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Paraphin [41]
2 years ago
10

A cellphone service provider charges $5.00 per month and $0.20 per minute per call. If a customer's current bill is $55, how man

y minutes did the customer use? (Round any intermediate calculations and your final answer to the nearest whole minute.) 300 minutes 275 minutes 250 minutes 225 minutes
Mathematics
1 answer:
andreev551 [17]2 years ago
3 0

Answer:

250 minutes

Step-by-step explanation:

Given,

Cell phone charges for a month = $ 5.00,

Additional charges per minute = $0.20,

Let the cell phone is used for x minutes for a month such that the total bill is $ 55,

⇒ Cell phone charge for the month + additional charges for x minutes = $ 55

⇒ 5.00 + 0.20x = 55

Subtracting 5 on both sides,

0.20x = 50

x = 250,

Hence, the cell phone is used for 250 minutes.

Third option is correct.

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(x-7)/(x+2)

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(x-7)(x+7)/(x+7)(x+2)

(x-7)/(x+2)

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Determine the value of A <br> Solve it please
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A= 30
Explanation: 88-58=30
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Is -30 - (-52) positive or negative?
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Read 2 more answers
The time intervals between successive barges passing a certain point on a busy waterway have an exponential distribution with me
lisov135 [29]

Answer:

a) <u>0.4647</u>

b) <u>24.6 secs</u>

Step-by-step explanation:

Let T be interval between two successive barges

t(t) = λe^λt where t > 0

The mean of the exponential

E(T) = 1/λ

E(T) = 8

1/λ = 8

λ = 1/8

∴ t(t) = 1/8×e^-t/8   [ t > 0]

Now the probability we need

p[T<5] = ₀∫⁵ t(t) dt

=₀∫⁵ 1/8×e^-t/8 dt

= 1/8 ₀∫⁵ e^-t/8 dt

= 1/8 [ (e^-t/8) / -1/8 ]₀⁵

= - [ e^-t/8]₀⁵

= - [ e^-5/8 - 1 ]

= 1 - e^-5/8 = <u>0.4647</u>

Therefore the probability that the time interval between two successive barges is less than 5 minutes is <u>0.4647</u>

<u></u>

b)

Now we find t such that;

p[T>t] = 0.95

so

t_∫¹⁰ t(x) dx = 0.95

t_∫¹⁰ 1/8×e^-x/8 = 0.95

1/8 t_∫¹⁰ e^-x/8 dx = 0.95

1/8 [( e^-x/8 ) / - 1/8 ]¹⁰_t  = 0.95

- [ e^-x/8]¹⁰_t = 0.96

- [ 0 - e^-t/8 ] = 0.95

e^-t/8 = 0.95

take log of both sides

log (e^-t/8) = log (0.95)

-t/8 = In(0.95)

-t/8 = -0.0513

t = 8 × 0.0513

t = 0.4104 (min)

so we convert to seconds

t = 0.4104 × 60

t = <u>24.6 secs</u>

Therefore the time interval t such that we can be 95% sure that the time interval between two successive barges will be greater than t is <u>24.6 secs</u>

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The decimals are equivalent to given expression are 50.020 and 50.02.
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