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Rashid [163]
3 years ago
12

Kala purchased a prepaid phone card for $30. Long distance calls cost 9 cents a minute using this card. Kala used her card only

once to make a long distance call. If the remaining credit on her card is$28.83 , how many minutes did her call last?
Mathematics
2 answers:
Romashka [77]3 years ago
8 0

Answer:

The call lasted 13 minutes

Step-by-step explanation:

In this question, we are told to find the number of minutes a call lasted.

The initial amount on the card is $30 and after making a phone call just once, the balance went down to $28.83

What was used would be 30 - 28.83 = $1.17 or simply 117 cents

Now these long calls come at a cost of 9 cents per minute and we have a total of 117 cents here; the number of minutes is thus 117/9 = 13

This means Kala spent 13 minutes on the call

ladessa [460]3 years ago
7 0

Answer: Her call lasted 13 minutes.

Step-by-step explanation: Kala purchased a prepaid card with $30 worth of talk time. For long distance calls, the card is charged (deduction) 9 cents per minute.

At the time Kala had only $28.83 left on her prepaid card, she had spent the following;

Call charge = 30.00 - 28.83

Call charge = 1.17

Since the calls are charged in cents, we shall convert the amount spent to the appropriate unit, which is cents.

Note that $1.17 equals 117 cents (1.17 x 100).

A minute’s call would cost 9 cents, and this means a call that cost 117 cents can be calculated as follows;

Charge per min = Total charge/9

Charge per min = 117/9

Charge per min = 13

Therefore her call lasted for 13 minutes.

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C = 5(f-32) / 9 in form ac + b / c
Lera25 [3.4K]

Answer: I don't know if you wanted me to solve for C or F. But, I Solved for both.

Step-by-step explanation:

<u>Solved for F </u>

  • f=(9C)/(5)+32

<u>Solved for C</u>

  • C=(5(f-32))/(9)
3 0
2 years ago
A dresser contains six pairs of shorts one each in the colors, red, black, blue, green, khaki, and gray. The dresser also contai
marissa [1.9K]
Answer: 1/24

---------------------------------------------------------------------
---------------------------------------------------------------------

Work Shown:

A = selects green pair of shorts
B = selects gray t-shirt

P(A) = probability of selecting green shorts
P(A) = (number of green shorts)/(number of shorts total)
P(A) = 1/6
P(B) = probability of selecting gray t-shirt
P(B) = (number of gray t-shirts)/(number of t-shirts total)
P(B) = 1/4

P(A and B) = probability of selecting green shorts AND gray t-shirt
P(A and B) = P(A)*P(B) ... since A and B are independent events
P(A and B) = (1/6)*(1/4)
P(A and B) = (1*1)/(6*4)
P(A and B) = 1/24

Note: The fraction 1/24 is approximately equal to 0.041667

5 0
3 years ago
The graph of an exponential function is given. Which of the following is the correct equation of the function?
katen-ka-za [31]

Answer:

If one of the data points has the form  

(

0

,

a

)

, then a is the initial value. Using a, substitute the second point into the equation  

f

(

x

)

=

a

(

b

)

x

, and solve for b.

If neither of the data points have the form  

(

0

,

a

)

, substitute both points into two equations with the form  

f

(

x

)

=

a

(

b

)

x

. Solve the resulting system of two equations in two unknowns to find a and b.

Using the a and b found in the steps above, write the exponential function in the form  

f

(

x

)

=

a

(

b

)

x

.

EXAMPLE 3: WRITING AN EXPONENTIAL MODEL WHEN THE INITIAL VALUE IS KNOWN

In 2006, 80 deer were introduced into a wildlife refuge. By 2012, the population had grown to 180 deer. The population was growing exponentially. Write an algebraic function N(t) representing the population N of deer over time t.

