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nikklg [1K]
3 years ago
12

A laptop computer was marked up 100% from an original cost of $740. Last Friday, Judy bought the laptop computer and paid an add

itional 10% in sales tax. What was her total cost?
Mathematics
1 answer:
lora16 [44]3 years ago
5 0

Answer:

$1628

Step-by-step explanation:

A laptop computer was marked up 100% from an original cost of $740.

Marked up price = 100% of $740

= $740

Current price of the Laptop = Initial price of the Laptop + Marked up price

= $740 + $740

= $1480

Last Friday, Judy bought the laptop computer and paid an additional 10% in sales tax. What was her total cost?

Sales tax = 10% of the cost of laptop

= 10% × $1480

Sales tax = $148

Judy's total cost = Price of the Laptop + Sales tax

= $1480 + $148

= $1628

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Find the value of x given that 25 x 8 = 22x
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Step-by-step explanation:

25*8=200 soooooooooo 200=22x divide 22 on both sides and you get x=100/11

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20÷1.5 since there is not X

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Michael multiplies 0.4 by a number. He records the product as 0.012. What number did Michael use? Michael used
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3 0
3 years ago
Tacoma's population in 2000 was about 200 thousand, and had been growing by about 9% each year. a. Write a recursive formula for
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Answer:

a) The recurrence formula is P_n = \frac{109}{100}P_{n-1}.

b) The general formula for the population of Tacoma is

P_n = \left(\frac{109}{100}\right)^nP_{0}.

c) In 2016 the approximate population of Tacoma will be 794062 people.

d) The population of Tacoma should exceed the 400000 people by the year 2009.

Step-by-step explanation:

a) We have the population in the year 2000, which is 200 000 people. Let us write P_0 = 200 000. For the population in 2001 we will use P_1, for the population in 2002 we will use P_2, and so on.

In the following year, 2001, the population grow 9% with respect to the previous year. This means that P_0 is equal to P_1 plus 9% of the population of 2000. Notice that this can be written as

P_1 = P_0 + (9/100)*P_0 = \left(1-\frac{9}{100}\right)P_0 = \frac{109}{100}P_0.

In 2002, we will have the population of 2001, P_1, plus the 9% of P_1. This is

P_2 = P_1 + (9/100)*P_1 = \left(1-\frac{9}{100}\right)P_1 = \frac{109}{100}P_1.

So, it is not difficult to notice that the general recurrence is

P_n = \frac{109}{100}P_{n-1}.

b) In the previous formula we only need to substitute the expression for P_{n-1}:

P_{n-1} = \frac{109}{100}P_{n-2}.

Then,

P_n = \left(\frac{109}{100}\right)^2P_{n-2}.

Repeating the procedure for P_{n-3} we get

P_n = \left(\frac{109}{100}\right)^3P_{n-3}.

But we can do the same operation n times, so

P_n = \left(\frac{109}{100}\right)^nP_{0}.

c) Recall the notation we have used:

P_{0} for 2000, P_{1} for 2001, P_{2} for 2002, and so on. Then, 2016 is P_{16}. So, in order to obtain the approximate population of Tacoma in 2016 is

P_{16} = \left(\frac{109}{100}\right)^{16}P_{0} = (1.09)^{16}P_0 = 3.97\cdot 200000 \approx 794062

d) In this case we want to know when P_n>400000, which is equivalent to

(1.09)^{n}P_0>400000.

Substituting the value of P_0, we get

(1.09)^{n}200000>400000.

Simplifying the expression:

(1.09)^{n}>2.

So, we need to find the value of n such that the above inequality holds.

The easiest way to do this is take logarithm in both hands. Then,

n\ln(1.09)>\ln 2.

So, n>\frac{\ln 2}{\ln(1.09)} = 8.04323172693.

So, the population of Tacoma should exceed the 400 000 by the year 2009.

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3 years ago
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