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3 years ago

Daddy's annual salary is $42603.00. If he gets the same salary

Mathematics

1 answer:

alexdok [17]3 years ago

5 0

Answer:

$4800.25

Step-by-step explanation:

$42603 is a yearly salary.

There are 12 months in 1 year.

Monthly salary:

$42603/12 = $3550.25

Monthly travelling allowance: $1250

Total amount earned in 1 month:

$3550.25 + $1250 = $4800.25

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25 times 4 =100
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Simplify 19 - 1.67 + (2.4)

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Explain the different ways it is possible to add two rational numbers and get a negative number

snow_lady [41]

Answer:
both numbers are negative
negative number is greater than the positive number

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In order to add two rational numbers and get a negative value, one of two scenarios should occur:
1- both numbers are negative.
In this case, the answer would definitely be negative as the two numbers are originally negative
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$\frac{-1}{2} + \frac{-3}{2} = \frac{-4}{2} = -2$

2- negative number is greater than the positive number
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3 years ago

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For the following amount at the given interest rate compounded continuously, find (a) the future value after 9 years, (b) the

Stella [2.4K]

Answer:

(a) The future value after 9 years is $7142.49.

(b) The effective rate is $r_E=4.759 \:{\%}$ .

(c) The time to reach $13,000 is 21.88 years.

Step-by-step explanation:

The definition of Continuous Compounding is

If a deposit of $P$ dollars is invested at a rate of interest $r$ compounded continuously for $t$ years, the compound amount is

$A=Pe^{rt}$

(a) From the information given

$P=4700$

$r=4.65\%=\frac{4.65}{100} =0.0465$

$t=9 \:years$

Applying the above formula we get that

$A=4700e^{0.0465\cdot 9}\\A=7142.49$

The future value after 9 years is $7142.49.

(b) The effective rate is given by

$r_E=e^r-1$

Therefore,

$r_E=e^{0.0465}-1=0.04759\\r_E=4.759 \:{\%}$

(c) To find the time to reach $13,000, we must solve the equation

$13000=4700e^{0.0465\cdot t}$

$4700e^{0.0465t}=13000\\\\\frac{4700e^{0.0465t}}{4700}=\frac{13000}{4700}\\\\e^{0.0465t}=\frac{130}{47}\\\\\ln \left(e^{0.0465t}\right)=\ln \left(\frac{130}{47}\right)\\\\0.0465t\ln \left(e\right)=\ln \left(\frac{130}{47}\right)\\\\0.0465t=\ln \left(\frac{130}{47}\right)\\\\t=\frac{\ln \left(\frac{130}{47}\right)}{0.0465}\approx21.88$

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3 years ago