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mario62 [17]
3 years ago
5

How long would it take to double your principal in an account that pays 6.5% annual interest compounded continuously? round your

answer to one decimal place?
Mathematics
2 answers:
prisoha [69]3 years ago
8 0
It would take 10.7 years.

The formula for continuously compounded interest is:
A=Pe^{rt}
where P is the principal, r is the interest rate as a decimal number, and t is the number of years.

Using our information we have:
A=Pe^{0.065t}

We want to know when it will double the principal; therefore we substitute 2P for A and solve for t:
2P=Pe^{0.065t}

Divide both sides by P:
\frac{2P}{P}=\frac{Pe^{0.065t}}{P}
\\
\\2=e^{0.065t}

Take the natural log, ln, of each side to "undo" e:
\ln{2}=\ln{e^{0.065t}}
\\
\\0.6931471806=0.065t

Divide both sides by 0.065:
\frac{0.6931471806}{0.065}=\frac{0.065t}{0.065}
\\
\\10.7\approx t
rjkz [21]3 years ago
3 0

Answer:

11 years

Step-by-step explanation:

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WHAT DOES THIS MEAN GIMME THE ANSWER
BARSIC [14]

Answer:

This is a geometric sequence and the ratio is equal to 4.

Step-by-step explanation:

Arithmetic means you can continue the sequence through addition or subtraction.

Geometric means you can continue the sequence through multiplication or division.

To get from 3 to 12, we could either add 9 (arithmetic) or multiply by 4 (geometric).

To get from 12 to 48, we could either add 36 or multiply by 4.

Since we are looking for a pattern, the sequence would be geometric. This is because the sequence can be continued by multiplying by 4 each time. It cannot be arithmetic because we cannot add by 9 each time.

On a side note, common difference is used when describing arithmetic while ratio is used to describe geometric.

3 0
3 years ago
what's 77787464564576876979875648366732783656938629312935o5193865435986581653195395619856589316589346531713683586982635 x 473853
Marizza181 [45]

Answer:

3.68599e+145

3 0
3 years ago
Read 2 more answers
Which expression is equivalent to t2 – 36?
defon

Answer:

(t + 6)(t - 6)

Step-by-step explanation:

(t - 6)(t + 6)

(t . t) + (t . (6) + ((-6) . t) + ((-6) . 6)

= t² - 6t + 6t - 36

= t² - 36

so,

t² - 36 = (t + 6)(t - 6)

6 0
3 years ago
I need to find the area of a semicircle with a radius of 2 units
MariettaO [177]

Answer:

6.2832 square units (rounded off to four decimal values)

Step-by-step explanation:

Area of a semicircle = \frac{1}{2} × π × r²

r (radius) = 2 units

For our semicircle the area = \frac{1}{2} × π × 2² = 6.28318530718  square units

Or 6.2832 square units (rounded off to four decimal values)

5 0
3 years ago
A computer programming team has 13 members. a. How many ways can a group of seven be chosen to work on a project? b. Suppose sev
Julli [10]

Answer:

1716 ;

700 ;

1715 ;

658 ;

1254 ;

792

Step-by-step explanation:

Given that :

Number of members (n) = 13

a. How many ways can a group of seven be chosen to work on a project?

13C7:

Recall :

nCr = n! ÷ (n-r)! r!

13C7 = 13! ÷ (13 - 7)!7!

= 13! ÷ 6! 7!

(13*12*11*10*9*8*7!) ÷ 7! (6*5*4*3*2*1)

1235520 / 720

= 1716

b. Suppose seven team members are women and six are men.

Men = 6 ; women = 7

(i) How many groups of seven can be chosen that contain four women and three men?

(7C4) * (6C3)

Using calculator :

7C4 = 35

6C3 = 20

(35 * 20) = 700

(ii) How many groups of seven can be chosen that contain at least one man?

13C7 - 7C7

7C7 = only women

13C7 = 1716

7C7 = 1

1716 - 1 = 1715

(iii) How many groups of seven can be chosen that contain at most three women?

(6C4 * 7C3) + (6C5 * 7C2) + (6C6 * 7C1)

Using calculator :

(15 * 35) + (6 * 21) + (1 * 7)

525 + 126 + 7

= 658

c. Suppose two team members refuse to work together on projects. How many groups of seven can be chosen to work on a project?

(First in second out) + (second in first out) + (both out)

13 - 2 = 11

11C6 + 11C6 + 11C7

Using calculator :

462 + 462 + 330

= 1254

d. Suppose two team members insist on either working together or not at all on projects. How many groups of seven can be chosen to work on a project?

Number of ways with both in the group = 11C5

Number of ways with both out of the group = 11C7

11C5 + 11C7

462 + 330

= 792

8 0
2 years ago
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