Question 22[select: , All parameters are correct except for the amplitude, which should be negative. The initial evaluation is still correct; the depth at 6 AM is 3 m., All parameters are except the phase shift: there should be a phase shift of -5. Therefore, the initial evaluation is incorrect; using the new equation, the correct depth at 6 AM is 4.1 m., All parameters are correct except the k-value; the k-value should be pi/2. The initial evaluation is still correct; the depth at 6 AM is 3 m. ] Student 3:I was asked to find the water depth at 1 PM.The equation I created was Since the minimum occurs at 5:00 AM, I used a cosine function with a shift to match the point.I found the depth at 1 PM to be 4.5 m.Answer 3 Question 22[select: , All parameters are correct except the k-value; the correct k-value is pi/4. The initial evaluation is still correct; the depth at 1 PM is 4.5 m., All parameters are correct except for the phase shift; there should be a phase shift of -3. Therefore, the initial evaluation is incorrect; using the new equation, the correct depth at 1 PM is 1.5 m.. , All parameters are correct and the student’s evaluation of the depth was correct; the depth at 1 PM is 4.5 m.]

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