a. Answer About 400 times the ad appear. Explanation Given information : The probability that an ad appears when a particular website is loaded is 0.2. The website gets 2000 hits a day . Calculation : Since, the probability that an ad appears when a particular website is loaded is 0.2. So, to find how many times the ad appears multiply the probability with the number of hits in a day. Therefore, 0.2 ⋅ 2000 = 400 Hence, About 400 times the ad appear. b. To determine To find : the numbers hits the website must have so that the ad appears 1000 times. Answer The website has 20,000 hits for the website so that the ad appears 1000 times. Explanation Given information : The fraction of number of cookies to bake with the number of cookies the recipe makes is 3 2 . Calculation : Since, the probability that an ad appears when a particular website is loaded is 0.05 and it appears 1000 times, so to find out number of hits divide the appearance of ad with the probability of appearing. 1000 0.05 = 20 , 000 Hence, The website has 20,000 hits for the website so that the ad appears 1000 times.

12th Edition

ISBN: 9780358403685

Author: Larson

Publisher: HM CC

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Question

Chapter 6.7, Problem 19E

To determine

To find : the number of times the ad appears.

Expert Solution

Answer to Problem 19E

About 400 times the ad appear.

Explanation of Solution

Given information :

The probability that an ad appears when a particular website is loaded is 0.2. The website gets 2000 hits a day.

Calculation :

Since, the probability that an ad appears when a particular website is loaded is 0.2. So, to find how many times the ad appears multiply the probability with the number of hits in a day. Therefore,

0.2⋅2000=400

Hence,

About 400 times the ad appear.

To determine

To find : the numbers hits the website must have so that the ad appears 1000 times.

Expert Solution

Answer to Problem 19E

The website has 20,000 hits for the website so that the ad appears 1000 times.

Explanation of Solution

Given information :

The fraction of number of cookies to bake with the number of cookies the recipe makes is 32 .

Calculation :

Since, the probability that an ad appears when a particular website is loaded is 0.05 and it appears 1000 times, so to find out number of hits divide the appearance of ad with the probability of appearing.

10000.05=20,000

Hence,

The website has 20,000 hits for the website so that the ad appears 1000 times.

Chapter 6.7, Problem 18E

Chapter 6.7, Problem 20E

Chapter 6 Solutions

EBK PRE-ALGEBRA

Ch. 6.1 - Prob. 1C Ch. 6.1 - Prob. 2C Ch. 6.1 - Prob. 3C Ch. 6.1 - Prob. 4C Ch. 6.1 - Prob. 1E Ch. 6.1 - Prob. 2E Ch. 6.1 - Prob. 3E Ch. 6.1 - Prob. 4E Ch. 6.1 - Prob. 5E Ch. 6.1 - Prob. 6E

