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madreJ [45]
4 years ago
5

If $3,000,000 is invested at 6% interest compounded continuously, how much

Mathematics
1 answer:
g100num [7]4 years ago
3 0

Answer:

  $24,498,509.74

Step-by-step explanation:

The formula for the value as a function of time is ...

  V(t) = P·e^(rt)

Filling in the numbers and doing the arithmetic, we have ...

  V(35) = 3,000,000·e^(0.06·35) ≈ 24,498,509.74

Compounded continuously for 35 years, the investment will be worth $24,498,509.74.

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A band consists of 3 woodwinds for every 7 members of the band. The independent quantity represents the total members of the ban
IgorC [24]

Answer: k=\frac{3}{7}

Step-by-step explanation:

For this exercise it is important to remember the following:

The constant of proportionality, which is represented as "k", describe  the constant ratio of two quantities that are proportional (the independent and dependent variables).

In this case we know that the independent quantity represents the total members of the band and the dependent quantity represents the woodwind members of the band.

Since the band consists of 3 woodwinds for every 7 members of the band, we can conclude that the constant of  proportionality for this relationship is:

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3 years ago
5x+2y=24 and 8x+2y=30
larisa86 [58]

Answer:

x = 2

y = 7

Step-by-step explanation:

5x+2y=24  -------------->equ 1

8x+2y=30 -----------> equ 2

equ 1 - equ 2 =====>         -3x = - 6

                                               x = -6 / -3  = 2

                                               x = 2

Put x = 2 in equ 1,  

 5*2 + 2y = 24

10 + 2y = 24

        2y = 24 - 10

         2y = 14

            y = 14/2

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10 snails is to be chosen from this population. Find the probability that the percentage of streaked-shelled snails in the sampl
Alex17521 [72]

Here is the full question:

The \  shell \   of  \ the \  land  \ snail \  Limocolaria  \ martensiana  \ has \  two  \ possible  \ colorforms:  \ streaked  \ and  \ pallid.  \ In  \  a \ certain \  population  \ of  \ these   \ snails,  \ 60\%  \ o f  \ theindividuals \  have  \ streaked  \ s hells.\ 16  \ Suppose  \ that  \ a  \ random \ sample  \ of \  10  \ snailsis \  to  \ be \ chosen  \ from  \ this \  population.  \ F ind  \ the  \ probabilit y \  that  \ the  \ percentage  \ of

streaked-shelled  \ snails   \ in  \ t he  \ sample \  will  \ be

(a) 50\%. \  (b) 60\%.  \ (c) 70\%.

Answer:

(a) 0.2007

(b) 0.2510

(c) 0.2150

Step-by-step explanation:

Given that:

The sample size = 10

Sample proportion= 60% 0.6

Let X represents the no of streaked-shell snails.

X \sim Binom (n =1 0, p = 0.60)

The Probability mass function (X) is:

P(X =x)= (^n_x) p^x(1-p)^{n-x}; x = 0,1,2,3...

The Binomial probability with mean μ = np

= 10 * 0.6

= 6

Standard deviation σ = \sqrt{np(1-p)}

= \sqrt{10*0.6*(1-0.6)}

= 1.55

The probability that the percentage of the streaked-shelled snails in the sample will be:

a)

P(X = 0.5) = ^nC_x p^x (1 -p) ^{(n-x)}

= ^{10}^C_5 * (0.6)^5(1-0.6)^{10-5}

= \dfrac{10!}{5!(10-5)!} * (0.6)^5(1-0.6)^{10-5}

= 0.2007

b)

P(X = 0.6) = ^nC_x p^x (1 -p) ^{(n-x)}

= ^{10}^C_6 * (0.6)^6(1-0.6)^{10-6}

= \dfrac{10!}{6!(10-6)!} * (0.6)^6(1-0.6)^{10-6}

= 0.2510

c)

P(X = 0.7) = ^nC_x p^x (1 -p) ^{(n-x)}

= ^{10}^C_7 * (0.6)^7(1-0.6)^{10-7}

= \dfrac{10!}{7!(10-7)!} * (0.6)^7(1-0.6)^{10-7}

= 0.2150

7 0
3 years ago
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