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adelina 88 [10]
3 years ago
9

Jimmy invests $4,000 in an account that pays 2% annual interest, compounded semi-annually. What is his balance, to the nearest c

ent, at the end of 10 years?
Mathematics
1 answer:
julia-pushkina [17]3 years ago
6 0

Answer:

$4,880.80

Step-by-step explanation:

A = P(1 + r/n)^nt

Where,

A = future value

P = principal = $4,000

r = interest rate = 2% = 0.02

n = number of periods = 2

t = time = 10

A = P(1 + r/n)^nt

= 4000( 1 + 0.02/2)^2*10

= 4000(1 + 0.01)^20

= 4000( 1.01 )^20

= 4000(1.2202)

= 4,880.8

A = $4,880.80 to the nearest cents

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15 x 3.24 =48.60
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Simplify -|-7 + 4|.<br><br> -11<br> -3<br> 3<br> 11
myrzilka [38]

|a| = -a for a < 0

|a| = a for a ≥ 0

examples:

|-1| = -(-1) = 1; |-4| = -(-4) = 4; |-0.1| = 0.1; |-109| = 109

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A total of 585 tickets were sold for the school play. They were either adult tickets or student tickets. There were 65 fewer stu
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Answer:

325

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Total Tickets = 585

Let the adult tickets be = x

If 65 fewer student tickets were sold than adult tickets then Student tickets = x-65

Student Tickets + Adult Tickets = 585

x + (x-65) = 585

x + x - 65 = 585

2x - 65 = 585

2x = 585 + 65

2x = 650

2x/2 = 650/2

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Answered by Gauthmath

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2 years ago
The expression (3 + 6i) + (5 - 10i) simplified is
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Answer:

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A survey conducted by a leading HMO found that of 2000 women, 340 were heavy smokers and 25 had emphysema. Of those who had emph
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Answer:

The events being a heavy smoker and having emphysema are not independent.

Step-by-step explanation:

We are given the following in the question:

Number of women = 2000

Number of heavy smoker = 340

Number of women who has emphysema = 25

Number of women who has emphysema and are heavy smoker = 21

\text{Probability} = \displaystyle\frac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}}

P(Smoker) =

P(S) = \dfrac{340}{2000}=0.17

P(emphysema) =

P(E) = \dfrac{25}{2000}= 0.0125

P(Smoker and emphysema) =

P(S\cap E) = \dfrac{21}{2000} = 0.0105

Two events A and B are said to be independent if

P(A\cap B)=P(A)\times P(B)

Checking conditions for independence:

P(S)\times P(E) = 0.17\times 0.0125=0.002125\\P(S)\times P(E) \neq 0.0105\\P(S)\times P(E)\neq P(S\cap E)

Thus, the events being a heavy smoker and having emphysema are not independent.

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3 years ago
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