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Flura [38]
3 years ago
5

2,87 × 30 com cálculo

Mathematics
1 answer:
Paraphin [41]3 years ago
5 0

Answer:

86.1 I believe because I'm assuming you meant 2.87

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Sara invests $1,200 each year in an IRA ( 12 years in an account that earned 5%
Sati [7]
A)

\bf \qquad \qquad \textit{Future Value of an ordinary annuity}
\\\\
A=pymnt\left[ \cfrac{\left( 1+\frac{r}{n} \right)^{nt}-1}{\frac{r}{n}} \right]
\\\\

\bf \begin{cases}
A=
\begin{array}{llll}
\textit{original amount}\\
\textit{already compounded}
\end{array}
\begin{array}{llll}

\end{array}\\
pymnt=\textit{periodic deposits}\to &1200\\
r=rate\to 5\%\to \frac{5}{100}\to &0.05\\
n=
\begin{array}{llll}
\textit{times it compounds per year}\\
\textit{each year, once}
\end{array}\to &1\\

t=years\to &12
\end{cases}

B)

let's say after 12years, she ended up with a value of say "P"
so.. now she's just sitting on P, making no more deposits to it
just taking whatever the compound 5% interest will give, thus

\bf \qquad \textit{Compound Interest Earned Amount}
\\\\
A=P\left(1+\frac{r}{n}\right)^{nt}
\qquad 
\begin{cases}
A=\textit{compounded amount}\\
P=\textit{original amount deposited}\to &\boxed{P}\\
r=rate\to 5\%\to \frac{5}{100}\to &0.05\\
n=
\begin{array}{llll}
\textit{times it compounds per year}\\
\textit{again, is only once}
\end{array}\to &1\\

t=years\to &11
\end{cases}


C)

from A) she made 1,200 every year, for 12 years that's 1200*12, that's how much she put out of pocket, if you got an amount P from A), then the interest is just the difference, or P - (1200*12)

from B), she started with an original amount of P, and ended up with a compounded amount of A after 11years, so the interest is just also the difference, or A - P

add those two folks together, and that's the total interest she got for the 23  years



4 0
4 years ago
What is the probability that the number chosen is greater than 39?​
andreyandreev [35.5K]

Answer:

Step-by-step explanation:

43

3 0
4 years ago
Read 2 more answers
Suppose A,B and C are 2 ×2 matrices , E,F and G are 3 × 3 matrices, H and K are 2×3 matrices and L and M are 3×2 matrices. For e
Alekssandra [29.7K]

Answer:

a) 2x2 b)3x3 c) Not possible d) 2x3 e) 3x2 f) 3x3 g)Not possible h)3x2

Step-by-step explanation:

-> we can only add matrices with same size

-> we can only multiply when columns of first equal to rows of second

-> nxm multiplied by pxq gives nxq eg 3x2 mul 2x2 gives 3x2

-> addition and mul by const number dont change the size

So considering the above facts :-

a) AB+C = [(2x2)(2X2)]+(2x2)=(2x2)+(2x2)=2x2

b) 3GF =3[(3x3)(3x3)]=3(3x3)=3x3

c)CK+B= [(2x2)(2x3)]+(2x2)=(2x3)+(2x2) which is not possible (see point 1)

d)CK+H =as CK is (2x3) so (2x3)+(2x3)=(2x3)

e)EMC=[(3x3)(3x2)](2x2)=(3x2)(2x2)=3x3

f)GLH=[(3x3)(3x2)](2x3)=(3x2)(2x3)=3x3

g)HLG=[(2x3)(3x2)](3x3)=(2x2)(3x3)= not possible(see point 2)

h)2EL+5MB=2[(3x3)(3x2)]+5[(3x2)(2x2)]=(3x2)+(3x2)=3x2

   

8 0
3 years ago
A rectangular jewelry box costs $125 to gold plate. Calculate the cost of gold plating a box which is similar in shape and that
scZoUnD [109]
The answer should be 1000$. I’m not positive but if it’s eight times the size, then you would multiply the price by 8.
4 0
3 years ago
Read 2 more answers
For Problems 9-10, simplify the expressions using properties of operations. 9. (4x – 7.2) + (-5.3x-8) 10. (t - 1) – (-7t + 2)
mafiozo [28]

The given expression is : (4x – 7.2) + (-5.3x-8)

Simplify :

Open the brackets :

(4x – 7.2) + (-5.3x-8) = 4x - 7.2 - 5.3x - 8

Arrange the variable and constant terms together :

(4x – 7.2) + (-5.3x-8) = 4x - 5.3x - 8 - 7.2

Solve the variable and constant terms together :

(4x – 7.2) + (-5.3x-8) = - 1.3x - 15.2

Since, the variable and constant term can not solve together so the above equation is the simplest form:

(4x – 7.2) + (-5.3x-8) = - 1.3x - 15.2

Answer : (4x – 7.2) + (-5.3x-8) = - 1.3x - 15.2

7 0
1 year ago
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