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Alisiya [41]
3 years ago
7

Customers of a phone company can choose between two service plans for long distance calls. The first plan has no monthly fee but

charges for each minute of calls. The second plan has a monthly fee and charges an additional for each minute of calls. For how many minutes of calls will the costs of the two plans be equal?
Mathematics
1 answer:
hram777 [196]3 years ago
5 0

Answer:

\\x= P/(c -d)[/tex],

Assume that the price of each minute in the first plan is $c and that the second plan charges a flat rate of $P and a charge  of additional $d for every minute.

Step-by-step explanation

Assume that the price of each minute in the first plan is $c and that the second plan charges a flat rate of $P and a charge  of additional $d for every minute.

Thus, the monthly cost of a customer who consumes x minutes in each plan is:

For the first plan: cx

and for the second plan: P + dx

Considering that the monthly costs must be the same in each plan, you have to:

cx = P + dx\\ transposing terms\\cx - dx = P\\   applying common factor\\(c -d)x = P\\ dividing by [tex]c - d

\\x= P/(c -d)[/tex].

For example if c = $2; d = $1 y P = $10, Then the number of minutes would be, x=10  and the total cost for each plan would be $20

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Answer:

  b.  x=1, y=2, z=3

Step-by-step explanation:

The system of equations ...

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has solution (x, y, z) = (1, 2, 3) . . . . matches choice B.

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While it is convenient to solve this using a graphing calculator or web site, one can easily solve the system by hand.

Subtract the second equation from 3 times the first:

  3(3x +2y +z) -(9x -6y +z) = 3(10) -(0)

  12y + 2z = 30 . . . . simplify

Dividing this result by 2 gives ...

  6y +z = 15 . . . . . . [eq4]

Subtract 3 times the third equation from the first:

  (3x +2y +z) -3(x -y -3z) = (10) -3(-10)

  5y +10z = 40 . . . . simplify

  y + 2z = 8 . . . . . . . divide by 5 . . . . . [eq5]

The two equations [eq4] and [eq5] can be solved any of the ways you usually solve two equations in two variables. Here, we'll use the first equation to write an expression for z that we can substitute into the second equation.

  z = 15 -6y . . . . . subtract 6y from [eq4]

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  -11y = -22 . . . . . . . subtract 30

  y = 2 . . . . . . . . . . . divide by the coefficient of y

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  x - 2 -3(3) = -10

  x -11 = -10 . . . . . . . . simplify

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<em>Comment on the problem statement</em>

Math is generally unforgiving of imprecision. The given system of equations has no variable "z", and some other typos are apparently involved. That is why we rewrote the system to the equations shown above.

It is very easy to mistake z for 2, or g for 9, or o for 0, or 1 for 7. There are other confusions that are possible, as well. Letters I (eye) and l (ell) are easily confused, and may be confused with 1 (one) as well. Sometimes y and 4, or 4 and 9, can also be written so as to be difficult to tell apart. Great care must be taken when handwriting these symbols.

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50 POINTS!!!!! PLEASE WILL MARK BRAINLIEST
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  • Q11 - 12 sides
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Step-by-step explanation:

<u>Use the formula for sum of interior angles of a regular polygon:</u>

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<u>Each angle A measures:</u>

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<u>Each angle is 150, then finding n:</u>

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<u>Each angle is 144, find n:</u>

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Assuming that boy and girl babies are equally likely, what would be Kathy's probability of having at most one daughter if she
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Step-by-step explanation:

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0.5 * 0.5 * 0.5 * 0.5 = 0.0625

<u>Outcome 2: first child is a girl, all others are boys</u>

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0.5 * 0.5 * 0.5 * 0.5 = 0.0625

<u>Outcome 3: second child is a girl, all others are boys</u>

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0.5 * 0.5 * 0.5 * 0.5 = 0.0625

<u>Outcome 4: third child is a girl, all others are boys</u>

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0.5 * 0.5 * 0.5 * 0.5 = 0.0625

<u>Outcome 5: fourth child is a girl, all others are boys</u>

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0.5 * 0.5 * 0.5 * 0.5 = 0.0625

Therefore,

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4 0
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