Find the distance between (-1,4) and (-1,12)

Anarel [89]

You are looking for the average. To find the average, one adds all terms and divides by the number of terms.
((1/2)+(5/6))/2
((3/6)+(5/6))/2
(8/6)/2
(4/6)
(2/3)
The answer is B) 2/3.

6 0

3 years ago

NEED ANSWERED ASAP WILL REWARD BRAINLIEST

Luda [366]

Answer:

The sum of the first 100 terms is 60400

Step-by-step explanation:

* Lets revise the arithmetic sequence

- There is a constant difference between each two consecutive

numbers

- Ex:

# 2 , 5 , 8 , 11 , ……………………….

# 5 , 10 , 15 , 20 , …………………………

# 12 , 10 , 8 , 6 , ……………………………

* General term (nth term) of an Arithmetic sequence:

- U1 = a , U2 = a + d , U3 = a + 2d , U4 = a + 3d , U5 = a + 4d

- Un = a + (n – 1)d, where a is the first term , d is the difference

between each two consecutive terms n is the position of the

number

- The sum of first n terms of an Arithmetic sequence is calculate from

Sn = n/2[a + l], where a is the first term and l is the last term

* Now lets solve the problem

- We will use method (1)

- From the table the terms of the sequence are:

10 , 22 , 34 , 46 , 58 , 82 , 94 , ............., where 10 is the first term

∵ an = a1 + (n - 1) d ⇒ explicit formula

∵ a1 = 10 and a2 = 22

∵ d = a2 - a1

∴ d = 22 - 10 = 12

- The 100th term means the term of n = 100

∴ a100 = 10 + (100 - 1) 12

∴ a100 = 10 + 99 × 12 = 10 + 1188 = 1198

∴ The 100th term is 1198

- Lets find the sum of the first 100 terms of the sequence

∵ Sn = n/2[a1 + an]

∵ n = 100 , a = 10 , a100 = 1198

∴ S100 = 100/2[10 + 1198] = 50[1208] = 60400

* The sum of the first 100 terms is 60400

7 0

3 years ago

Helpppp!!!!!!!!! 1.Which of the following is in the solution set of

Flauer [41]

The answers are D, D, 93, A, A.

4 0

4 years ago

Choose the correct answer below. A. Increasing the level of confidence has no effect on the interval. B. Increasing the level of

murzikaleks [220]

Answer:

$ME = t_{\alpha/2} \frac{s}{\sqrt{n}}$

And the width for the confidence interval is given by:

$Width =2ME=2t_{\alpha/2} \frac{s}{\sqrt{n}}$

And we want to see the effect if we increase the confidence level for a interval. On this case if we increase the confidence level then the critical value for the confidence interval $t_{\alpha/2}$ would be higher and then the width of the interval would increase. So then the best answer for this case would be:

B. Increasing the level of confidence widens the interval.

Step-by-step explanation:

Let's assume that we have a parameter of interest $\mu$ who represent for example the true mean for a population. And we can construct a confidence interval in order to estimate this parameter if we know the distribution for the statistic let's say $\bar X$ and for this particular example the confidence interval is given by:

$\bar X \pm ME$

Where ME represent the margin of error for the estimation and this margin of error is given by:

$ME = t_{\alpha/2} \frac{s}{\sqrt{n}}$

And the width for the confidence interval is given by:

$Width =2ME=2t_{\alpha/2} \frac{s}{\sqrt{n}}$

And we want to see the effect if we increase the confidence level for a interval. On this case if we increase the confidence level then the critical value for the confidence interval $t_{\alpha/2}$ would be higher and then the width of the interval would increase. So then the best answer for this case would be:

B. Increasing the level of confidence widens the interval.

3 0

3 years ago

Which of the following is a trinomial with a constant term? A. 4x4 + 2x2 - 2x B. -3 + 4x4 C. 2x2 - 5 + x D. 4x4 - 3x2 + x

castortr0y [4]

C. 2x² - 5 + x is a trinomial with a constant term
That polynomial has three terms: (2x²), (-5) and (x)
Term (-5) contains no variable, so this is a constant term.

7 0

3 years ago