SOLUTION

We let our independent variable t be the number of years after 2006. Thus, the information given in the problem can be written as input-output pairs: (0, 80) and (6, 180). Notice that by choosing our input variable to be measured as years after 2006, we have given ourselves the initial value for the function, a = 80. We can now substitute the second point into the equation  

N

(

t

)

=

80

b

t

to find b:

⎧

⎪

⎪

⎪

⎪

⎪

⎪

⎪

⎪

⎪

⎨

⎪

⎪

⎪

⎪

⎪

⎪

⎪

⎪

⎪

⎩

N

(

t

)

=

80

b

t

180

=

80

b

6

Substitute using point  

(

6

,

180

)

.

9

4

=

b

6

Divide and write in lowest terms

.

b

=

(

9

4

)

1

6

Isolate  

b

using properties of exponents

.

b

≈

1.1447

Round to 4 decimal places

.

NOTE: Unless otherwise stated, do not round any intermediate calculations. Then round the final answer to four places for the remainder of this section.

The exponential model for the population of deer is  

N

(

t

)

=

80

(

1.1447

)

t

. (Note that this exponential function models short-term growth. As the inputs gets large, the output will get increasingly larger, so much so that the model may not be useful in the long term.)

We can graph our model to observe the population growth of deer in the refuge over time. Notice that the graph below passes through the initial points given in the problem,  

(

0

,

8

0

)

and  

(

6

,

18

0

)

. We can also see that the domain for the function is  

[

0

,

∞

)

, and the range for the function is  

[

80

,

∞

)

.

Graph of the exponential function, N(t) = 80(1.1447)^t, with labeled points at (0, 80) and (6, 180).If one of the data points has the form  

(

0

,

a

)

, then a is the initial value. Using a, substitute the second point into the equation  

f

(

x

)

=

a

(

b

)

x

, and solve for b.

If neither of the data points have the form  

(

0

,

a

)

, substitute both points into two equations with the form  

f

(

x

)

=

a

(

b

)

x

. Solve the resulting system of two equations in two unknowns to find a and b.

Using the a and b found in the steps above, write the exponential function in the form  

f

(

x

)

=

a

(

b

)

x

.

EXAMPLE 3: WRITING AN EXPONENTIAL MODEL WHEN THE INITIAL VALUE IS KNOWN

In 2006, 80 deer were introduced into a wildlife refuge. By 2012, the population had grown to 180 deer. The population was growing exponentially. Write an algebraic function N(t) representing the population N of deer over time t.

SOLUTION

We let our independent variable t be the number of years after 2006. Thus, the information given in the problem can be written as input-output pairs: (0, 80) and (6, 180). Notice that by choosing our input variable to be measured as years after 2006, we have given ourselves the initial value for the function, a = 80. We can now substitute the second point into the equation  

N

(

t

)

=

80

b

t

to find b:

⎧

⎪

⎪

⎪

⎪

⎪

⎪

⎪

⎪

⎪

⎨

⎪

⎪

⎪

⎪

⎪

⎪

⎪

⎪

⎪

⎩

N

(

t

)

=

80

b

t

180

=

80

b

6

Substitute using point  

(

6

,

180

)

.

9

4

=

b

6

Divide and write in lowest terms

.

b

=

(

9

4

)

1

6

Isolate  

b

using properties of exponents

.

b

≈

1.1447

Round to 4 decimal places

.

NOTE: Unless otherwise stated, do not round any intermediate calculations. Then round the final answer to four places for the remainder of this section.

The exponential model for the population of deer is  

N

(

t

)

=

80

(

1.1447

)

t

. (Note that this exponential function models short-term growth. As the inputs gets large, the output will get increasingly larger, so much so that the model may not be useful in the long term.)

We can graph our model to observe the population growth of deer in the refuge over time. Notice that the graph below passes through the initial points given in the problem,  

(

0

,

8

0

)

and  

(

6

,

18

0

)

. We can also see that the domain for the function is  

[

0

,

∞

)

, and the range for the function is  

[

80

,

∞

)

.

Graph of the exponential function, N(t) = 80(1.1447)^t, with labeled points at (0, 80) and (6, 180).

Step-by-step explanation:

4 0
2 years ago
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