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Yanka [14]
3 years ago
10

UPS charges $7 for the first pound, and $0.20 for each additional pound. FedEx charges $5 for the first pound and $0.30 for each

additional pound.
What equation can be used to find how many pounds it will it take for UPS and FedEx to cost the same?
Mathematics
1 answer:
hjlf3 years ago
6 0

Answer:

The costs will be the same at 21 pounds.

Step-by-step explanation:

Let p be the number of additional pounds.  (Total weight would be p+1).

UPS cost = 7 + 0.20p

FedEx cost = 5 + 0.30p

We need these to be equal.

7 + 0.20p = 5 + 0.30p

0.10p = 2

p = 20

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Negative 5.4 + x = 12.2. Minus 5.4. x = 6.8.<br> What error, if any, did she make?
lisabon 2012 [21]

Answer:

The answer is correct x =6.8, if she made an error, it might have been the steps she took.

Step-by-step explanation:

5.4 + x = 12.2

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3 0
3 years ago
If discount rate is 12%, the present value of Rs X recieved at the end of each year for the next five years is equal to
myrzilka [38]

Answer:

Present Value = X [\frac{1}{(1 + 0.12)^{1} }  + \frac{1}{(1 + 0.12)^{2} }  + \frac{1}{(1 + 0.12)^{3} }  + \frac{1}{(1 + 0.12)^{4} }  + \frac{1}{(1 + 0.12)^{5} } ]

Step-by-step explanation:

To find - If discount rate is 12%, the present value of Rs X received at the end of each year for the next five years is equal to .... ?

Solution -

We know that, formula for finding the Present vale is given by

Present value = Future value / (1 + r)ⁿ

where r is the rate of interest

and n is Number of periods

Now,

Here in the question, we have

r = 12% = 12/100 = 0.12

n = 5

Also, Given that, we have received Rs X at the end of each year

So,

Present Value = \frac{X}{(1 + 0.12)^{1} }  + \frac{X}{(1 + 0.12)^{2} }  + \frac{X}{(1 + 0.12)^{3} }  + \frac{X}{(1 + 0.12)^{4} }  + \frac{X}{(1 + 0.12)^{5} }

                        = X [\frac{1}{(1 + 0.12)^{1} }  + \frac{1}{(1 + 0.12)^{2} }  + \frac{1}{(1 + 0.12)^{3} }  + \frac{1}{(1 + 0.12)^{4} }  + \frac{1}{(1 + 0.12)^{5} } ]

⇒Present Value = X [\frac{1}{(1 + 0.12)^{1} }  + \frac{1}{(1 + 0.12)^{2} }  + \frac{1}{(1 + 0.12)^{3} }  + \frac{1}{(1 + 0.12)^{4} }  + \frac{1}{(1 + 0.12)^{5} } ]

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