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lara31 [8.8K]
3 years ago
7

Tina sends 12 text messages each day in June. A mobile phone company offers two packages. Package A: The first 100 texts each mo

nth cost 3 paisa each, the second 100 texts cost 2p each and after this the cost of each text is 1paisa. Package B: All texts cost 2paisa. Which package offers her the better value for money in 30 days? Justify your answer.
Mathematics
1 answer:
Mnenie [13.5K]3 years ago
6 0

Two options are given to Tina. We find how much she pays for each option, and the one in which she pays less offers the better value.

Tina sends 12 text messages each day in June

June has 30 days, so in June, Tina sent 12*30 = 360 text messages.

Package A:

First 100 messages cost 3p, the next 100 cost $2 and after that each costs $1.

She sends 360 messages, with:

The first 100 costing 3p.

The next 100 costing 2p.

The final 360 - 200 = 160 costing 1p.

She pays:

100*3 + 100*2 + 160*1 = 660

Package B:

2p for each message, 360 messages, so:

360*2p = 720p.

Which package is better?

Package A costs less, thus, it offers her the better value for money in 30 days.

Another example of a problem in which a person has to choose between two packages is given in brainly.com/question/10693932

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Place the three functions in order from the fastest decreasing average rate of change to the slowest decreasing average rate of
victus00 [196]

Answer:

g(x), f(x) and h(x)

Step-by-step explanation:

Given

Interval: (0,3)

See attachment for functions f(x), g(x) and h(x)

Required

Order from fastest to slowest decreasing average rate of change

The average rate of change is calculated as:

Rate = \frac{f(b) - f(a)}{b - a}

In this case:

(a,b) = (0,3)

i.e.

a = 0\\b=3

For f(x)

f(x) = 16(\frac{1}{2})^x

Rate = \frac{f(b) - f(a)}{b - a}

Rate = \frac{f(3) - f(0)}{3 - 0}

Rate = \frac{f(3) - f(0)}{3}

Calculate f(3) and f(0)

f(x) = 16(\frac{1}{2})^x

f(3) = 16(\frac{1}{2})^3 = 16 * \frac{1}{8} = 2

f(0) = 16(\frac{1}{2})^0 = 16 * 1 = 16

So:

Rate = \frac{f(3) - f(0)}{3}

Rate = \frac{2 - 16}{3}

Rate = -\frac{14}{3}

For g(x)

Rate = \frac{g(b) - g(a)}{b - a}

Rate = \frac{g(3) - g(0)}{3 - 0}

Rate = \frac{g(3) - g(0)}{3}

From the table of g(x)

g(3) = 1

g(1) = 27

So:

Rate = \frac{1 - 27}{3}

Rate = -\frac{26}{3}

For h(x)

Rate = \frac{h(b) - h(a)}{b - a}

Rate = \frac{h(3) - h(0)}{3 - 0}

Rate = \frac{h(3) - h(0)}{3}

From the graph of h(x)

h(3) = -3

h(0) = 4

So:

Rate = \frac{-3 - 4}{3}

Rate = -\frac{7}{3}

So, the calculated rates of change are:

f(x) = -\frac{14}{3} = -4.67

g(x) = -\frac{26}{3} =-8.67

h(x) = -\frac{7}{3} =-2.33

By comparison:

From the fastest decreasing to slowest, the order is: <em>g(x), f(x) and h(x)</em>

4 0
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