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horsena [70]
3 years ago
5

At the beginning of the spelling bee there were 60 participants. One-third of the participants were eliminated during the first

round. One-fourth of the remaining participants were eliminate during the second round. One-sixth of the remaining participants were eliminated during the third round. How many participants were still cmpeting at the beginning of the fourth round?
Mathematics
1 answer:
lesya [120]3 years ago
3 0
Participants = 60
.
First round:
1/3 of 60 = 20
or
60 x 1/3
= 60/3
= 20
60 - 20 = 40
There are 20 participants that are eliminated on the first sound and 40 participants to move on to the second sound.
.
Participants = 40
Second round:
1/4 of 40 = 10
or
40 x 1/4
= 40/4
= 10
40 - 10 = 30
There are 10 participants that are eliminated on the second round and 30 participants that is moving on to the next round.
.
Participants = 30
Third round:
1/6 of 30 = 5
or
30 x 1/6
= 30/6
= 5
30 - 5 = 25
There are 5 participants that are eliminated on the third round and 25 participants to move on to the fourth round.
:)
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When is a zero not significant?
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Answer: A. When it comes after the decimal point, as in 4.000

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The sequence$$1,2,1,2,2,1,2,2,2,1,2,2,2,2,1,2,2,2,2,2,1,2,\dots$$consists of $1$'s separated by blocks of $2$'s with $n$ $2$'s i
kicyunya [14]

Consider the lengths of consecutive 1-2 blocks.

block 1 - 1, 2 - length 2

block 2 - 1, 2, 2 - length 3

block 3 - 1, 2, 2, 2 - length 4

block 4 - 1, 2, 2, 2, 2 - length 5

and so on.


Recall the formula for the sum of consecutive positive integers,

\displaystyle \sum_{i=1}^j i = 1 + 2 + 3 + \cdots + j = \frac{j(j+1)}2 \implies \sum_{i=2}^j = \frac{j(j+1) - 2}2

Now,

1234 = \dfrac{j(j+1)-2}2 \implies 2470 = j(j+1) \implies j\approx49.2016

which means that the 1234th term in the sequence occurs somewhere about 1/5 of the way through the 49th 1-2 block.

In the first 48 blocks, the sequence contains 48 copies of 1 and 1 + 2 + 3 + ... + 47 copies of 2, hence they make up a total of

\displaystyle \sum_{i=1}^48 1 + \sum_{i=1}^{48} i = 48+\frac{48(48+1)}2 = 1224

numbers, and their sum is

\displaystyle \sum_{i=1}^{48} 1 + \sum_{i=1}^{48} 2i = 48 + 48(48+1) = 48\times50 = 2400

This leaves us with the contribution of the first 10 terms in the 49th block, which consist of one 1 and nine 2s with a sum of 1+9\times2=19.

So, the sum of the first 1234 terms in the sequence is 2419.

8 0
2 years ago
Solve <br>|-4b-8|+|-1-b^2|+2b^3. b= -2<br>show steps please .
evablogger [386]
Ok, first put in the -2 for each b. That gives:
|-4(-2)-8|+|-1(-(-2))^2|+2(-2)^3
Let's do each section.
The first section is |-4(-2)-8)|
-4 times -2 is 8, minus 8 is 0. The absolute value of 0 is still 0.
Now we move on to |-1(-(-2))^2)|
First we do exponents
-(-2) is 2, and 2^2 is 4. 4 times -1 is -4. The absolute value of -4 is 4
Now the last section, 2(-2)^3
Exponents first: (-2)^3 is -2 * -2 * -2, which is -8.
-8*2=-16.
0+4+(-16)=-12
7 0
3 years ago
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