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denis23 [38]
1 year ago
5

You want to calculate the value of your car each year when it depreciates by 15%. What does the term y mean in the exponential f

unction y=9500*0.85^x?

Mathematics
2 answers:
insens350 [35]1 year ago
4 0

Answer:

Value of the car after x years

Step-by-step explanation:

Exponential functions are generally expressed in the form: y=a(b)^x and since this function is decreasing, meaning it has an exponential decay, it can be expressed as: y=a(1-r)^x where r=rate of decay.

In this case the a represents the initial value or y-intercept, since when x=0, (1-r)^0 = 1, so we just have y=a(1) or y=a

The r represents the rate of decay

the x usually represents the time in seconds, days, months, or whatever unit is being used

The y value represents the value of whatever your measuring, and in this context it represents the value of the car after x years.

This is because a represents the initial value

After one year it's only 85% worth it's initial value or 15% less which is represented as 0.85(a)

After two years it's only 85% worth the previous year or 15% less the previous year, which is represented as 0.85(0.85(a))

This will continue to decrease the number of years increase

Alex73 [517]1 year ago
4 0

Answer:

the value of your car after x years

Step-by-step explanation:

The given function represents a geometric sequence with initial term $9500 and common factor 0.85.  Note that every time x increases by 1, we get the newest term by multiplying the previous one by 0.85.  The given function y=9500*0.85^x represents the value of your car after x years (which matches the 3rd answer choice).

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Step-by-step explanation:

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A cellular phone company monitors monthly phone usage. The following data represent the monthly phone use of one particular cust
Fiesta28 [93]

SOLUTION

Given the question in the image, the following are the solution steps to answer the question.

STEP 1: Write the given set of values

321,397,559,454,475,324,482,558,369,513,385,360,459,403,498,477,361,366,372,320

STEP 2: Write the formula for calculating the Standard deviation of a set of numbers

\begin{gathered} S\tan dard\text{ deviation=}\sqrt[]{\frac{\sum^{}_{}(x_i-\bar{x})^2}{n-1}} \\ where\text{ }x_i\text{ are data points,} \\ \bar{x}\text{ is the mean} \\ \text{n is the number of values in the data set} \end{gathered}

STEP 3: Calculate the mean

\begin{gathered} \bar{x}=\frac{\sum ^{}_{}x_i}{n} \\ \bar{x}=\frac{\sum ^{}_{}(321,397,559,454,475,324,482,558,369,513,385,360,459,403,498,477,361,366,372,320)}{20} \\ \bar{x}=\frac{8453}{20}=422.65 \end{gathered}

STEP 4: Calculate the Standard deviation

\begin{gathered} S\tan dard\text{ deviation=}\sqrt[]{\frac{\sum^{}_{}(x_i-\bar{x})^2}{n-1}} \\ \sum ^{}_{}(x_i-\bar{x})^2\Rightarrow\text{Sum of squares of differences} \\ \Rightarrow10332.7225+657.9225+18591.3225+982.8225+2740.52251+9731.8225+3522.4225+18319.6225+2878.3225 \\ +8163.1225+1417.5225+3925.0225+1321.3225+386.1225+5677.6225+2953.9225+3800.7225 \\ +3209.2225+2565.4225+10537.0225 \\ \text{Sum}\Rightarrow108974.0275 \\  \\ S\tan dard\text{ deviation}=\sqrt[]{\frac{111714.55}{20-1}}=\sqrt[]{\frac{111714.55}{19}} \\ \Rightarrow\sqrt[]{5879.713158}=76.67928767 \\  \\ S\tan dard\text{ deviation}\approx76.68 \end{gathered}

Hence, the standard deviation of the given set of numbers is approximately 76.68 to 2 decimal places.

STEP 5: Calculate the First and third quartile

\begin{gathered} \text{IQR}=Q_3-Q_1 \\  \\ To\text{ get }Q_1 \\ We\text{ first arrange the data in ascending order} \\ \mathrm{Arrange\: the\: terms\: in\: ascending\: order} \\ 320,\: 321,\: 324,\: 360,\: 361,\: 366,\: 369,\: 372,\: 385,\: 397,\: 403,\: 454,\: 459,\: 475,\: 477,\: 482,\: 498,\: 513,\: 558,\: 559 \\ Q_1=(\frac{n+1}{4})th \\ Q_1=(\frac{20+1}{4})th=\frac{21}{4}th=5.25th\Rightarrow\frac{361+366}{2}=\frac{727}{2}=363.5 \\  \\ To\text{ get }Q_3 \\ Q_3=(\frac{3(n+1)}{4})th=\frac{3\times21}{4}=\frac{63}{4}=15.75th\Rightarrow\frac{477+482}{2}=\frac{959}{2}=479.5 \end{gathered}

STEP 6: Find the Interquartile Range

\begin{gathered} IQR=Q_3-Q_1 \\ \text{IQR}=479.5-363.5 \\ \text{IQR}=116 \end{gathered}

Hence, the interquartile range of the data is 116

3 0
1 year ago
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