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riadik2000 [5.3K]

3 years ago

15

Which statement is true of z11.7?

1 answer:

Alexxx [7]3 years ago

7 0

Answer

z11.7 is between 1 and 2 standard deviations of the mean.

Step-by-step explanation:

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Ben Teal lost his weighted

11Alexandr11 [23.1K]

15 pounds each and 16 watermelons= 240 pounds. (15x16=240)

7 0

3 years ago

Read 2 more answers

Which number is irrational? A. Syntax error from line 1 column 53 to line 1 column 60. Unexpected '0'. B. 1 over 9 C. square roo

DerKrebs [107]

C. because the decimal is never ending, so it is irrational.

8 0

3 years ago

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Lim. √2x -√3x-a/√x-√a<br>x approaches a

Ludmilka [50]

Answer: $\sqrt{2}a-\sqrt{3}a-2\sqrt{a}$

Step-by-step explanation:

$\lim _{x\to \:a}\left(\sqrt{2}x-\sqrt{3}x-\frac{a}{\sqrt{x}}-\sqrt{a}\right)$

$=\sqrt{2}a-\sqrt{3}a-\frac{a}{\sqrt{a}}-\sqrt{a}$

$=\sqrt{2}a-\sqrt{3}a-2\sqrt{a}$

3 0

3 years ago

Rewrite in simplest terms: 3h - 2(-0.1h - 0.8)<br>please help

levacccp [35]

Answer:

3.2h+1.6

Step-by-step explanation:

3h-2(-0.1h-0.8)

3h+0.2h+1.6

3.2h+1.6

8 0

3 years ago

Determine formula of the nth term 2, 6, 12 20 30,42

nalin [4]

Check the forward differences of the sequence.

If $\{a_n\} = \{2,6,12,20,30,42,\ldots\}$ , then let $\{b_n\}$ be the sequence of first-order differences of $\{a_n\}$ . That is, for n ≥ 1,

$b_n = a_{n+1} - a_n$

so that $\{b_n\} = \{4, 6, 8, 10, 12, \ldots\}$ .

Let $\{c_n\}$ be the sequence of differences of $\{b_n\}$ ,

$c_n = b_{n+1} - b_n$

and we see that this is a constant sequence, $\{c_n\} = \{2, 2, 2, 2, \ldots\}$ . In other words, $\{b_n\}$ is an arithmetic sequence with common difference between terms of 2. That is,

$2 = b_{n+1} - b_n \implies b_{n+1} = b_n + 2$

and we can solve for $b_n$ in terms of $b_1=4$ :

$b_{n+1} = b_n + 2$

$b_{n+1} = (b_{n-1}+2) + 2 = b_{n-1} + 2\times2$

$b_{n+1} = (b_{n-2}+2) + 2\times2 = b_{n-2} + 3\times2$

and so on down to

$b_{n+1} = b_1 + 2n \implies b_{n+1} = 2n + 4 \implies b_n = 2(n-1)+4 = 2(n + 1)$

We solve for $a_n$ in the same way.

$2(n+1) = a_{n+1} - a_n \implies a_{n+1} = a_n + 2(n + 1)$

Then

$a_{n+1} = (a_{n-1} + 2n) + 2(n+1) \\ ~~~~~~~= a_{n-1} + 2 ((n+1) + n)$

$a_{n+1} = (a_{n-2} + 2(n-1)) + 2((n+1)+n) \\ ~~~~~~~ = a_{n-2} + 2 ((n+1) + n + (n-1))$

$a_{n+1} = (a_{n-3} + 2(n-2)) + 2((n+1)+n+(n-1)) \\ ~~~~~~~= a_{n-3} + 2 ((n+1) + n + (n-1) + (n-2))$

and so on down to

$a_{n+1} = a_1 + 2 \displaystyle \sum_{k=2}^{n+1} k = 2 + 2 \times \frac{n(n+3)}2$

$\implies a_{n+1} = n^2 + 3n + 2 \implies \boxed{a_n = n^2 + n}$

6 0

2 years ago