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2 years ago

A jar contains 444 red marbles, 444 green marbles, and 555 blue marbles.

Mathematics

1 answer:

ziro4ka [17]2 years ago

4 0

human woohoo trook troomhi

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Assume a bank loan requires an interest payment of $85 per year and a principal payment of $1,000 at the end of the loan's eight

Alex

Answer:

$919.98

Step-by-step explanation:

To solve using a financial calculator do

N=8

I/Y=10

PMT=85

FV=1000

CMPT PV get 919.98

To do by hand find the present value of the interst payments

85*(1-(1/1.1)^8)

which is 453.4687

Find the present value of the final ballon payment

1000*(1/1.1)^8

which is 466.507

take the sum

466.507+453.4678= 919.98

3 0

3 years ago

"A line that goes through (4, 7) and (-2, 1)"

Olenka [21]

Y= x + 3
1. Find the gradient(slope)

1 - 7 / - 2 - 4
=1

2. Find the y. Intercept ( c )
Y = x + c
Replace by any of the two coordinates given (4 , 7 )
7 = 4 + c
C= 3

Equation => Y = x + 3

6 0

2 years ago

If Free Yer Assets Bank (FYAB) will give you 0.25% compounded quarterly and I.M.A.Q.T.π bank will give you 0.23% interest compou

ivann1987 [24]

Answer:

FYAB gives a better deal.

Step-by-step explanation:

Compound interest:

$A = P(1 + \dfrac{r}{n})^{nt}$

Continuously compounded interest:

$A = Pe^{rt}$

For the quarterly compounded interest, r = 0.25%, and n = 4.

$2P = P(1 + \dfrac{0.0025}{4})^{4t}$

$1.000625^{4t} = 2$

$\log (1.000625^{4t}) = \log 2$

$4t(\log 1.000625) = \log 2$

$t = \dfrac{\log 2}{4\log 1.000625}$

$t = 277$

For the continuously compounded interest, r =0.23%

$A = Pe^{rt}$

$2P = Pe^{0.0023t}$

$2 = e^{0.0023t}$

$\ln 2 = \ln e^{0.0023t}$

$\ln 2 = 0.0023t$

$t = \dfrac {\ln 2}{0.0023}$

$t = 301$

The quarterly compounded doubles in 277 years.

The continuously compounded doubles in 301 years.

Answer: FYAB gives a better deal.

7 0

2 years ago

PLZ HELP!!!!!!!!!!!!!!

Mashutka [201]

The answer would be B

7 0

3 years ago

The ratio of dogs to cats at an animal shelter is 3 to

Morgarella [4.7K]

20 dogs are at the shelter.

8 0

2 years ago