Find the two square roots of 1/16

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

A digital camcorder repair service has set a goal not to exceed an average of 5 working days from the time the unit is brought i

garri49 [273]

Answer:

The degrees of freedom first given by:

$df=n-1=12-1=11$

Then we can find the critical value taking in count the degrees of freedom and the alternative hypothesis and then we need to find a critical value who accumulates 0.05 of the area in the right tail and we got:

$t_{\alpha}= 1.796$

And for this case the rejection region would be:

b) Reject H0 if tcalc >1.7960

Step-by-step explanation:

Information given

5, 7, 4, 6, 7, 5, 5, 6, 4, 4, 7, 5.

System of hypothesis

We want to test if the true mean is higher than 5, the system of hypothesis are :

Null hypothesis: $\mu \leq 5$

Alternative hypothesis: $\mu > 5$

The statistic is given by:

$t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}$ (1)

The degrees of freedom first given by:

$df=n-1=12-1=11$

Then we can find the critical value taking in count the degrees of freedom and the alternative hypothesis and then we need to find a critical value who accumulates 0.05 of the area in the right tail and we got:

$t_{\alpha}= 1.796$

And for this case the rejection region would be:

b) Reject H0 if tcalc >1.7960

6 0

4 years ago

Eli's Electronics has 60-inch television on sale this week for $798. The television is regularly priced for $1050. By what perce

umka2103 [35]

Answer: that would be 76% hope that helps

Step-by-step explanation:

3 0

3 years ago

The diagram shows a row of five identical tiles each in the shape of an isosceles trapezoid what is the area of one tile

Misha Larkins [42]

Get the total amount of the 5 tiles and divide it by 5.

5 0

4 years ago

How do I find the simplest form,for this problem 4-2(3+6)+2-2=

levacccp [35]

Try using PEMDAS,
Parentheses
Exponents
Multiplication
Division
Addition
Subtraction

So just do the problem in that order.

5 0

4 years ago

Find the two square roots of 1/16​

Find the two square roots of 1/16