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cricket20 [7]
1 year ago
13

Jacques deposited $1,900 into an account that earns 4% interest compounded semiannually. After t years, Jacques has $3,875.79 in

the account. Assuming he made no additional deposits or withdrawals, how long was the money in the account?
Compound interest formula:mc007-1.jpg

t = years since initial deposit
n = number of times compounded per year
r = annual interest rate (as a decimal)
P = initial (principal) investment
V(t) = value of investment after t years


2 years
9 years
18 years
36 years
Mathematics
2 answers:
nika2105 [10]1 year ago
4 0
T=(log(3,875.79÷1,900)÷log(1+0.02))÷2
T=18 years
Mnenie [13.5K]1 year ago
3 0

Answer:

Option 3 is correct. After 18 years the amount will $3,875.79.

Step-by-step explanation:

The compound interest formula is

v(t)=p(1+\frac{r}{n})^{nt}

Where, t = years since initial deposit


n = number of times compounded per year


r = annual interest rate (as a decimal)


P = initial (principal) investment


V(t) = value of investment after t years.

The initial amount is $1,900. Interest rate is 4% and interest compounded semiannually. It means interest compounded 2 times in a year. The amount after t years is  $3,875.79.

3,875.79=1900(1+\frac{0.04}{2})^{2t}

3,875.79=1900(1.02)^{2t}

\frac{3,875.79}{1900}=(1.02)^{2t}

log(\frac{3,875.79}{1900})=log(1.02)^{2t}

\frac{0.309606636902}{log(1.02)}=2t

t=18

After 18 years the amount will $3,875.79.

Therefore the option 3 is correct.

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Step-by-step explanation:

(1, 0.80) represents the unit rate.

(25, 20) represents the price of a necklace with 25 beads.

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Point (1, r) → (1, 0.80) represents the unit rate.

25(0.80) = 20; thus, (25, 20) represents the price of 25 beads.

Hope this helps

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The earth has a mass of approximately 6\cdot 10^{24}6⋅10 24 6, dot, 10, start superscript, 24, end superscript kilograms (\text{
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Answer:

0.02

Step-by-step explanation:

The volume of the earth's oceans is approximately 1.34\cdot 10^{9}1.34⋅10

9

1, point, 34, dot, 10, start superscript, 9, end superscript cubic kilometers (\text{km}^3)(km

3

)left parenthesis, start text, k, m, end text, cubed, right parenthesis, and ocean water has a mass of about 1.03 \cdot 10^{12}\,\dfrac{\text{kg}}{\text{km}^3}1.03⋅10

12

 

km

3

kg

​

1, point, 03, dot, 10, start superscript, 12, end superscript, start fraction, start text, k, g, end text, divided by, start text, k, m, end text, cubed, end fraction .

To simplify, we will use the product of powers property of exponents that says that x^a\cdot x^b = x^{a+b}x

a

⋅x

b

=x

a+b

x, start superscript, a, end superscript, dot, x, start superscript, b, end superscript, equals, x, start superscript, a, plus, b, end superscript.

\qquad 1.34\cdot 10^{9}\,\cancel{\text{km}^3} \cdot 1.03 \cdot 10^{12}\,\dfrac{\text{kg}}{\cancel{\text{km}^3}} = 1.3802 \cdot 10^{21}\,\text{kg}1.34⋅10

9

 

km

3

⋅1.03⋅10

12

 

km

3

kg

​

=1.3802⋅10

21

kg1, point, 34, dot, 10, start superscript, 9, end superscript, start cancel, start text, k, m, end text, cubed, end cancel, dot, 1, point, 03, dot, 10, start superscript, 12, end superscript, start fraction, start text, k, g, end text, divided by, start cancel, start text, k, m, end text, cubed, end cancel, end fraction, equals, 1, point, 3802, dot, 10, start superscript, 21, end superscript, start text, k, g, end text

Hint #2

Next we want to know what portion of the earth's mass this represents. We have:

\qquad \begin{aligned} \dfrac{\text{mass of the oceans}}{\text{total mass of the earth}} &= \dfrac{1.3802 \cdot 10^{21}\,\text{kg}}{6\cdot 10^{24}\,\text{kg}} \\\\ &= \dfrac{1.3802}{6\cdot 10^{3}} \\\\ &= \dfrac{1.3802}{6000} \\\\ &= 0.0002300\overline{3} \end{aligned}

total mass of the earth

mass of the oceans

​

​

 

=

6⋅10

24

kg

1.3802⋅10

21

kg

​

=

6⋅10

3

1.3802

​

=

6000

1.3802

​

=0.0002300

3

​

To convert this to a percent, we multiply by 100100100, so the oceans represent 0.02300\overline{3}\%0.02300

3

%0, point, 02300, start overline, 3, end overline, percent of the earth's total mass, according to these figures.

Hint #3

To the nearest hundredth of a percent, 0.020.020, point, 02 percent of the earth's mass is from oceans.

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