All categories

3 years ago

7400 dollars is placed in an account with an annual interest

Mathematics

1 answer:

Anastasy [175]3 years ago

6 0

Answer:

Step-by-step explanation:

We have the following equation for compounding annual interest

AV=PV(1+i)ⁿ

We have

19200=7400(1+.06)ⁿ

Solve for n

divide both sides by 7400

2.594=(1.06)ⁿ

then use the following rule for exponents

$x^y=z\\log_xz=y$

which means that

$log_{1.06}2.594=n$

Solve and get

16.362

which rounds to 17

You might be interested in

Ill give a brainlist ! :>

bekas [8.4K]

Answer:

Answer is given above in the image 

Step-by-step explanation:

I hope this will help you 

 HAVE A NICE DAY!

THANKS FOR GIVING ME THE OPPORTUNITY TO ANSWER YOUR QUESTION. 

ANSWER IS (c+8)(c+3)

7 0

3 years ago

What does asap mean like rn and yassss

rjkz [21]

As Soon As Possible is what asap means or asap yams rip or asap ferg and rocky

6 0

3 years ago

How to add the integers -7 and -5.

Rzqust [24]

-7 + -5
When you add negative, it would be the same as subtracting
Turns into: -7-5
-(7+5) = -12
The solutions is -12

8 0

2 years ago

A. a teacher wants a large poster of the Declaration of Independence that is three times its actual size. What are the dimension

bonufazy [111]

1.25 or 0.50
because you rounded to a decimal

5 0

3 years ago

Read 2 more answers

An automobile firm wants to purchase either machine A or machine B. The intial cost of machines A and B are $40,000 and $50,000,

Fynjy0 [20]

Answer:

C. $118,000 and $110,000, respectively.

Step-by-step explanation:

Please consider the complete question.

An automobile firm wants to purchase either machine A or machine B. The intial cost of machines A and B are $40,000 and $50,000, respectively. The power consumption per year of machines A and B are 120 MWh and 100 MWh, respectively. The power cost is $40/MWh. The estimated operating and maintenance cost over 10 years is $30,000 for machine A and $20,000 for machine B.

First of all, we will find the amount spent on power consumption for 10 years by each machine as:

$\text{Machine A}=10\cdot 120(\$40)\\\text{Machine A}=\$48,000$