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How to display the prime numbers that are less than the (limit) in ascending order by using the function showPrimes() and parameter (limit) [closed]

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<script type = "text/javascript">
function showPrimes(limit) {
for (let number = 2; number <= limit; ++number) {
let isPrime = true;
}
}
</script>
Above the code, I am currently working on so far. Not sure how to continue.

function getPrimes(limit) {
var sieve = [], i, j, primes = [];
for (i = 2; i <= limit; ++i) {
if (!sieve[i]) {
primes.push(i);
for (j = i << 1; j <= limit; j += i) {
sieve[j] = true;
}
}
}
return primes;
}
console.log(getPrimes(100));

Uncaught SyntaxError: missing ; after for-loop initializer, but I can't see where it's missing [closed]

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I just started learning javascript, and I want to make an online lcm hcf interactive teaching tool, but I keep getting this error pointed out below, and I can't see where it's missing
$("document").ready(function(){
function start() {
var firstNum = document.getElementById("first-num");
var secondNum = document.getElementById("second-num");
var primeList1 = [];
var primeList2 = [];
var primes = [];
var maxPrime = math.max(firstNum, secondNum) / 2 + 1;
**for (int num = 2; num < maxPrime; num++) {** <--- this line has the error
for (int i = 2; num < i; i++)
if (num % i == 0) {
break;
} else {
primes.add(num);
};
};
};

There is no int keyword in JavaScript.
You need to use var or let to declare and initialize your num and i variables
for (let num = 2; num < maxPrime; num++) {
for (let i = 2; num < i; i++)

How can I get the average of an array using a function? [closed]

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Quiz instruction:
Code a function that takes an array of numbers as a parameter and returns the average of them as the result.
Examples: average([2,2,5,8]) returns 4.25. average([59,60,40]) returns 53.
Code so far:
function average(array) {
var denominator = array.length;
var numbers = array;
for( var i = 0; i < array.length; i++) {
var total = numbers[0] + numbers[1] + numbers[2];
}
var avg = total / denominator;
return avg;
console.log( avg );
}
average([1,2,3]);
//Returning error. Not sure where the mistake is. Also, what would the code be if you were to expand the array and still get the average, no matter the inputs in the array parameter.
//Please help. Thanks!

Here is your average function;
function average(numbers) {
let denominator = numbers.length;
let sum = 0;
for (let i = 0; i < denominator; i++) {
sum = sum + numbers[i];
}
return sum / denominator;
}
console.log(average([1, 2, 3]));
console.log(average([2, 2, 5, 8]));
console.log(average([59, 60, 40]));

let average = (array) => array.reduce((a, b) => a + b) / array.length;
console.log(average([5,10,30]));

Please refer below url for calculating sum. and then find average by dividing this sum amount with array.length
https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/Array/reduce

You can do something like that:
function average(nums) {
return nums.reduce((a, b) => (a + b)) / nums.length;
}
average([1,2,3]);

JavaScript Odd Number Sorter Coding Challenge [closed]

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Ok so I'm working on learning JavaScript and I came across a Coding Challenge question which I can not solve. It gives me the outline/skeleton of the format I should use and wants me to find the odd numbers from 1-5000.
Starting with the basic function given below, write a function called sumOddNumbers that will print to the console and return the sum of all the odd numbers from 1 to 5000. Consider using a loop, and don't forget to call the function afterwards!
~ Format ~
function sumOddNumbers() {
var sum = 0;
// Your code here
console.log(sum);
return sum;
}

There is no need for a loop:
console.log(5000**2/4);
If 5000 is a dynamic input to your function, then the formula is as follows:
function sumOddNumbers(n) {
return (n + n%2)**2/4;
}
console.log(sumOddNumbers(5000));
If really it has to be done with a loop, and according to the template (which is not very nice BTW):
function sumOddNumbers() {
var sum = 0;
for (let i = 1; i <= 5000; i+=2) {
sum += i;
}
console.log(sum);
return sum;
}
sumOddNumbers();

let sum = 0;
for (let i = 1; i < 5000; i += 2) {
sum += i;
}

let sum = 0;
for (let i = 0; i < 5000; i++) {
if (i % 2 !== 0) {
sum += i;
}
}

function sumOddNumbers(max) {
let sum = 0;
for(let i = 1; i <= max; i+=2){
sum += i;
}
console.log(sum);
return sum;
}
sumOddNumbers(5000);

Find the next highest number in array [closed]

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I have an array of numbers and a value. I'd like to find the numbers around the value. The most important is the highest value but I need both.
var array = [2,4,5,8,11];
var value = 9;
The result I need should be : 8 and 11.
How can I do that?
Thanks.

Just write a simple for-loop searching for each of the values.
So for example:
function findNumbers(arr, num){
var low = num + 1; //largest number lower than num, set initial value above num
var high = num - 1; //smallest number higher than num, set initial value below num
for(var i = 0; i < arr.length; i ++){
if(arr[i] > low && arr[i] < num){
low = arr[i];
}
if(arr[i] < high && arr[i] > num){
high = arr[i];
}
}
//optional check to see if low/high exists
if(low > num){
low = -1; //not found
}
if(high < num){
high = -1; //not found
}
return [low,high];
}
and there you go. This works whether or not arr is sorted.

This help you :
<!DOCTYPE html>
<html>
<head>
</head>
<body>
<script>
var array = [2,4,5,8,11];
var value = 9;
var min,max;
fun();
function fun() {
var sortArray = array.sort(function(a,b){return a-b});
for(var i=0; i<sortArray.length;i++) {
if (sortArray[i] < value)
min = sortArray[i];
if (sortArray[i] > value)
max = sortArray[i];
}
alert(min + " , " + max);
}
</script>
</body>
</html>

Looking at your question looks like you are still learning jquery.However I am sure you will get good suggestions here.
Here is an example with your test case,I am leaving the explanation on you so that you can study each line and learn.
var myArray = [2, 4, 5, 8, 11];
var myValue = 9;
function BiggerThan(inArray) {
return inArray > myValue;
}
function LesserThan(inArray) {
return inArray < myValue;
}
var arrBiggerElements = myArray.filter(BiggerThan);
var nextElement = Math.min.apply(null, arrBiggerElements);
alert(nextElement);
var arrLesserElements = myArray.filter(LesserThan);
var prevElement = Math.max.apply(null, arrLesserElements);
alert(prevElement);