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Wewaii [24]
3 years ago
14

Two students were asked to find the value of a $1000 item after 3 years. The item was depreciating (losing value) at a rate of 4

0% per year. Which is incorrect? Explain the error. Student 1: `1000(0.6)^{3}=$216` Student 2: `1000(0.4\right)^{3}=\$64`
Mathematics
1 answer:
Yuki888 [10]3 years ago
8 0

Answer:

<em>Student 2 is incorrect because he didn't use the formula properly</em>

Step-by-step explanation:

The exponential function is often used to model natural growing or decaying processes, where the change is proportional to the actual quantity.

An exponential decaying function is expressed as:

C(t)=C_o\cdot(1-r)^t

Where:

C(t) is the actual value of the function at time t

Co is the initial value of C at t=0

r is the decaying rate, expressed in decimal

The initial value of the item is Co=$1000, the rate of decay is r=40%=0.4, and the time is t=3 years.

Substituting into the formula:

C(3)=\$1000\cdot(1-0.4)^3

C(3)=\$1000\cdot0.6^3

C(3)=$216

Student 2 is incorrect because he didn't use the formula properly

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Part 3 - Discussion/Explanation Question
SpyIntel [72]

Step-by-step explanation:

Vertical asymptote can be Identites if there is a factor only in the denominator. This means that the function will be infinitely discounted at that point.

For example,

\frac{1}{x - 5}

Set the expression in the denominator equal to 0, because you can't divide by 0.

x - 5 = 0

x = 5

So the vertical asymptote is x=5.

Disclaimer if you see something like this

\frac{(x - 5)(x + 3)}{(x - 5)}

x=5 won't be a vertical asymptote, it will be a hole because it in the numerator and denominator.

Horizontal:

If we have a function like this

\frac{1}{x}

We can determine what happens to the y values as x gets bigger, as x gets bigger, we will get smaller answers for y values. The y values will get closer to 0 but never reach it.

Remember a constant can be represent by

a \times  {x}^{0}

For example,

1 = 1 \times  {x}^{0}

2 =  2 \times {x}^{0}

And so on,

and

x =  {x}^{1}

So our equation is basically

\frac{1 \times  {x}^{0} }{ {x}^{1} }

Look at the degrees, since the numerator has a smaller degree than the denominator, the denominator will grow larger than the numerator as x gets larger, so since the larger number is the denominator, our y values will approach 0.

So anytime, the degree of the numerator < denominator, the horizontal asymptote is x=0.

Consider the function

\frac{3 {x}^{2} }{ {x}^{2}  + 1}

As x get larger, the only thing that will matter will be the leading coefficient of the leading degree term. So as x approach infinity and negative infinity, the horizontal asymptote will the numerator of the leading coefficient/ the leading coefficient of the denominator

So in this case,

x =  \frac{3}{1}

Finally, if the numerator has a greater degree than denominator, the value of horizontal asymptote will be larger and larger such there would be no horizontal asymptote instead of a oblique asymptote.

8 0
2 years ago
Michaela works in a bank she is paid £19200 per annum how much does she get per month
Step2247 [10]

in a year she earns=19200

So in a month she earns=19200/12

=1600

So she earns 1600 every month.

If the answer is correct then make me brain list.please


4 0
3 years ago
What is 2 over 3 + 4 over3
Fynjy0 [20]
2/3 +4/3=6/3=2.... Self explanatory
8 0
3 years ago
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3x + 3 = x - 5<br> Solve for x
Mars2501 [29]

Answer:

x = -4

Step-by-step explanation:

plz vote brainliest :)

4 0
3 years ago
Read 2 more answers
(URGENT) Madelina wants to make a prediction using a linear model with a correlation coefficient of 0.55.
alekssr [168]

Answer:

<h2>Her prediction will be somewhat likely to be close to the actual value.</h2>

Step-by-step explanation:

A correlation coefficient of 0.55 implies a weak positive correlation, because it's close to 0.50. Remember that strong correlations are 1 or -1, positive or negative, and the middle of the interval represents no correlation.

Having said that, 0.55 indicates some correlation, but weak. Therefore, the right answer is C.

3 0
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