answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
natulia [17]
1 year ago
10

Demand for Tablet Computers The quantity demanded per month, x, of a certain make of tablet computer is related to the average u

nit price, p (in dollars), of tablet computers by the equation x = f(p) = 100 9 810,000 − p2 It is estimated that t months from now, the average price of a tablet computer will be given by p(t) = 400 1 + 1 8 t + 200 (0 ≤ t ≤ 60) dollars. Find the rate at which the quantity demanded per month of the tablet computers will be changing 25 months from now. (Round your answer to one decimal place.) tablet computers/month
Mathematics
1 answer:
soldier1979 [14.2K]1 year ago
8 0

x = f ( p ) = \frac { 100 } { 9 } \sqrt { 810,000 - p ^ { 2 } } } \\\\ \qquad { p ( t ) = \dfrac { 400 } { 1 + \dfrac { 1 } { 8 } \sqrt { t } } + 200 \quad ( 0 \leq t \leq 60 ) }

Answer:

12.0 tablet computers/month

Step-by-step explanation:

The average price of the tablet 25 months from now will be:

p ( 25) = \dfrac { 400 } { 1 + \dfrac { 1 } { 8 } \sqrt { 25 } } + 200 \\= \dfrac { 400 } { 1 + \dfrac { 1 } { 8 } \times 5 } + 200\\\\=\dfrac { 400 } { 1 + \dfrac { 5 } { 8 } } + 200\\p(25)=\dfrac { 5800 } {13}

Next, we determine the rate at which the quantity demanded changes with respect to time.

Using Chain Rule (and a calculator)

\dfrac{dx}{dt}= \dfrac{dx}{dp}\dfrac{dp}{dt}

\dfrac{dx}{dp}= \dfrac{d}{dp}\left[{ \dfrac { 100 } { 9 } \sqrt { 810,000 - p ^ { 2 } } }\right] =-\dfrac{100}{9}p(810,000-p^2)^{-1/2}

\dfrac{dp}{dt}=\dfrac{d}{dt}\left[\dfrac { 400 } { 1 + \dfrac { 1 } { 8 } \sqrt { t } } + 200 \right]=-25\left[1 + \dfrac { 1 } { 8 } \sqrt { t } \right]^{-2}t^{-1/2}

Therefore:

\dfrac{dx}{dt}= \left[-\dfrac{100}{9}p(810,000-p^2)^{-1/2}\right]\left[-25\left[1 + \dfrac { 1 } { 8 } \sqrt { t } \right]^{-2}t^{-1/2}\right]

Recall that at t=25, p(25)=\dfrac { 5800 } {13} \approx 446.15

Therefore:

\dfrac{dx}{dt}(25)= \left[-\dfrac{100}{9}\times 446.15(810,000-446.15^2)^{-1/2}\right]\left[-25\left[1 + \dfrac { 1 } { 8 } \sqrt {25} \right]^{-2}25^{-1/2}\right]\\=12.009

The quantity demanded per month of the tablet computers will be changing at a rate of 12 tablet computers/month correct to 1 decimal place.

You might be interested in
Steve is trying to increase his average pace per mile by running hills. The hill on 1st Avenue rises 3 vertical feet for each ho
marissa [1.9K]
1st Avenue would be more difficult because it’s rise and run is for every one foot forward it is 3 feet up. Meanwhile avenue 16th would start at (3,1) and the rise and run would be for every 3 feet it would go up 1 foot.
7 0
2 years ago
Read 2 more answers
Which of the following statement is true about k-NN algorithm?
BaLLatris [955]

The answer is D. All of the above.

The computational complexity of K-NN increases as the size of the training data set increase and the algorithm gets significantly slower as the number of examples and independent variables increase.

Also, K-NN is a non-parametric machine learning algorithm and as such makes no assumption about the functional form of the problem at hand.

The algorithm works better with data  of the same scale, hence normalizing the data prior to applying the algorithm is recommended.

6 0
2 years ago
A square with an area A2 of is enlarged to a square with an area of 25A2. How was the side of the smaller square changed?
makvit [3.9K]
Given:
a square with an area of a² is enlarged to a square with an area of 25a².

The side length of the smaller square was changed when The side length was multiplied by 5.

