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tester [92]
3 years ago
5

Your cell phone plan costs $39.99 per month plus $.15 for each text message you send or receive. You have at most $45 to spend o

n your cell phone bill. What is the maximum number of text messages that you can send or receive next month?
-What information do you know? What information do you need?
-What inequality can you use to find the maximum number of text messages that you can send or receive?
-What are the solutions of the inequality? Are they reasonable?
Mathematics
2 answers:
Andreas93 [3]3 years ago
7 0
I think the aswer is 300 text mesages
babymother [125]3 years ago
7 0
Pretty sure its 33 with a little left over
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NO LINKS!! Jacob just won $32000 on the new game show, "The Wall,". He invested his winnings at an interest rate of 4.5%, compou
mixer [17]

Answers:

a) See the table below

b) The equation is y = 32000(1.01125)^x

c) $40,024.02

d) See the graph below

=========================================================

Explanations:

a)

Start with part (b) where I detail how to get the equation.

Once the equation is found, plug in x = 0 to get

y = 32000(1.01125)^x\\\\y = 32000(1.01125)^0\\\\y = 32000\\\\

Repeat for x = 1

y = 32000(1.01125)^x\\\\y = 32000(1.01125)^1\\\\y = 32360\\\\

Repeat for x = 2, x = 3, x = 4 and x = 20 to get the table shown below.

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

b)

The template for any exponential equation is y = a(b)^x

a = starting amount = 32000

b = growth factor

The annual interest rate is 4.5%

We compound quarterly, so the quarterly rate is (4.5%)/4 = 1.125% which converts to the decimal form 0.01125; adding one to this leads to the growth factor of b = 1.01125

We go from y = a(b)^x to y = 32000(1.01125)^x

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

c)

Plug in x = 20 to represent 20 quarters have elapsed (aka 20/4 = 5 years)

y = 32000(1.01125)^x\\\\y = 32000(1.01125)^{20}\\\\y \approx 32000(1.25075052084381)\\\\y \approx 40024.016667002\\\\y \approx 40024.02\\\\

The investment would be worth $40,024.02 after five years, aka twenty quarters.

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

d)

See below for the graph. I'm using GeoGebra to make the graph. Another option is Desmos. It's preferable to use technology than to graph by hand. If you wanted to graph by hand, then you'd plot each of the points found in the table. Then draw a curve through all those points.

8 0
2 years ago
How do I find the area?
maksim [4K]
Area of square = 5 x 5 = 25
area of trapezoid = 1/2(2+5)(4) = 28/2 = 14
area of figure = 25 + 14 = 39

answer
39 in^2
5 0
3 years ago
Please help! What is the product in the form ax^2+bx+c​
Firlakuza [10]

Answer:

a=2, b=2, c=-12

Step-by-step explanation:

Since it's in factor form, you would first distribute using foil to get 2x^{2} -4x+6x-12 and then combine like terms when necessary. In this case, the fully simplified version is 2x^{2} +2x-12. Therefore your a is 2, b is 2, and c is -12.

3 0
3 years ago
A sample of 12 measurements has a mean of 16.5, and a sample of 15 measurements has a mean of 18.6. Find the mean of all 27 meas
Wewaii [24]

Answer: 17.67

Step-by-step explanation:

Given

Sample  of 12 measurements has a mean of 16.5 and

a sample of 15 measurements has a mean of 18.6

Take \bar{x_1},n_1 be the mean and no of measurements

and \bar{x_2},n_2 be the mean and no of measurements in second case

\therefore \bar{x_1}=\dfrac{\sum a_1}{n_1}\\\\\Rightarrow \sum a_1=\bar{x_1}\times n_1\\\\\Rightarrow \sum a_1=198\quad \ldots(1)

Similarly,

\therefore \bar{x_2}=\dfrac{\sum a_2}{n_2}\\\\\Rightarrow \sum a_2=\bar{x_2}\times n_2\\\\\Rightarrow \sum a_2=279\quad \ldots(2)

Mean of 27 measurements

\Rightarrow \bar{x_3}=\dfrac{\sum a_1+\sum a_2}{n_1+n_2}\\\\\Rightarrow \bar{x_3}=\dfrac{198+279}{12+15}\\\\\Rightarrow \bar{x_3}=\dfrac{477}{27}\\\\\Rightarrow \bar{x_3}=17.67

4 0
2 years ago
Which is equivalent
murzikaleks [220]

Answer:

A. 9m³

Step-by-step explanation:

5 0
3 years ago
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