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3 years ago

Explain how to calculate the percent of change.

1 answer:

3 0

<span>work out the difference (increase) between the two numbers you are comparing.Increase = New Number - Original Number.Then: divide the increase by the original number and multiply the answer by 100.% increase = Increase ÷ Original Number × 100.</span>

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What are benchmarks when using rounding?

cricket20 [7]

Answer:

Benchmark can be defined as the standard or reference point against which something can be measured, compared, or assessed.

Step-by-step explanation:

7 0

3 years ago

If (a,1),(-2,b).(c,-3) are points on the straight line 2x + y = 3x - y +1. Find the values of a, b and c. THANK YOU FOR THR PERS

Lisa [10]

Answer:

it’s g ! i just turned it in and it’s g :)

6 0

2 years ago

Determine whether the sequences converge.

Alik [6]

$a_n=\sqrt{\dfrac{(2n-1)!}{(2n+1)!}}$

Notice that

$\dfrac{(2n-1)!}{(2n+1)!}=\dfrac{(2n-1)!}{(2n+1)(2n)(2n-1)!}=\dfrac1{2n(2n+1)}$

So as $n\to\infty$ you have $a_n\to0$ . Clearly $a_n$ must converge.

The second sequence requires a bit more work.

$\begin{cases}a_1=\sqrt2\\a_n=\sqrt{2a_{n-1}}&\text{for }n\ge2\end{cases}$

The monotone convergence theorem will help here; if we can show that the sequence is monotonic and bounded, then $a_n$ will converge.

Monotonicity is often easier to establish IMO. You can do so by induction. When $n=2$ , you have

$a_2=\sqrt{2a_1}=\sqrt{2\sqrt2}=2^{3/4}>2^{1/2}=a_1$

Assume $a_k\ge a_{k-1}$ , i.e. that $a_k=\sqrt{2a_{k-1}}\ge a_{k-1}$ . Then for $n=k+1$ , you have

$a_{k+1}=\sqrt{2a_k}=\sqrt{2\sqrt{2a_{k-1}}\ge\sqrt{2a_{k-1}}=a_k$

which suggests that for all $n$ , you have $a_n\ge a_{n-1}$ , so the sequence is increasing monotonically.

Next, based on the fact that both $a_1=\sqrt2=2^{1/2}$ and $a_2=2^{3/4}$ , a reasonable guess for an upper bound may be 2. Let's convince ourselves that this is the case first by example, then by proof.

We have

$a_3=\sqrt{2\times2^{3/4}}=\sqrt{2^{7/4}}=2^{7/8}$
$a_4=\sqrt{2\times2^{7/8}}=\sqrt{2^{15/8}}=2^{15/16}$

and so on. We're getting an inkling that the explicit closed form for the sequence may be $a_n=2^{(2^n-1)/2^n}$ , but that's not what's asked for here. At any rate, it appears reasonable that the exponent will steadily approach 1. Let's prove this.

Clearly, $a_1=2^{1/2}$ . Let's assume this is the case for $n=k$ , i.e. that $a_k$ . Now for $n=k+1$ , we have

$a_{k+1}=\sqrt{2a_k}$

and so by induction, it follows that $a_n$ for all $n\ge1$ .

Therefore the second sequence must also converge (to 2).

4 0

3 years ago

As Sally was shopping for a turkey tree for Thanksgiving, she looked at 48 trees. Of those, she found that 7/8 of them were too

djverab [1.8K]

Answer:

<h2>42 trees</h2>

Step-by-step explanation:

<h3>48/1 x 7/8 = 6/1 x 7/1 (Simplify 48 and 8 because they have the common factor of 8)=42/1= 42</h3>

8 0

3 years ago

Read 2 more answers

A sample of 2,000 union members was selected, and a survey recorded their opinions regarding a proposed management union contrac

zzz [600]

Answer:

Step-by-step explanation:

6 0

2 years ago