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alexira [117]
4 years ago
11

I NEED HELP QUICKLY PLEASE PLEASE PLEASE

Mathematics
2 answers:
bija089 [108]4 years ago
6 0

Answer:

The compound interest formula is :

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

Now, we have to find p.

A = 2000

r = 2.8% or 0.028

n = 1

t = 3

Putting the values in formula we get;

2000=p(1+\frac{0.028}{1})^{3}

=> 2000=p(1.028)^{3}

=> 2000=p(1.08637)

=> 2000/1.08637=p

So, p = $1840.99 rounded to $1841.

Part A: John should invest $1841 now to have $2000 in 3 years.

------------------------------------------------------------------------------------------------

Part B:

Now p = 1841

r = 0.028

n = 1

t = 5

A=1841(1+\frac{0.028}{1})^{5}

=> A=1841(1.028)^{5}

=> A=1841(1.1480)

A = $2113.46

Hence, John will have $2113.46 in 5 years.

3241004551 [841]4 years ago
3 0
A. you plug in to a calculator, which will give 1840.986 so you need to round up to 1840.99. if you truncate it to .98 then he won't reach 2000 in 3 years

b. for this one if you look at the equation given to find the principle it is principle = result (1+rate) ^ -time

if you re arrange this you get result=principle (1+rate)^time
so result = 1840.99(1.028)^5
= 2113.57
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Answer:

The probability that it will choose food #2 on the second trial after the initial trial = 0.3125

Step-by-step explanation:

Given - A lab animal may eat any one of three foods each day. Laboratory records show that if the animal chooses one food on one trial, it will choose the same food on the next trial with a probability of 50%, and it will choose the other foods on the next trial with equal probabilities of 25%.

To find - If the animal chooses food #1 on an initial trial, what is the probability that it will choose food #2 on the second trial after the initial trial?

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By the given information, we get the stohastic matrix

H = \left[\begin{array}{ccc}0.5&0.25&0.25\\0.25&0.5&0.25\\0.25&0.25&0.5\end{array}\right]

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Let

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x_{0} = \left[\begin{array}{ccc}1\\0\\0\end{array}\right]

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Now,

x_{1} = Hx_{0} \\ = \left[\begin{array}{ccc}0.5&0.25&0.25\\0.25&0.5&0.25\\0.25&0.25&0.5\end{array}\right]\left[\begin{array}{ccc}1\\0\\0\end{array}\right] \\= \left[\begin{array}{ccc}0.5\\0.25\\0.25\end{array}\right]

∴ we get

x_{1} = \left[\begin{array}{ccc}0.5\\0.25\\0.25\end{array}\right]

Now,

x_{2} = Hx_{1} \\ = \left[\begin{array}{ccc}0.5&0.25&0.25\\0.25&0.5&0.25\\0.25&0.25&0.5\end{array}\right]\left[\begin{array}{ccc}0.5\\0.25\\0.25\end{array}\right] \\= \left[\begin{array}{ccc}0.25+0.0625+0.0625\\0.125+0.125+0.0625\\0.125+0.0625+0.125\end{array}\right]\\= \left[\begin{array}{ccc}0.375\\0.3125\\0.3125\end{array}\right]

∴ we get

x_{2} = \left[\begin{array}{ccc}0.375\\0.3125\\0.3125\end{array}\right]

∴ we get

The probability that it will choose food #2 on the second trial after the initial trial = 0.3125

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