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daser333 [38]
2 years ago
7

The value of a car t years after it is purchased is given by the decreasing function V, where V(t) is measured in dollars. The r

ate of change of the car’s value in dollars per year is proportional to the car’s value. Which of the following differential equations could be used to model the value of the car, where k is a constant?
Mathematics
1 answer:
Lana71 [14]2 years ago
7 0

Answer:

dV(t)/dt = kV(t)

Step-by-step explanation:

Since the rate of change of the car’s value in dollars per year dV(t)/dt is proportional to the car’s value V(t), ten

dV(t)/dt ∝ V(t)

dV(t)/dt = kV(t)

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If the equation is in y = k*x form, then we have a direct proportional relationship between x and y. In this case, y = (1/5)*x is in the form y = k*x where k = 1/5. So this equation is proportional. The constant of proportionality is k = 1/5

In terms of a graph, you can tell if it has the following properties:
1) The graph goes through the origin (0,0) which is where the x and y axis cross
2) The graph is a straight line

You should find that graphing y = (1/5)x will satisfy both properties above, so that will visually confirm you have the right answer. The graph is shown in the attached image. The red line represents the graph of the equation. The red line goes through (0,0) and (5,1), which are point A and point B respectively.

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A and B will take the same 10-question examination. Each question will be answered correctly by A with probability .7, independe
MatroZZZ [7]

Answer:

a) The expected number of questions that are answered correctly by both A and B = 11 (7 + 4).

b) The Variance of the number of questions that are answered correctly by either A or B = 2.25.

Step-by-step explanation:

Number of questions in the examination = 10

Probability of A's answer being correct = 0.7

Probability of B's answer being correct = 0.4

The expected number of questions that are answered correctly by both A and B:

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        Correct Answer       Value          Variance

A              0.7                     7 (0.7 * 10)    2.25

B              0.4                     4 (0.4 * 10)    2.25

Total expected value =    11

           Mean =                  5.5                2.25

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A designer is adding a border around the edge of a rectangular swimming pool. He measures the pool and finds that the length of
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Tire pressure monitoring systems (TPMS) warn the driver when the tire pressure of the vehicle is 28% below the target pressure.
Alona [7]

Answer:

Step-by-step explanation:

Hello!

The monitoring system warn the driver when the tire pressure of the vehicle is 28% below target pressure.

Be X: target tire pressure of a certain car (pounds per square inch)

a)

X= 28 psi

If the monitoring system will warn the driver when the pressure is 28% below the target pressure: X-0.28X

First step, you have to calculate the 28% of 28psi

28*0.28= 7.84

Second step, is to subtract the calculated 28% to the target pressure:

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The TPMS will trigger a warning at 20.16 psi.

b)

If X~N(μ;σ²)

μ= 28psi (since the average is on target, then the target pressure for the car will be the average value of the distribution)

σ= 3psi

P(X≤20.16)

The standard normal distribution is tabulated. Any value of any random variable X with normal distribution can be "converted" by subtracting the variable from its mean and dividing it by its standard deviation.

So to calculate each of the asked probabilities, you have to first, "transform" the value of the variable to a value of the standard normal distribution Z, then you use the standard normal tables to reach the corresponding probability.

Z= (X-μ)/σ= (20.16-28)/3= -2.61

Now you have to look for the corresponding value of probability using the Z-table. Since the value is negative you have to the use the left entry of the Z-table, in the first column you'll find the integer and first decimal of the value -2.6- and in the first row you'll find the second decimal value -.-1

The value of probability that corresponds to -2.61 is:

P(Z≤-2.61)= 0.005

c)

You have to calculate the probability of inspecting a tire at random and it being inflated within recommended range, symbolically this is:

P(30≤X≤26)= P(X≤30)-P(X≤26)

Calculate both Z values:

Z= (30-28)/3= 0.67

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P(Z≤0.67)-P(Z≤-0.67)= 0.749 - 0.251= 0.498

The probability of the tire being inflated within recommended inflation range is 0.498.

I hope this helps!

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