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Zarrin [17]
3 years ago
6

A phone company offers two packages of service. Package A is $35.00 per month with an additional charge of $0.15 per minute for

long distance. Package B is $45.00 per month with an additional charge of $0.10 per minute for long distance. How many minutes have to be used for the costs of both plans to be the same?
Mathematics
1 answer:
ella [17]3 years ago
3 0

Answer: 200 minutes have to be used for the costs of both plans to be the same.

Step-by-step explanation:

Let x represent the number of minutes that have to be used for the costs of both plans to be the same.

Package A is $35.00 per month with an additional charge of $0.15 per minute for long distance. This means that the cost of using package A for x minutes in a month would be

35 + 0.15x

Package B is $45.00 per month with an additional charge of $0.10 per minute for long distance. This means that the cost of using package A for x minutes in a month would be

45 + 0.1x

For both costs to be the same, it means that

35 + 0.15x = 45 + 0.1x

0.15x - 0.1x = 45 - 35

0.05x = 10

x = 10/0.05

x = 200

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Answer:

We can do it with envelopes with amounts $1,$2,$4,$8,$16,$32,$64,$128,$256 and $489

Step-by-step explanation:

  • Observe that, in binary system, 1023=1111111111. That is, with 10 digits we can express up to number 1023.

This give us the idea to put in each envelope an amount of money equal to the positional value of each digit in the representation of 1023. That is, we will put the bills in envelopes with amounts of money equal to $1,$2,$4,$8,$16,$32,$64,$128,$256 and $512.

However, a little modification must be done, since we do not have $1023, only $1,000. To solve this, the last envelope should have $489 instead of 512.

Observe that:

  1. 1+2+4+8+16+32+64+128+256+489=1000
  2. Since each one of the first 9 envelopes represents a position in a binary system, we can represent every natural number from zero up to 511.
  3. If we want to give an amount "x" which is greater than $511, we can use our $489 envelope. Then we would just need to combine the other 9 to obtain x-489 dollars. Since x-489\leq511, by 2) we know that this would be possible.

4 0
3 years ago
8, 2, 0, 2, 8, 18<br> What are the next three numbers in the sequence?<br> What is the pattern?
Tom [10]

Answer:

22,40,62

Step-by-step explanation:

0+2=2  2+6=8 8+10=18

2 6 10 14 18 22 four in between

so you will add these numbers each time

18+14=22  22+18=40  40+22=62

4 0
3 years ago
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galina1969 [7]

Answer/Step-by-step explanation:

Recall: x^{-a} = \frac{1}{x^a}

a. 4^{-3} = \frac{1}{4^3} = \frac{1}{64}

b. 13^{-2} = \frac{1}{13^2} = \frac{1}{169}

c. (-3)^{-2} = \frac{1}{-3^2} = \frac{1}{9}

6 0
3 years ago
I travelled at 60km/h and took 2 hours for a certain journey. How long would it have taken me if I had travelled at 50km/h?​
Marina86 [1]

Answer:

2 hours and 24 minutes

Step-by-step explanation:

2 hours at 60 km/h means you have travelled 2*60=120 km

120 km at 50 km/h takes 120/50 = 2.4 hours

2.4 hours is 2 hours and 0.4*60 = 24 minutes.

7 0
4 years ago
Work out b(4½-3⅔)+1⅔​
Zolol [24]

Answer:

2 \frac{1}{2}

Step-by-step explanation:

Change the mixed numbers to improper fractions

(\frac{9}{2} - \frac{11}{3} ) + \frac{5}{3} ← the LCM of 2 and 3 is 6

= \frac{9(3)}{2(3)} - \frac{11(2)}{3(2)} + \frac{5(2)}{3(2)}

= \frac{27}{6} - \frac{22}{6} + \frac{10}{6}

= \frac{5}{6} + \frac{10}{6}

= \frac{15}{6}

= \frac{5}{2}

= 2 \frac{1}{2}

6 0
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