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sertanlavr [38]
4 years ago
5

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

charges $ 0.14 for each minute of calls. The second plan has a $ 22 monthly fee and charges an additional $ 0.10 for each minute of calls. For how many minutes of calls will the costs of the two plans be equal?
Mathematics
1 answer:
MAVERICK [17]4 years ago
5 0
Hey there!

We'll define x as the amount of minutes for a call.

The monthly fee is the initial value, while the cost per call is te constant. The cost per call is the coefficient of x because you're multiplying the cost/call times the number of calls.

Now, we'll look at the first company, that has no monthly fee. However, it has 14 cents/minute, so we have:

y = .14x

For the second one, we have a 22 dollar upfront fee, along with 10 cents per call. In this problem, the 10 cents is the cost per call, or the coefficient of x.

We have:

y = 22 + .10x

Now, to see when the minutes of calls will equal to when the costs are equal, we set both equations equal to each other because we want to see the value of x that works on the left and right side of the equation:

22 + .10x = .14x

Subtract .10x from both sides:

22 = .04x

Divide both sides by .04:

x = 550

If we plug it back in, we get:

22 + .10(550) = .14(550)

77 = 77

Therefore, you would need 550 calls.

Hope this helps!


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The pie must stay for 60 - x more minutes must the pie bake

<h3>For how many more minutes must the pie bake?</h3>

Let the amount of time spent in the oven be x

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Total amount of time = 60 minutes

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Evaluate

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Answer:

87.5%

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4 years ago
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A box contains 15 resistors. twelve of them are labelled 50ω and the other three are labeled 100ω. what is the probability that
Igoryamba

Answer : P(second resistor is 100ω , given that the first resistor is 50ω) is given by

\frac{1}{5}

Explanation :

Since we have given that

Total number of resistors =15

Number of resistors labelled with 50ω = 12

Number of resistors labelled with 100ω =3

Let A: Event getting resistor with 50ω

B: Event getting resistor with 100ω

Since A and B are independent events .

So,

P(A\cap B)=P(A).P(B)

Now, According to question , we can get that

P(A)= \frac{12}{15}=\frac{4}{5}\\\\P(B)=\frac{3}{15}=\frac{1}{5}

So,

P(A\cap B)=P(A).P(B)\\\\P(A\cap B)=\frac{4}{5}\times \frac{1}{5}\\\\P(A\cap B)=\frac{4}{25}

So, by using the conditional probability , which state that

P(B\mid A)=\frac{P(A\cap B)}{P(A)}

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So, P(second resistor is 100ω , given that the first resistor is 50ω) is given by

\frac{1}{5}


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