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igor_vitrenko [27]
3 years ago
5

Need help please please

Mathematics
2 answers:
ludmilkaskok [199]3 years ago
6 0

Step-by-step explanation:

Three and one half

= 3 \frac{1}{2}

Less than symbol ( < )

Five and one half

=  5\frac{1}{2}

So, equation will be,

3 \frac{1}{2}  < 5 \frac{1}{2}

<em><u>Hence</u></em><em><u>,</u></em><em><u> </u></em><em><u>your</u></em><em><u> </u></em><em><u>chosen</u></em><em><u> </u></em><em><u>option</u></em><em><u> </u></em><em><u>is</u></em><em><u> </u></em><em><u>correct</u></em><em><u> </u></em>

Lera25 [3.4K]3 years ago
3 0

Answer:

3 1/2<5 1/2

Step-by-step explanation:

The first answer is correct

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4 0
3 years ago
B) Alex invests $2000 in an account that has a 6% annual rate of growth. To the nearest year, when
wlad13 [49]

Answer:

part 1) 10 years

part 2) 10 years

Step-by-step explanation:

<u><em>The correct question is:</em></u>

Part 1) Alex invests $2000 in an account that has a 6% annual rate of growth compounded annually. To  the nearest year, when will the investment be worth $3600?

Part 2) Alex invests $2000 in an account that has a 6% annual rate of growth compounded continuously. To  the nearest year, when will the investment be worth $3600?

Part 1) we know that    

The compound interest formula is equal to  

A=P(1+\frac{r}{n})^{nt}  

where  

A is the Final Investment Value  

P is the Principal amount of money to be invested  

r is the rate of interest  in decimal

t is Number of Time Periods  

n is the number of times interest is compounded per year

in this problem we have  

P=\$2,000\\A=\$3,600\\ r=6\%=6/100=0.06\\n=1  

substitute in the formula above

3,600=2,000(1+\frac{0.06}{1})^{t}  

1.8=(1.06)^{t}  

Apply log both sides

log(1.8)=log[(1.06)^{t}]  

Applying property of exponents

log(1.8)=(t)log(1.06)  

t=log(1.8)/log(1.06)  

t=10.09\ years

Round to the nearest year

t=10\ years

Part 2) we know that

The formula to calculate continuously compounded interest is equal to

A=P(e)^{rt}  

where  

A is the Final Investment Value  

P is the Principal amount of money to be invested  

r is the rate of interest in decimal  

t is Number of Time Periods  

e is the mathematical constant number

we have  

P=\$2,000\\A=\$3,600\\ r=6\%=6/100=0.06  

substitute in the formula above

3,600=2,000(e)^{0.06t}  

1.8=[e]^{0.06t}  

Apply ln both sides

ln(1.8)=ln[e]^{0.06t}  

ln(1.8)=(0.06t)ln[e]  

t=ln(1.8)/0.06  

t=9.80\ years

Round to the nearest year

t=10\ years

6 0
2 years ago
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