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2 years ago

6

The graph and table shows the relationship between y, the number of words Jean has typed for her essay and x, the number of minu

tes she has been typing on the computer.
A 2-column table with 9 rows. The first column is labeled x with entries 5, 10, 15, 20, 25, 30, 35, 40, 45. The second column is labeled y with entries 271, 464, 820, 965, 1124, 1501, 1718, 2076, 2257. A graph shows the horizontal axis numbered 20 to 80 and the vertical axis numbered 1000 to 3000. A line increases from 0 to 80.
According to the line of best fit, about how many words will Jean have typed when she completes 60 minutes of typing?

2,500
2,750
3,000
3,250

1 answer:

Tasya [4]2 years ago

3 0

Answer:

c) 3,000

Step-by-step explanation:

i just took the test and passed with 100%

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Alexandra has a stand in the marketplace where she sells her cumin for 7.20 per kilogram. If she sells 90 kilograms of cumin her

Drupady [299]

Answer:

$234

Step-by-step explanation:

First we need to define profits. Profits are Income minus Expenses:

P = I - E

We know profits are $414, so:

414 = I - E

We also can calculate income, as it is equal to price by the sales:

I = p*Q

Here she sold 90 kgs at $7.20 b kg. So:

I = p*Q = 7.20 * 90 = 648

So, replacing in profits equation:

414 = I - E

414 = 648 - E

If we sum E in both sides:

414 + E = 648 - E + E = 648

414 + E = 648

Now, subtracting 414 in both sides:

414 + E - 414 = 648 - 414

E = 234

So, her expenses are $234

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3 years ago

X-5/2 =9 <br><br> solve for x

Liono4ka [1.6K]

Best Answer:<span> </span><span>(x - 5)^2 = 9
x - 5 = ±√9
x = 5 ± 3
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3 years ago

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Find the slope/rate of change of the line on the graph.

ohaa [14]

Answer:

y=-1

From where the line crosses the y axis, it goes up 1 and over 1 from each plane although its moving to the left from 0 every time. This means the slope is negative.

In the equation y=mx+b, b is the y intercept and m is the slope

plugin the info and you get y=-1x+0 or

y=-1x

Step-by-step explanation:

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3 years ago

Simplify:<br> 4-5-9<br> PLEASE HURRY!!! THANK U

Masja [62]

Answer:

The answer is -10.

Step-by-step explanation:

4 - 5 - 9

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3 years ago

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2,17,82,257,626,1297 next one please ?

In-s [12.5K]

The easy thing to do is notice that 1^4 = 1, 2^4 = 16, 3^4 = 81, and so on, so the sequence follows the rule $n^4+1$ . The next number would then be fourth power of 7 plus 1, or 2402.

And the harder way: Denote the <em>n</em>-th term in this sequence by $a_n$ , and denote the given sequence by $\{a_n\}_{n\ge1}$ .

Let $b_n$ denote the <em>n</em>-th term in the sequence of forward differences of $\{a_n\}$ , defined by

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

for <em>n</em> ≥ 1. That is, $\{b_n\}$ is the sequence with

$b_1=a_2-a_1=17-2=15$

$b_2=a_3-a_2=82-17=65$

$b_3=a_4-a_3=175$

$b_4=a_5-a_4=369$

$b_5=a_6-a_5=671$

and so on.

Next, let $c_n$ denote the <em>n</em>-th term of the differences of $\{b_n\}$ , i.e. for <em>n</em> ≥ 1,

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

so that

$c_1=b_2-b_1=65-15=50$

$c_2=110$

$c_3=194$

$c_4=302$

etc.

Again: let $d_n$ denote the <em>n</em>-th difference of $\{c_n\}$ :

$d_n=c_{n+1}-c_n$

$d_1=c_2-c_1=60$

$d_2=84$

$d_3=108$

etc.

One more time: let $e_n$ denote the <em>n</em>-th difference of $\{d_n\}$ :

$e_n=d_{n+1}-d_n$

$e_1=d_2-d_1=24$

$e_2=24$

etc.

The fact that these last differences are constant is a good sign that $e_n=24$ for all <em>n</em> ≥ 1. Assuming this, we would see that $\{d_n\}$ is an arithmetic sequence given recursively by

$\begin{cases}d_1=60\\d_{n+1}=d_n+24&\text{for }n>1\end{cases}$

and we can easily find the explicit rule:

$d_2=d_1+24$

$d_3=d_2+24=d_1+24\cdot2$

$d_4=d_3+24=d_1+24\cdot3$

and so on, up to

$d_n=d_1+24(n-1)$

$d_n=24n+36$

Use the same strategy to find a closed form for $\{c_n\}$ , then for $\{b_n\}$ , and finally $\{a_n\}$ .

$\begin{cases}c_1=50\\c_{n+1}=c_n+24n+36&\text{for }n>1\end{cases}$

$c_2=c_1+24\cdot1+36$

$c_3=c_2+24\cdot2+36=c_1+24(1+2)+36\cdot2$

$c_4=c_3+24\cdot3+36=c_1+24(1+2+3)+36\cdot3$

and so on, up to

$c_n=c_1+24(1+2+3+\cdots+(n-1))+36(n-1)$

Recall the formula for the sum of consecutive integers:

$1+2+3+\cdots+n=\displaystyle\sum_{k=1}^nk=\frac{n(n+1)}2$

$\implies c_n=c_1+\dfrac{24(n-1)n}2+36(n-1)$

$\implies c_n=12n^2+24n+14$

$\begin{cases}b_1=15\\b_{n+1}=b_n+12n^2+24n+14&\text{for }n>1\end{cases}$

$b_2=b_1+12\cdot1^2+24\cdot1+14$

$b_3=b_2+12\cdot2^2+24\cdot2+14=b_1+12(1^2+2^2)+24(1+2)+14\cdot2$

$b_4=b_3+12\cdot3^2+24\cdot3+14=b_1+12(1^2+2^2+3^2)+24(1+2+3)+14\cdot3$

and so on, up to

$b_n=b_1+12(1^2+2^2+3^2+\cdots+(n-1)^2)+24(1+2+3+\cdots+(n-1))+14(n-1)$

Recall the formula for the sum of squares of consecutive integers:

$1^2+2^2+3^2+\cdots+n^2=\displaystyle\sum_{k=1}^nk^2=\frac{n(n+1)(2n+1)}6$

$\implies b_n=15+\dfrac{12(n-1)n(2(n-1)+1)}6+\dfrac{24(n-1)n}2+14(n-1)$

$\implies b_n=4n^3+6n^2+4n+1$

$\begin{cases}a_1=2\\a_{n+1}=a_n+4n^3+6n^2+4n+1&\text{for }n>1\end{cases}$

$a_2=a_1+4\cdot1^3+6\cdot1^2+4\cdot1+1$

$a_3=a_2+4(1^3+2^3)+6(1^2+2^2)+4(1+2)+1\cdot2$

$a_4=a_3+4(1^3+2^3+3^3)+6(1^2+2^2+3^2)+4(1+2+3)+1\cdot3$

$\implies a_n=a_1+4\displaystyle\sum_{k=1}^3k^3+6\sum_{k=1}^3k^2+4\sum_{k=1}^3k+\sum_{k=1}^{n-1}1$

$\displaystyle\sum_{k=1}^nk^3=\frac{n^2(n+1)^2}4$

$\implies a_n=2+\dfrac{4(n-1)^2n^2}4+\dfrac{6(n-1)n(2n)}6+\dfrac{4(n-1)n}2+(n-1)$

$\implies a_n=n^4+1$

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3 years ago