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CIW 1D0-735 Exam - Topic 4 Question 69 Discussion

Actual exam question for CIW's 1D0-735 exam
Question #: 69
Topic #: 4
[All 1D0-735 Questions]

Consider the following code:

What change should be made to ensure that it correctly displays the value of name in all uppercase letters?

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Suggested Answer: B

by Dante at Sep 13, 2024, 04:00 AM

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Jade
5 months ago
D is just wrong, it's "string" not "sting"!
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Willie
5 months ago
Wait, is it really "new sting"? That seems off.
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Tammara
5 months ago
C won't work, NAME isn't defined.
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Alishia
5 months ago
I think B is incorrect, prototype isn't needed here.
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Malcom
6 months ago
A is definitely the right choice!
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Willis
6 months ago
Option C looks off to me; I don't recall using uppercase like that in document.write.
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Refugia
6 months ago
I remember a similar question where we had to manipulate strings, and I think changing line 2 is the right approach.
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Shawn
6 months ago
I'm not entirely sure, but option B seems wrong because prototype methods are not called like that.
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Mitsue
6 months ago
I think option A makes the most sense since it directly modifies the variable to be uppercase.
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Arlene
6 months ago
This looks like a good opportunity to demonstrate my understanding of string manipulation in JavaScript. I think option A is the correct answer, as it's the most straightforward way to convert the `name` variable to uppercase.
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Marshall
6 months ago
Okay, let's see. The issue is that the `name` variable is not being converted to uppercase before it's displayed. I think option A is the right solution - we just need to call `toUpperCase()` on the `name` variable.
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My
6 months ago
I'm a bit confused by this question. The code seems to be displaying the value of `name`, but it's not clear why it's not in uppercase. I'll need to think this through carefully.
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Maybelle
6 months ago
Hmm, this looks like a straightforward question about string manipulation. I think the answer is A, since the `toUpperCase()` method is the standard way to convert a string to uppercase.
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Tamala
7 months ago
I'm not sure about this one. The code seems simple enough, but I'm not confident I understand the difference between the options. I'll need to review the string methods in JavaScript to make sure I'm choosing the right solution.
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Zack
7 months ago
Okay, I've got this. The key is to leverage existing resources like CVDs, connect all assets on the network, and use the data to drive business outcomes. That's the most comprehensive approach to engaging customers and developing IoT opportunities.
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Annelle
11 months ago
I'm cracking up at the typo in Option D. 'sting' instead of 'string'? Somebody needs to proofread these exam questions!
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William
11 months ago
User 3: Definitely, that typo could confuse someone during the exam.
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Kati
11 months ago
User 2: I know right, they really need to proofread these questions better.
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Monroe
11 months ago
User 1: Option D is hilarious, 'sting' instead of 'string'!
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Alana
12 months ago
Option C looks promising, but I'm not sure if document.write() is the best way to display the value. Might be better to use innerHTML or textContent.
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Rashad
11 months ago
Option C looks promising, but I'm not sure if document.write() is the best way to display the value. Might be better to use innerHTML or textContent.
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Marlon
11 months ago
B) Line 2 should be changed to name .prototype . toUpperCase () ;
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Lynette
11 months ago
A) Line 2 should be changed to name = name . toUpperCase () ;
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Meghan
12 months ago
Haha, Option B is clearly a joke. Why would we need to call toUpperCase() on the prototype? That's just silly.
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Selma
11 months ago
Kimberlie: I think so too, calling toUpperCase() directly on the string variable makes more sense.
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Yesenia
11 months ago
User 3: Option A seems like the correct answer to me.
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Kimberlie
11 months ago
User 2: Yeah, it's definitely not the right choice.
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Tasia
11 months ago
User 1: I agree, Option B doesn't make any sense.
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Alecia
1 year ago
I think Option D is the way to go. We need to create a new String object to ensure the name is properly defined.
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Sena
11 months ago
User 2: Line 1 should be changed to var name = new String(\'Jaccb\').
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Micah
11 months ago
User 1: I think Option D is the way to go. We need to create a new String object to ensure the name is properly defined.
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Tanja
1 year ago
Option A is the correct answer. We need to call the toUpperCase() method on the name variable to convert it to uppercase.
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Daisy
1 year ago
I'm not sure, but I think the answer might be C.
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Lyndia
1 year ago
I disagree, I believe the correct answer is B.
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Marsha
1 year ago
I think the answer is A.
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