PLEASE HELP I WILL FAIL WITHOUT THIS
1 answer:
Answer:
In the numbers mumbo jumbo, 76 is the minimum and 89 is the maximum. So it can look like this. (Fig A)
If you understood Fig A, the rest is easy.
There are one 76s, one 77s, four 78s, no 79s, four 80s, five 81s, no 82s, five 84s, four 85s, one 86s, two 87s, one 88s, and two 89s in the mumbo jumbo.
So the completed dot plot would look like Fig B. (I drew this in the computer and the dots aren't presicely balanced, but you'll get the idea)
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Answer and Step-by-step explanation:
Suppose that we have a number y which is a positive integer and that:
y =
Where;
= digit at 10⁰ => one's place (units place)
= digit at 10¹ => 10's place (tens place)
= digit at 10² => 100's place (hundreds place)
= digit at 10³ => 1000's place (thousands place)
.
.
.
= digit at 10ⁿ place
Then;
y = * 10⁰ +
* 10¹ +
* 10² +
* 10³ +
* 10⁴ +
* 10⁵ + . . . +
* 10ⁿ
<em>Since 10⁰ = 1, let's rewrite y as follows;</em>
y = +
* 10¹ +
* 10² +
* 10³ +
* 10⁴ +
* 10⁵ + . . . +
* 10ⁿ
Now, to test if y is divisible by 11, replace 10 in the equation above by -1. Since 10 divided by 11 gives -1 (mod 11) [mod means modulus]
y = +
* (-1)¹ +
* (-1)² +
* (-1)³ +
* (-1)⁴ +
* (-1)⁵ + . . . +
* (-1)ⁿ
=> y = -
+
-
+
-
+ . . . +
(-1)ⁿ (mod 11)
Therefore, it can be seen that, y is divisible by 11 if and only if alternating sum of its digits -
+
-
+
-
+ . . . +
(-1)ⁿ is divisible by 11
<em>Let's take an example</em>
Check if the following is divisible by 11.
i. 1859
<em>Solution</em>
1859 is divisible by 11 if and only if the alternating sum of its digit is divisible by 11. i.e if (1 - 8 + 5 - 9) is divisible by 11.
1 - 8 + 5 - 9 = -11.
Since -11 is divisible by 11 so is 1859
ii. 31415
<em>Solution</em>
31415 is divisible by 11 if and only if the alternating sum of its digit is divisible by 11. i.e if (3 - 1 + 4 - 1 + 5) is divisible by 11.
3 - 1 + 4 - 1 + 5 = 10.
Since 10 is not divisible by 11 so is 31415 not divisible.