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2 years ago

6

Look at the table on texts assigned to students. A 4-column table with 3 rows titled texts assigned to students. The first colum

n is blank with entries fiction, non-fiction, total. The second column is labeled poetry with entries 0. 2, a, 0. 18. The third column is labeled prose with entries 0. 8, 0. 9, 0. 82. The fourth column is labeled total with entries 1. 0, 1. 0, 1. 0. Which value for a completes the conditional relative frequency table by row? 0. 01 0. 02 0. 1 0. 2.

1 answer:

Crazy boy [7]2 years ago

5 0

Answer:maybe 9 but if not is 8 and the seond is 321

Step-by-step explanation:

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Jake began the day with 88 dollars in his wallet. Then he deposited some money into the bank from what was in his wallet. After

Arada [10]

Answer:

If you're asking how much he deposited at the bank the answer would be 65.

Explanation:

88 - 23 = 65

7 0

3 years ago

An employee joined a company in 2009 with a starting salary of $50,000. Every year this employee receives a raise of $1000 plus

stepladder [879]

Answer:

(a) The required recurrence relation for the salary of the employee of n years after 2009 is $a_n=1.05a_{n-1}+1000$ .

(b)The salary of the employee will be $83421.88 in 2017.

(c) $\therefore a_n=70,000 . \ 1.05^n-20,000$

Step-by-step explanation:

Summation of a G.P series

$\sum_{i=0}^n r^i= \frac{r^{n+1}-1}{r-1}$

(a)

Every year the salary is increasing 5% of the salary of the previous year plus $1000.

Let $a_n$ represents the salary of the employee of n years after 2009.

Then $a_{n-1}$ represents the salary of the employee of (n-1) years after 2009.

Then $a_n= a_{n-1}+5\%.a_{n-1}+1000$

$=a_{n-1}+0.05a_{n-1}+1000$

$=(1+0.05)a_{n-1}+1000$

$=1.05a_{n-1}+1000$

The required recurrence relation for the salary of the employee of n years after 2009 is $a_n=1.05a_{n-1}+1000$ .

(b)

Given, $a_0=\$50,000$

$a_n=1.05a_{n-1}+1000$

Since 2017 is 8 years after 2009.

So, n=8.

∴ $a_8$

$=1.05 a_7+1000$

$=1.05(1.05a_6+1000)+1000$

$=1.05^2a_6+1.05\times 1000+1000$

$=1.05^2(1.05a_5+1000)+1.05\times 1000+1000$

$=1.05^3a_5+1.05^2\times 1000+1.05\times 1000+1000$

$=1.05^3(1.05a_4+1000)+1.05^2\times 1000+1.05\times 1000+1000$

$=1.05^4a_4+1.05^3\times 1000+1.05^2\times 1000+1.05\times 1000+1000$

$=1.05^4(1.05a_3+1000)+1.05^3\times 1000+1.05^2\times 1000+1.05\times 1000+1000$

$=1.05^5a_3+1.05^4\times1000+1.05^3\times 1000+1.05^2\times 1000+1.05\times 1000+1000$

$=1.05^5(1.05a_2+1000)+1.05^4\times1000+1.05^3\times 1000+1.05^2\times 1000+1.05\times 1000+1000$

$=1.05^6a_2+1.05^51000+1.05^4\times1000+1.05^3\times 1000+1.05^2\times 1000+1.05\times 1000+1000$

$=1.05^6(1.05a_1+1000)+1.05^51000+1.05^4\times1000+1.05^3\times 1000+1.05^2\times 1000+1.05\times 1000+1000$

$=1.05^7a_1+1.05^6\times1000+1.05^51000+1.05^4\times1000+1.05^3\times 1000+1.05^2\times 1000+1.05\times 1000+1000$

$=1.05^7(1.05a_0+1000)+1.05^6\times1000+1.05^51000+1.05^4\times1000+1.05^3\times 1000+1.05^2\times 1000+1.05\times 1000+1000$

$=1.05^8a_0+1.05^7\times1000+1.05^6\times1000+1.05^51000+1.05^4\times1000+1.05^3\times 1000+1.05^2\times 1000+1.05\times 1000+1000$

$=1.05^8a_0+(1.05^7+1.05^6+1.05^5+1.05^4+1.05^3+1.05^2+1.05+1)1000$

$=1.05^8 \times 50,000+\frac{1.05^8-1}{1.05-1}\times 1000$

$=1.05^8\times 50,000+20,000(1.58^8-1)$

$=70,000\times 1.05^8-20,000$

≈$83421.88

The salary of the employee will be $83421.88 in 2017.

(c)

Given, $a_0=\$50,000$

$a_n=1.05a_{n-1}+1000$

We successively apply the recurrence relation

$a_n=1.05a_{n-1}+1000$

$=1.05^1a_{n-1}+1.05^0.1000$

$=1.05^1(1.05a_{n-2}+1000)+1.05^0.1000$

$=1.05^2a_{n-2}+1.05^1.1000+1.05^0.1000$

$=1.05^2(1.05a_{n-3}+1000)+(1.05^1.1000+1.05^0.1000)$

$=1.05^3a_{n-3}+(1.05^2.1000+1.05^1.1000+1.05^0.1000)$

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

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

$=1.05^na_{n-n}+\sum_{i=0}^{n-1}1.05^i.1000$

$=1.05^na_0+1000\sum_{i=0}^{n-1}1.05^i$

$=1.05^n.50,000+1000.\frac{1.05^n-1}{1.05-1}$

$=1.05^n.50,000+20,000.(1.05^n-1)$

$=(50,000+20,000)1.05^n-20,000$

$=70,000 . \ 1.05^n-20,000$

$\therefore a_n=70,000 . \ 1.05^n-20,000$

6 0

3 years ago

Sebastian and Caroline obtained a 30-year, fixed rate mortgage for $154,850 on a home that cost them $169,900. If the interest r

kotegsom [21]

Answer:

$318,972.30

Step-by-step explanation:

The total of 360 payments will be $318,972.30.

___

A financial calculator or app is useful for answering questions like this. Most spreadsheets have functions that will serve the same purpose.

_____

You may be given a formula for computing the monthly payment (A) from the principal (P), the annual interest rate (r) and the number of years (t). It might look like this:

A = P·(r/12)/(1 -(1 +r/12)^(-12·t))

Filling in the given numbers, you get ...

A = 154,850·(0.0557/12)/(1 -(1 +0.0557/12)^-360) ≈ 886.034154

Then the value of 360 such payments is ...

P+I = 360·886.034154 ≈ 318,972.30

___

In practice, the final payment will be different from the rest, to make up for the fact that the value of each payment was rounded down to the nearest cent. Rather than try to figure the amount of that payment, we have used a more precise value for the monthly payment and have assumed that doing so will give approximately the same total as the sequence of actual payments. (Our number here is about $2.34 low.)

6 0

3 years ago

You have two accounts with balances $450 and $35. you want to bring up the balance in the second account but do not want the bal

kakasveta [241]

Original balance minus Minimum required= Transfer amount

$450 - $215= $235

You can transfer $235.
The answer is C) $235.

Hope this helps! :)

5 0

3 years ago

8 × 1 + 5 × ( 1/100) + 9 ×( 1/1000) = 8.059?? did I do this correctly???

Sati [7]

Answer:

Yes you did it correctly

Step-by-step explanation:

the answer is

$8.059$

5 0

4 years ago

Read 2 more answers