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Nadusha1986 [10]
3 years ago
5

Which of the following is most likely the next step in the series?

Mathematics
2 answers:
user100 [1]3 years ago
7 0

Answer:  The correct option is (A) E20e.

Step-by-step explanation:  We are to find the most likely next step in the following series:

A4a,  b8B,  C12c,  d16D,   .   .   ..

The first letters are A, b, C, d,  .  .  .. So, the next letter in this series will be E.

The second numbers are 4, 8, 12, 16,  .  .  ..

That is, 4 × 1, 4 × 2, 4 × 3, 4 × 4, .  .  . . So, the next number in this series will be 4 × 5 = 20.

The third terms are a, B, c, D,  .  .  .. So, the next letter in this series will be e.

Thus, the next step in the given series is

E20e.

Option (A) is CORRECT.

Delicious77 [7]3 years ago
6 0

it is E20e not e20E

so the answer is A

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Ashley is preparing for a horse riding competition. She has trained her horse for 5 hours and has completed 195 rounds. At what
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3 years ago
Please need help explanation need it
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2 years ago
If Tanisha has ​$1000 to invest at 7​% per annum compounded semiannually​, how long will it be before she has ​$1600​? If the co
Sphinxa [80]

Answer:

Using continuous interest 6.83 years before she has ​$1600​.

Using continuous compounding, 6.71 years.

Step-by-step explanation:

Compound interest:

The compound interest formula is given by:

A(t) = P(1 + \frac{r}{n})^{nt}

Where A(t) is the amount of money after t years, P is the principal(the initial sum of money), r is the interest rate(as a decimal value), n is the number of times that interest is compounded per unit year and t is the time in years for which the money is invested or borrowed.

Continuous compounding:

The amount of money earned after t years in continuous interest is given by:

P(t) = P(0)e^{rt}

In which P(0) is the initial investment and r is the interest rate, as a decimal.

If Tanisha has ​$1000 to invest at 7​% per annum compounded semiannually​, how long will it be before she has ​$1600​?

We have to find t for which A(t) = 1600 when P = 1000, r = 0.07, n = 2

A(t) = P(1 + \frac{r}{n})^{nt}

1600 = 1000(1 + \frac{0.07}{2})^{2t}

(1.035)^{2t} = \frac{1600}{1000}

(1.035)^{2t} = 1.6

\log{1.035)^{2t}} = \log{1.6}

2t\log{1.035} = \log{1.6}

t = \frac{\log{1.6}}{2\log{1.035}}

t = 6.83

Using continuous interest 6.83 years before she has ​$1600​

If the compounding is​ continuous, how long will it​ be?

We have that P(0) = 1000, r = 0.07

Then

P(t) = P(0)e^{rt}

1600 = 1000e^{0.07t}

e^{0.07t} = 1.6

\ln{e^{0.07t}} = \ln{1.6}

0.07t = \ln{1.6}

t = \frac{\ln{1.6}}{0.07}

t = 6.71

Using continuous compounding, 6.71 years.

7 0
3 years ago
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