Ch. 6.1 - Prob. 7E Ch. 6.1 - Prob. 8E Ch. 6.1 - Prob. 9E Ch. 6.1 - Prob. 10E Ch. 6.1 - Prob. 11E Ch. 6.1 - Prob. 12E Ch. 6.1 - Prob. 13E Ch. 6.1 - Prob. 14E Ch. 6.1 - Prob. 15E Ch. 6.1 - Prob. 16E Ch. 6.1 - Prob. 17E Ch. 6.1 - Prob. 18E Ch. 6.1 - Prob. 19E Ch. 6.1 - Prob. 20E Ch. 6.1 - Prob. 21E Ch. 6.1 - Prob. 22E Ch. 6.1 - Prob. 23E Ch. 6.1 - Prob. 24E Ch. 6.1 - Prob. 25E Ch. 6.1 - Prob. 26E Ch. 6.1 - Prob. 27E Ch. 6.1 - Prob. 28E Ch. 6.1 - Prob. 29E Ch. 6.1 - Prob. 30E Ch. 6.1 - Prob. 31E Ch. 6.1 - Prob. 32E Ch. 6.1 - Prob. 33E Ch. 6.1 - Prob. 34E Ch. 6.1 - Prob. 35E Ch. 6.1 - Prob. 36E Ch. 6.1 - Prob. 37E Ch. 6.1 - Prob. 38E Ch. 6.1 - Prob. 39E Ch. 6.1 - Prob. 40E Ch. 6.1 - Prob. 41E Ch. 6.1 - Prob. 42E Ch. 6.1 - Prob. 43E Ch. 6.1 - Prob. 44E Ch. 6.1 - Prob. 45E Ch. 6.1 - Prob. 46E Ch. 6.1 - Prob. 47E Ch. 6.1 - Prob. 48E Ch. 6.1 - Prob. 49E Ch. 6.1 - Prob. 50E Ch. 6.1 - Prob. 51E Ch. 6.1 - Prob. 52E Ch. 6.1 - Prob. 53E Ch. 6.1 - Prob. 54E Ch. 6.1 - Prob. 55E Ch. 6.1 - Prob. 56E Ch. 6.2 - 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Prob. 18E Ch. 6.7 - Prob. 19E Ch. 6.7 - Prob. 20E Ch. 6.7 - Prob. 21E Ch. 6.7 - Prob. 22E Ch. 6.7 - Prob. 23E Ch. 6.7 - Prob. 24E Ch. 6.7 - Prob. 25E Ch. 6.7 - Prob. 26E Ch. 6.7 - Prob. 27E Ch. 6.7 - Prob. 28E Ch. 6.7 - Prob. 29E Ch. 6.7 - Prob. 30E Ch. 6.7 - Prob. 31E Ch. 6.7 - Prob. 32E Ch. 6.8 - Prob. 2C Ch. 6.8 - Prob. 1E Ch. 6.8 - Prob. 2E Ch. 6.8 - Prob. 3E Ch. 6.8 - Prob. 4E Ch. 6.8 - Prob. 5E Ch. 6.8 - Prob. 6E Ch. 6.8 - Prob. 7E Ch. 6.8 - Prob. 8E Ch. 6.8 - Prob. 9E Ch. 6.8 - Prob. 10E Ch. 6.8 - Prob. 11E Ch. 6.8 - Prob. 12E Ch. 6.8 - Prob. 13E Ch. 6.8 - Prob. 14E Ch. 6.8 - Prob. 15E Ch. 6.8 - Prob. 16E Ch. 6.8 - Prob. 17E Ch. 6.8 - Prob. 18E Ch. 6.8 - Prob. 19E Ch. 6.8 - Prob. 20E Ch. 6.8 - Prob. 21E Ch. 6.8 - Prob. 22E Ch. 6.8 - Prob. 23E Ch. 6.8 - Prob. 24E Ch. 6.8 - Prob. 25E Ch. 6.8 - Prob. 26E Ch. 6.8 - Prob. 27E Ch. 6.8 - Prob. 28E Ch. 6.8 - Prob. 29E Ch. 6.8 - Prob. 30E Ch. 6 - Prob. 1PSQ Ch. 6 - Prob. 2PSQ Ch. 6 - Prob. 3PSQ Ch. 6 - Prob. 4PSQ Ch. 6 - Prob. 5PSQ Ch. 6 - 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Prob. 29CR Ch. 6 - Prob. 30CR Ch. 6 - Prob. 31CR Ch. 6 - Prob. 32CR Ch. 6 - Prob. 33CR Ch. 6 - Prob. 34CR Ch. 6 - Prob. 35CR Ch. 6 - Prob. 1CT Ch. 6 - Prob. 2CT Ch. 6 - Prob. 3CT Ch. 6 - Prob. 4CT Ch. 6 - Prob. 5CT Ch. 6 - Prob. 6CT Ch. 6 - Prob. 7CT Ch. 6 - Prob. 8CT Ch. 6 - Prob. 9CT Ch. 6 - Prob. 10CT Ch. 6 - Prob. 11CT Ch. 6 - Prob. 12CT Ch. 6 - Prob. 13CT Ch. 6 - Prob. 14CT Ch. 6 - Prob. 15CT Ch. 6 - Prob. 16CT Ch. 6 - Prob. 17CT Ch. 6 - Prob. 18CT Ch. 6 - Prob. 19CT Ch. 6 - Prob. 20CT Ch. 6 - Prob. 21CT Ch. 6 - Prob. 22CT Ch. 6 - Prob. 23CT Ch. 6 - Prob. 1CST Ch. 6 - Prob. 2CST Ch. 6 - Prob. 3CST Ch. 6 - Prob. 4CST Ch. 6 - Prob. 5CST Ch. 6 - Prob. 6CST Ch. 6 - Prob. 7CST Ch. 6 - Prob. 8CST Ch. 6 - Prob. 9CST

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