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UkoKoshka [18]
3 years ago
7

Thang decided to borrow ​$7000 from his local bank to help pay for a car. His loan was for 3 years at a simple interest rate of

8​%. How much interest will Thang​ pay?
Mathematics
2 answers:
jekas [21]3 years ago
6 0
You need to find 24% of $7000 because that's the total interest over the course of 3 years. The answer is $1680

leonid [27]3 years ago
5 0
Total interest=principal*interest rate*time,,,
I=prt
I=7000*0.08*3=??
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Solve the inequality: 5x + 14 greater than or equal too 54
White raven [17]

Answer:

5*8+14=54

Step-by-step explanation:

5*x+14= 54

so

54-14= 40

40/5=8

so x=8

7 0
3 years ago
Given the pay rate and hours worked, determine the gross earnings, Federal taxes (assuming 18% of gross earnings), state taxes (
ella [17]

Answer:

Let us assume that the pay rate per hour = x

no. of hours worked = n

Gross earnings = x*n

Federal taxes = 18% of gross earnings

= 0.18(x*n)

State taxes = 4% of gross earnings

= 0.04(x*n)

Social security deduction = 7.05% of gross earnings

= 0.0705(x*n)

Total deductions = Federal taxes + State taxes +SSD

= 0.18(x*n) + 0.04(x*n) + 0.0705(x*n)

= 0.2905(x*n)

Net pay = Gross earnings - Total Deduction

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3 0
3 years ago
Can someone help me on this question? Harry and Tim both made New Years resolutions. Harry made 5 more resolutions than Tim. Tog
solniwko [45]

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6 0
3 years ago
How do you figure this out? I am having a hard time. The variables x and y are related to proportionally. When x = 8, y = 20. Fi
kvv77 [185]

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5 0
3 years ago
A worker was paid a salary of $10,500 in 1985. Each year, a salary increase of 6% of the previous year's salary was awarded. How
Mazyrski [523]
Note that 6% converted to a decimal number is 6/100=0.06. Also note that 6% of a certain quantity x is 0.06x.

Here is how much the worker earned each year:


In the year 1985 the worker earned <span>$10,500. 

</span>In the year 1986 the worker earned $10,500 + 0.06($10,500). Factorizing $10,500, we can write this sum as:

                                            $10,500(1+0.06).



In the year 1987 the worker earned

$10,500(1+0.06) + 0.06[$10,500(1+0.06)].

Now we can factorize $10,500(1+0.06) and write the earnings as:

$10,500(1+0.06) [1+0.06]=$10,500(1.06)^2.


Similarly we can check that in the year 1987 the worker earned $10,500(1.06)^3, which makes the pattern clear. 


We can count that from the year 1985 to 1987 we had 2+1 salaries, so from 1985 to 2010 there are 2010-1985+1=26 salaries. This means that the total paid salaries are:

10,500+10,500(1.06)^1+10,500(1.06)^2+10,500(1.06)^3...10,500(1.06)^{26}.

Factorizing, we have

=10,500[1+1.06+(1.06)^2+(1.06)^3+...+(1.06)^{26}]=10,500\cdot[1+1.06+(1.06)^2+(1.06)^3+...+(1.06)^{26}]

We recognize the sum as the geometric sum with first term 1 and common ratio 1.06, applying the formula

\sum_{i=1}^{n} a_i= a(\frac{1-r^n}{1-r}) (where a is the first term and r is the common ratio) we have:

\sum_{i=1}^{26} a_i= 1(\frac{1-(1.06)^{26}}{1-1.06})= \frac{1-4.55}{-0.06}= 59.17.



Finally, multiplying 10,500 by 59.17 we have 621.285 ($).


The answer we found is very close to D. The difference can be explained by the accuracy of the values used in calculation, most important, in calculating (1.06)^{26}.


Answer: D



4 0
3 years ago
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