The frequency table was made using a box containing slips of paper. Each slip of paper was numbered 0, 1, 2, 3, or 4

Solve the following absolute value equation.

cestrela7 [59]

Answer:

x= 8

x= -8

Step-by-step explanation:

|x+1| /3 = 3

|x+1| = 9

|x| + |1| = 9

|x| = 8

x= 8

x= -8

4 0

2 years ago

All of the following have the same unit price except _____.

Aliun [14]

2 1/2 for $6.25 is the answer

5 0

3 years ago

Read 2 more answers

Juan bought 4 pounds of chocolate

Andru [333]

Answer:

Step-by-step explanation:

This is a system of equations. The 2 equations are:

4c + 2p = 16

2c + 3p = 12

Let's get rid of the p's first. Do this by multiplying the first equation by -3 and the second equation by 2 to get a new system that looks like this:

-12x - 6p = -48

4c + 6p = 24

Now add to get

-8c = -24 and

c = 3

That means that chocolate is $3 per pound. Now sub that back in to either one of the original equations to get the price for peanuts:

4(3) + 2p = 16 and

12 + 2p = 16 and

2p = 4 so

p = 2

The price for peanuts is $2 per pound.

6 0

3 years ago

4 / 7 + −1 / 3 <br> HELP PLAESEEEEE

sasho [114]

Answer:

are those fractions or division signs

Step-by-step explanation:

.........................

6 0

3 years ago

Read 2 more answers

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