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3 years ago

Plz I need help!! what is 1,2,3????!??????????

Mathematics

1 answer:

STALIN [3.7K]3 years ago

4 0

1: m
2:16
3:6
hoped ihelped

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Pls help will give brainless to the person help picture is the question

Sedbober [7]

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2 years ago

Hello Please help me.

olga55 [171]

Answer:

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2 years ago

An investor plans to invest $500 a year and expects to get a 10.5% return. If the investor makes these contributions at the end

taurus [48]

The present value of the investment is $6000.

According to the statement

Principal amount = $500

and Return amount = 10.5%

Time period = 20 years.

Now we find the present value of money then

By the formula

PV = P[1-(1+r)^n]/r

PV = 500[1-(1+0.10)^20]/0.10

PV = 6000

So, The present value of the investment is $6000.

Learn more about Return Amount here brainly.com/question/2456547

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1 year ago

Round to the nearest whole number.<br><br><br><br> 568.3

aalyn [17]

Answer:

568.3 into the nearest hundred

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1 year ago

Read 2 more answers

A particular isotope has a half-life of 74 days. If you start with 1 kilogram of this isotope, how much will remain after 150

shtirl [24]

$\bf \stackrel{150~days}{\textit{Amount for Exponential Decay using Half-Life}} \\\\ A=P\left( \frac{1}{2} \right)^{\frac{t}{h}}\qquad \begin{cases} A=\textit{accumulated amount}\\ P=\textit{initial amount}\dotfill &1\\ t=\textit{elapsed time}\dotfill &150\\ h=\textit{half-life}\dotfill &74 \end{cases} \\\\\\ A=1\left( \frac{1}{2} \right)^{\frac{150}{74}}\implies A=1\left( \frac{1}{2} \right)^{\frac{75}{37}}\implies \boxed{A\approx 0.24536}$

$\bf \stackrel{300~days}{\textit{Amount for Exponential Decay using Half-Life}} \\\\ A=P\left( \frac{1}{2} \right)^{\frac{t}{h}}\qquad \begin{cases} A=\textit{accumulated amount}\\ P=\textit{initial amount}\dotfill &1\\ t=\textit{elapsed time}\dotfill &300\\ h=\textit{half-life}\dotfill &74 \end{cases} \\\\\\ A=1\left( \frac{1}{2} \right)^{\frac{300}{74}}\implies A=1\left( \frac{1}{2} \right)^{\frac{150}{37}}\implies \boxed{A\approx 0.060202}$

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2 years ago