Area = (1a)² = a²
Area = 1a * 5 = 5a ⇒ (5a)² = 25a²
3 0
2 years ago
Read 2 more answers
The random variable X is exponentially distributed, where X represents the waiting time to see a shooting star during a meteor s
faltersainse [42]

Answer:

The parameters of the exponential distribution is 0.0133.

Step-by-step explanation:

Exponential distribution is a continuous probability distribution.

The density function of exponential distribution is,  

f _{X}(x)=\theta \cdot e^{-\theta\cdot x},\ x\geq 0

Here the parameter θ is the reciprocal of the mean of the random variable <em>X</em>.

The random variable <em>X</em> has an average value of 75 seconds.

Compute the parameters of the exponential distribution as follows:

\theta=\frac{1}{\mu}\\\\

  =\frac{1}{75}\\\\=0.013333333333\\\\\approx 0.0133

Thus, the parameters of the exponential distribution is 0.0133.

3 0
1 year ago
The lifespans of lions in a particular zoo are normally distributed. The average lion lives 12.512.512, point, 5 years; the stan
zaharov [31]

This question was not written properly

Complete Question

The lifespans of lions in a particular zoo are normally distributed. The average lion lives 12.5 years; the standard deviation is 2.4 years. Use the empirical rule (68-95-99.7\%)(68−95−99.7%) to estimate the probability of a lion living between 5.3 to 10. 1 years.

Answer:

Thehe probability of a lion living between 5.3 to 10. 1 years is 0.1585

Step-by-step explanation:

The empirical rule formula states that:

1) 68% of data falls within 1 standard deviation from the mean - that means between μ - σ and μ + σ.

2) 95% of data falls within 2 standard deviations from the mean - between μ – 2σ and μ + 2σ.

3) 99.7% of data falls within 3 standard deviations from the mean - between μ - 3σ and μ + 3σ.

Mean is given in the question as: 12.5

Standard deviation : 2.4 years

We start by applying the first rule

1) 68% of data falls within 1 standard deviation from the mean - that means between μ - σ and μ + σ.

μ - σ

12.5 -2.4

= 10.1

We apply the second rule

2) 95% of data falls within 2 standard deviations from the mean - between μ – 2σ and μ + 2σ.

μ – 2σ

12.5 - 2 × 2.4

12.5 - 4.8

= 7.7

We apply the third rule

3)99.7% of data falls within 3 standard deviations from the mean - between μ - 3σ and μ + 3σ.

μ - 3σ

= 12.5 - 3(2.4)

= 12.5 - 7.2

= 5.3

From the above calculation , we can see that

5.3 years corresponds to one side of 99.7%

Hence,

100 - 99.7%/2 = 0.3%/2

= 0.15%

And 10.1 years corresponds to one side of 68%

Hence

100 - 68%/2 = 32%/2 = 16%

So,the percentage of a lion living between 5.3 to 10. 1 years is calculated as 16% - 0.15%

= 15.85%

Therefore, the probability of a lion living between 5.3 to 10. 1 years

is converted to decimal =

= 15.85/ 100

= 0.1585

8 0
2 years ago
Other questions:
  • stilt scored 5 points less than twice the number scored by dunk. together they scored 43 points. how many points were scored by
    10·1 answer
  • Consider the equation 3p – 7 + p = 13. What is the resulting equation after the first step in the solution?
    8·2 answers
  • In a shipment of 850 widgets, 32 are found to be defective. At this rate, how many defective widgets could be expected in 22,000
    10·1 answer
  • The midpoint of MN is point P at (–4, 6). If point M is at (8, –2), what are the coordinates of point N?
    5·2 answers
  • Explain what it means to have a correlation in a scatterplot.
    9·2 answers
  • For a school drama performance, student tickets cost $5 each and adult tickets cost $10 each. The sellers collected $3,570 from
    14·2 answers
  • He table shows the price paid per concert ticket on a popular online auction site. What was the average price paid per ticket? A
    7·2 answers
  • What would the inter-rater reliability be for a 50-item measure in which the number of agreements between Rater 1 and Rater 2 wa
    6·1 answer
  • In a large corporate computer network, user log-ons to the system can be modeled as a Poisson RV with a mean of 25 log-ons per h
    5·1 answer
  • Sana deals in designer handbags. She bought 5 handbags at a price of 18,000 each and sold them all at a profit
    5·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!