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horsena [70]
3 years ago
7

A man invests $5000 in an account that pays 8.5% interest per year, compounded quarterly

Mathematics
2 answers:
vredina [299]3 years ago
7 0
Oh this is easy
5,000x.085=425
That is the amount he made in the year from intrest
Now cut that into a 4th
106.25
He makes 106.25 quarterly
notka56 [123]3 years ago
3 0
5000(1+.085/4)^n=1000
to find n use log:
1.02125^n 
N X log1.02125
when computed=32.9
round to 33 quarters compounded
33/4
equals to 8 years 3 months

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Ed's new car has a value of $23,500, but it is expected to depreciate at a rate of 7.5% per year. If you were to write an expone
Delicious77 [7]

Answer:

0.925 or 92.5%

Step-by-step explanation:

y=a(1-b)^x

decay factor is 1-b <- and the part we care about

b is 7.5%=0.075

1-0.075

=0.925

I'm a little rusty on exponential rates, but I hope this helps!

3 0
3 years ago
Solve for x: 4(x + 2) = 3(x − 2)<br><br>A) −2<br>B)−4<br>C) −10<br>D) −14
BARSIC [14]

4(x+2)=3(x-2)

Multiply the first bracket by 4

Multiply the second bracket by 3

4x+8=3x-6

Move 3x to the left hand side, whenever moving a number with a letter the sign changes ( positive 3x to negative 3x)

4x-3x+8=3x-3x-6

x+8=-6

Move positive 8 to the right hand side

x+8-8=-6-8

x=-14

Check answer by using substitution method

Use x=-14 into both of the equations

4(-14+2)=3(-14-2)

-56+8=-42-6

-48=-48

Answer is -14- D)

6 0
3 years ago
What is the truth value for the following conditional statement?
Studentka2010 [4]

Answer:

TT→T

Step-by-step explanation:

If p is false, then ~p is true.

If q is false, then ~q is true.

Now note that

  • If a and b are both true, then a→b is true.
  • If a is true, b is false, then a→b is false.
  • If a is false, b is true, then a→b is true.
  • If a and b are both false, then a→b is true.

In your case, both~p and ~q are true, then ~p→~q is true too (or TT→T)

3 0
3 years ago
Spam e-mail containing a virus is sent to 1000 e-mail addresses. After 1 second, a recipient machine broadcasts 10 new spam e-ma
pickupchik [31]

Answer:

Answer explained below

Step-by-step explanation:

Spam e-mail containing a virus is sent to 1000 e-mail addresses. After 1 second, a recipient machine broadcasts 10 new spam e-mails containing the same virus, after which the virus disables itself on that machine. (1) Write a recursive definition (i.e. recurrence relation) to show how many spam emails will be sent out after n seconds. (2) Solve the recurrence relation. (3) How many e-mails are sent at the end of 20 seconds

1.START T=0.....

1000 EMAILS SENT AND RECEIVED BY 1000 M/CS.

T=1.....

EACH OF THE M/C SENDS 10 NEW MAILS ....

.................................

LET M[N] BE THE NUMBER OF MAILS SENT OUT AFTER N SECONDS.

SO , EACH OF THESE M/CS WILL SEND 10 MAILS IN NEXT 1 SECOND.

HENCE NUMBER OF MAILS SENT IN N+1 SECONDS=M[N+1]=

M[N+1]=10*M[N].......................1

THIS IS THE RECURRENCE RELATION.....

2.SOLUTION .....

M[N+1]=10M[N]=10*10M[N-1]=10*10*10M[N-2]=........

M(N+1)=[10^1][M(N)]=[10^2][M(N-1)]=[10^3][M(N-2)]=..........=[10^N][M(1)]=[10^(N+1)][M(0)]

M[N+1]=[10^(N+1)][1000]=[10^(N+1)][10^3]=[10^(N+4)]......................................2

THIS IS THE NUMBER OF MAILS SENT AFTER N+1 SECONDS .....OR ....

M[N]=[10^(N+3)].............................................3

..................IS THE SOLUTION FOR NUMBER OF MAILS SENT AFTER N SECONDS.....

3.AFTER N=20 SECONDS , THE ANSWER IS ....

M[20]=10^(20+3)=10^23

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