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Lubov Fominskaja [6]
3 years ago
8

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Mathematics
2 answers:
mojhsa [17]3 years ago
7 0
I- please ask a question. This is not a question
daser333 [38]3 years ago
4 0
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$15,000 is invested for 15 years with an APR of 4.2% and monthly compounding.
Phoenix [80]

Answer:

51-54: Simple Interest. Calculate the amount of money you will have in the following accounts after 5 years, assuming that you eam simple interest 51. You deposit $ 700 in an account with an annual interest rate of 4% 52. You deposit $1200 in an account with an annual interest rate of 3% 53. You deposit $3200 in an account with an annual interest rate of 3.5% 54. You deposit $1800 in an account with an annual interest rate of 3.8% 55-56: Simple versus Compound Interest. Complete the following tables, which show the performance of two investments over a 5-year period. Round all figures to the nearest dollar. 55 Suzanne deposits $3000 in an account that earns simple interest at an annual rate of 2.5%. Derek deposits $3000 in an account that earns compound interest at an annual rate of 2.5%. Suzanne's Suzanne's Derek's Annual | Derek's Year Annual Interest Balance Interest Balance rest formula to the stated pe 57-62: Compound Interest. Use the compound interest form compute the balance in the following accounts after the state riod of time, assuming interest is compounded annually. 57. $10,000 is invested at an APR of 4% for 10 years. 58. $10,000 is invested at an APR of 2.5% for 20 years. 59. $15,000 is invested at an APR of 3.2% for 25 years. 60. $3000 is invested at an APR of 1.8% for 12 years. 61. 55000 is invested at an APR of 3.1% for 12 years. 62. $ 40,000 is invested at an APR of 2.8% for 30 years. 63-70: Compounding More Than Once a Year. Use the appropriate compound interest formula to compute the balance in the following accounts after the stated period of time. 63. $10,000 is invested for 10 years with an APR of 2% and quarterly compounding. 64. $2000 is invested for 5 years with an APR of 3% and daily compounding 65. $25,000 is invested for 5 years with an APR of 3% and daily compounding 66. $10,000 is invested for 5 years with an APR of 2.75% and monthly compounding. 67. $2000 is invested for 15 years with an APR of 5% and monthly compounding 68. $30,000 is invested for 15 years with an APR of 4.5% ana daily compounding. 69. $25,000 is invested for 30 years with an APR of 3.7% quarterly compounding. 70. $15,000 is invested for 15 years with an APR of 4.2% monthly compounding. 71-74. Annual.

Hope this helps

7 0
2 years ago
It is important to re-evaluate financial goals periodically. In which of the following situations would it be necessary to chang
icang [17]

Answer:

Your answer should be A. You fell sharply behind your expected schedule with regard to saving

Step-by-step explanation:

8 0
3 years ago
Read 2 more answers
You place a billiard cue against a billiard table. The foot of the billiard cue is 15 cm away from the table and the top of the
Luba_88 [7]

Answer:

  • -4

Step-by-step explanation:

<u>The billiard cue is 15 cm away from the table</u>

  • This refers to coordinates of (15, 0)

<u>The billiard cue touches the table at a point 60 cm off the ground</u>

  • This refers to coordinates of (0, 60)

<u>The slope is as per slope formula;</u>

  • m = (60 - 0)/(0-15) = 60/ -15 = -4

Thea answer is -4

6 0
3 years ago
Read 2 more answers
There are x number of students at helms. If the number of students increases by 7.8% each year, how many students will be there
vodomira [7]

There will be 1.078x students next year and equation is number of students in next year = x + 7.8% of x

<h3><u>Solution:</u></h3>

Given, There are "x" number of students at helms.  

The number of students increases by 7.8% each year which means if there "x" number of students in present year, then the number of students in next year will be x + 7.8% of x

Number of students in next year = number of students in present year + increased number of students.

\begin{array}{l}{\text { Number of students in next year }=x+7.8 \% \text { of } x} \\\\ {\text { Number of students in next year }=x\left(1+\frac{7.8}{100}\right)} \\\\ {\text { Number of students in next year }=x(1+0.078)=1.078 x}\end{array}

Thus there will be 1.078x students in next year

3 0
3 years ago
000 000
olga55 [171]

Answer:

9+10 =19

Step-by-step explanation:is that wat u wanted

4 0
2 years ago
Read 2 more answers
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