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sweet [91]
2 years ago
13

Harper invested $3,200 in an account paying an interest rate of 7% compounded continuously. Assuming no deposits or withdrawals

are made, how much money, to the nearest ten dollars, would be in the account after 6 years?
Mathematics
1 answer:
andreyandreev [35.5K]2 years ago
8 0
3200(1.06)^⁶


I forgot how to explain

Mark brainliest please
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P is the midpoint of XY x p equals 8 x - 2 + p y equals 12x - 30 find the value of x​
disa [49]

Answer:

but please give me and please mark me as

7 0
3 years ago
farouk scored 110% in his last mathematics test.Explain how this is possible.Give an actual mark allocations that will prove you
irakobra [83]

Answer:

Grades higher than 100% can be achieved by doing extra credit and/or answering bonus questions.

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8 0
2 years ago
A sample of 260 students enrolled at a university were asked what they intended to do after graduation. The results are shown be
vovikov84 [41]
ANSWER

30\%


EXPLANATION

The total number of students can be calculated by adding all the frequencies.


From the graph the total frequency is,

39+117+78+26 =260

The number of students who intend to go into state or local government after graduation is 78.


The percentage of students who intend to go into state or local government after graduation

=  \frac{78}{260}  \times 100\%

=  \frac{3}{10}  \times 100\%

= 30\%

4 0
3 years ago
Read 2 more answers
Sarah already had 45 stamps in her collection and she gets 7 new stamps each month how long will it take before she has 129 stam
tatuchka [14]
The answer is:  " 12 months " .
________________________________
Explanation:
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(129 − 45) / 7 = 84 / 7 = 12 mos.
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3 0
3 years ago
Read 2 more answers
(c) The total number of one centimeter lines in the first n diagrams is given by the expression
trapecia [35]

Answer:

I) f + g = 10/3

II) 4f + 2g = 20/3

III) f = 2 and g = 4/3

Step-by-step explanation:

From the chart,

P = 25

q = 40

The total number of one centimeter lines in the first n diagrams is given by the expression 

2/3n^3 + fn^2 + gn. 

When n = 1, the total number of line = 4. So,

2/3(1)^3 + f(1)^2 + g(1) = 4

2/3 + f + g = 4

Make f+g the subject of formula

f + g = 4 - 2/3

f + g = (12 - 2)/3

f + g = 10/3 ......(1)

When n = 2

Total number of line = 12

2/3(2)^3 + f(2)^2 + g(2) = 12

2/3×8 + 4f + 2g = 12

16/3 + 4f + 2g = 12

4f + 2g = 12 - 16/3

4f + 2g = (36 - 16)/3

4f + 2g = 20/3 ......(2)

(iii) To find the values of f and g, solve equation 1 and 2 simultaneously

f + g = 10/3 × 2

4f + 2g = 32/3

2f + 2g = 20/3

4f + 2g = 32/3

- 2f = - 12/3

f = 12/6

f = 2

Substitutes f in equation 1

f + g = 10/3

2 + g = 10/3

g = 10/3 - 2

g = (10 - 6)/3

g = 4/3

4 0
3 years ago
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