According to a survey conducted by Netpop Research, 65% of new car buyers use online search engines as part of their car-buying experience. Another study reported that 11% of new car buyers skip the test drive. Suppose 7% of new car buyers use online search engines as part of their car-buying experience and skip the test drive. If a new car buyer is randomly selected, what is the probability that:

(Round your answers to 2 decimal places.)

a. the buyer used an online search engine as part of the car-buying experience or skipped the test drive?

b. the buyer did not use an online search engine as part of the car-buying experience or did skip the test drive?

c. the buyer used an online search engine as part of the car-buying experience or did not skip the test drive?

(a) 0.76

(b) 0.46

(c) 0.58

Step-by-step explanation:

Let's first list the probabilities of each event:

Buyers using online search engine: 0.65

Buyers skipping test drive = 0.11

Buyers that use search engines AND skip test drives = 0.65 * 0.11 = 0.07

(a) To find this probability, we simply need to add the probabilities of the above conditions that meet the criteria of the question. Since all of the above conditions meet the criteria set by the question, we have:

Probability = 0.65 + 0.11

Probability = 0.76

(b) The probability of buyers that do not use search engines = 1 - 0.65 = 0.35

Probability of buyers that do not use search engines OR do skip the test drive = 0.35 + 0.11 = 0.46

(c) Probability of buyers buyers that do not skip test drive = 1 - 0.11 = 0.89

Probability of buyers using online search engine AND not skipping the test drive = 0.65 * 0.89 = 0.58

(It must be noted that question c is phrased wrong. Since both events are independent, i.e buying a car OR skipping the test drive would mean 0.65 + 0.89 = 1.54 (not possible) )  

geometric series converge the common ratio r has magnitude less than one.

which is equivalent to

for the example we need

|2-x| < 1

-1 < 2-x   and 2-x < 1

x < 3 and x > 1

we can write that

answer: 1 < x < 3

1a. common ratio r=250/25=10 > 1     diverges

1b.   r=50/100=1/2 converges

when these converge they converge to

where a is the first term.

answer: 200

1c. r=-1   so it's not the case that |r| < 1.     diverges

2a. x is the common ratio so we converge when |x|< 1

2b.   r=-10x / -2 = 5x.

we converge when

|5x| < 1

|x| < 1/5

2c.   r=3-2x

-1 < 3-2x and 3-2x < 1

-4 < -2x and 2 < 2x

x < 2 and 1 < x

1 < x < 2

2d. [/tex] r=-48x^3/12 = 4x^3[/tex]

|4x^3| < 1

|x^3| < \frac 1 4

|x| < 4^{-1/3}

need points to ask question

step-by-step explanation: