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ArbitrLikvidat [17]

2 years ago

11

The initial value of a car is $27,000. After one year, the value of the car is $20,250. What exponential function models the exp

ected value of the car? Estimate the value of the car after 3 years.

2 answers:

Sidana [21]2 years ago

8 0

11390.63.... Hope this helps

Schach [20]2 years ago

5 0

<span>You are given the initial value of a car of $27,000 and after one year, the value of the car is $20,250. The exponential function models the expected value of the car after 3 years is $11,390.63. </span>

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This table represents a quadratic function. x y 0 -3 1 -3.75 2 -4 3 -3.75 4 -3 5 -1.75 What is the value of a in the function’s

serg [7]

Answer:

a = 0.25

Step-by-step explanation:

Our strategy to solve this problem will be to use the information given in the table to obtain first the value of c in the quadratic equation which has the form ax^2 + bx + c and then form a system of 2 linear equations and solve for the coefficients a and b as follow:

from x=0 and y=-3 a*0 + b* 0 + c = - 3

c = -3

from x=1 and y= -3.75

a*(1) + *(1) + (-3) = -3.75

a + b = -3.75 + 3 = -0.75

from x= 2 and y = -4

a* (2)^2 + b*(2) + (-3) = -4

4a + 2b = -4 +1

4a + 2b = -1

Now we can solve the system f equations by elimination:

a + b = -0.75

4a + 2b = -1

multiply first equation by -2 and add to the second and get

-2a - 2b = +1.50

4a + 2b = -1

2a = 0.50 and substituting into any of the equations get b = -1

so the quadratic equation has a= 0.25 b= -1 c= -3

we can even plug the any of the other values for x given in the table and check the answer.

8 0

2 years ago

Read 2 more answers

A machine produces parts that are either defect free (90%), slightly defective (3%), or obviously defective (7%). Prior to shipm

AURORKA [14]

Answer:

(a) 0.0686

(b) 0.9984

(c) 0.0016

Step-by-step explanation:

Given that a machine produces parts that are either defect free (90%), slightly defective (3%), or obviously defective (7%).

Let A, B, and C be the events of defect-free, slightly defective, and the defective parts produced by the machine.

So, from the given data:

P(A)=0.90, P(B)=0.03, and P(C)=0.07.

Let E be the event that the part is disregarded by the inspection machine.

As a part is incorrectly identified as defective and discarded 2% of the time that a defect free part is input.

So, $P\left(\frac{E}{A}\right)=0.02$

Now, from the conditional probability,

$P\left(\frac{E}{A}\right)=\frac{P(E\cap A)}{P(A)}$

$\Rightarrow P(E\cap A)=P\left(\frac{E}{A}\right)\times P(A)$

$\Rightarrow P(E\cap A)=0.02\times 0.90=0.018\cdots(i)$

This is the probability of disregarding the defect-free parts by inspection machine.

Similarly,

$P\left(\frac{E}{A}\right)=0.40$

and $\Rightarrow P(E\cap B)=0.40\times 0.03=0.012\cdots(ii)$

This is the probability of disregarding the partially defective parts by inspection machine.

$P\left(\frac{E}{A}\right)=0.98$

and $\Rightarrow P(E\cap C)=0.98\times 0.07=0.0686\cdots(iii)$

This is the probability of disregarding the defective parts by inspection machine.

(a) The total probability that a part is marked as defective and discarded by the automatic inspection machine

$=P(E\cap C)$

$= 0.0686$ [from equation (iii)]

(b) The total probability that the parts produced get disregarded by the inspection machine,

$P(E)=P(E\cap A)+P(E\cap B)+P(E\cap C)$

$\Rightarrow P(E)=0.018+0.012+0.0686$

$\Rightarrow P(E)=0.0986$

So, the total probability that the part produced get shipped

$=1-P(E)=1-0.0986=0.9014$

The probability that the part is good (either defect free or slightly defective)

$=\left(P(A)-P(E\cap A)\right)+\left(P(B)-P(E\cap B)\right)$

$=(0.9-0.018)+(0.03-0.012)$

$=0.9$

So, the probability that a part is 'good' (either defect free or slightly defective) given that it makes it through the inspection machine and gets shipped

$=\frac{\text{Probabilily that shipped part is 'good'}}{\text{Probability of total shipped parts}}$

$=\frac{0.9}{0.9014}$

$=0.9984$

(c) The probability that the 'bad' (defective} parts get shipped

=1- the probability that the 'good' parts get shipped

=1-0.9984

=0.0016

5 0

2 years ago

Water is poured into a conical paper cup at the rate of 3/2 in3/sec (similar to Example 4 in Section 3.7). If the cup is 6 inche

aliya0001 [1]

Answer:

The water level rising when the water is 4 inches deep is $\frac{3}{8\times \pi} inch/s$ .

Step-by-step explanation:

Rate of water pouring out in the cone = R= $\frac{3}{2} inch^3/s$

Height of the cup = h = 6 inches

Radius of the cup = r = 3 inches

$\frac{r}{h}=\frac{3 inch}{6 inch}=\frac{1}{2}$

r = h/2

Volume of the cone = $V=\frac{1}{3}\pi r^2h$

$V=\frac{1}{3}\pi r^2h$

$\frac{dV}{dt}=\frac{d(\frac{1}{3}\pi r^2h)}{dt}$

$\frac{dV}{dt}=\frac{d(\frac{1}{3}\pi (\frac{h}{2})^2h)}{dt}$

$\frac{dV}{dt}=\frac{1}{3\times 4}\pi \times \frac{d(h^3)}{dt}$

$\frac{dV}{dt}=\frac{1\pi }{12}\times 3h^2\times \frac{dh}{dt}$

$\frac{3}{2} inch^3/s=\frac{1\pi }{12}\times 3h^2\times \frac{dh}{dt}$

h = 4 inches

$\frac{3}{2} inch^3/s=\frac{1\pi }{12}\times 3\times (4inches )^2\times \frac{dh}{dt}$

$\frac{3}{2} inch^3/s=\pi\times 4\times \frac{dh}{dt} inches^2$

$\frac{dh}{dt}=\frac{3}{8\times \pi} inch/s$

The water level rising when the water is 4 inches deep is $\frac{3}{8\times \pi} inch/s$ .

6 0

2 years ago

During a college fun-fair, the entrance fee for students was $12 and for teachers (and their family members) it was $30. 75 peop

solmaris [256]

Answer:

25 Students attended and 50 teachers and their family members attended

Step-by-step explanation:

50x30=1500

25x12=300

1500+300=1800

Hope this helped :)

3 0

2 years ago

A local amusement park has 30 rides that park visitors can go on. The following table shows the relative frequency distribution

erica [24]

Answer:

5.83

Step-by-step explanation:

Find the mean:

10(.1) + 15(.2) + 20(.2) + 25(.4) + 3-).1)

u = 21

Find the standard deviation:

$\sqrt{(10 -21)x^{2} (.1) + (15 -21)x^{2} (.2)+(20-21)x^{2} (.2)+(25-21)x^{2} (.4)+(30-21)x^{2} (.1)}$ = 5.83

5 0

2 years ago

Read 2 more answers