A consumer is considering two different purchasing options for the car of their choice. The first option, which is leasing, is described by the equation 250x - y + 4000 = 0 where x represents the number of months of ownership and y represents the total paid for the car after ‘x' months. The second option, which is the financing option, will cost $400 for 0 months of ownership, (0,400), and $4400 for 10 months of ownership, (10, 4400). Part A: Find the equations, in slope/y-intercept form, for each of the purchasing options. Explain the significance of the slope and y-intercept for each purchasing option. Part B: Graph each equation on the same set of axes and compare. Under what conditions is each purchasing option the best choice? Be sure to provide a thorough answer. Use detailed information from your graph to back up your choices.

Step-by-step explanation:

Given that:

First Option (Leasing)

250x - y + 4000 = 0

Expressing the equation in the Slope-Intercept Form y=mx+b, we have:

y=250x+4000

Second Option (Financing)

$400 for 0 months of ownership, (0,400), and $4400 for 10 months of ownership, (10, 4400).

First, we determine the slope of the line joining (0,400) and (10,4400)

We have:

y=400x+b

When y=400, x=0

400=400(0)+b

b=400

Therefore, the Slope-Intercept Form of the second option is:

y=400x+400

Significance

Part B

We notice from the graph that after 24 months, the cost for leasing and financing becomes the same ($10,000). Therefore, a consumer will be better off financing since the downpayment for leasing is higher.

i.e

ad = ae + ed = y + 7

similar triangles therefor

ac/ab = ad/ae

6/4 = (y + 7) / y

cross multiply

6y = 4(y + 7)

6y = 4y + 28

2y = 28

y = 14

step-by-step explanation:

step-by-step explanation:

i found the life tables for the year 2012. 

life table of males at age 30. probability of dying between ages x and x+1

total population:  

29–30 . . . . . . . . . . . . 0.001030

30–31 . . . . . . . . . . . . 0.001061

males:  

29–30 . . . . . . . . . . . . 0.001420

30–31 . . . . . . . . . . . . 0.001448

white population

29–30 . . . . . . . . . . . . 0.001005

30–31 . . . . . . . . . . . . 0.001037

white males

29–30 . . . . . . . . . . . . 0.001372

30–31 . . . . . . . . . . . . 0.001399

non hispanic white population

29–30 . . . . . . . . . . . . 0.001083

30–31 . . . . . . . . . . . . 0.001116

non hispanic white males

29–30 . . . . . . . . . . . . 0.001477

30–31 . . . . . . . . . . . . 0.001503

jacob, non hispanic white male. 30 years old.

probability that jacob lives to be 30 years old. 

since the given probabilities are probabilities of dying, i think we simply subtract it from 1 to get the probability of living. we will use the probability of non hispanic white male.

                                          probability of living to be 30 y.o.

29 - 30 : 1 - 0.001477 = 0.998523

30 - 31 : 1 - 0.001503 = 0.998497