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Elodia [21]
3 years ago
7

Advertisers contract with Internet service providers and search engines to place ads on websites. They pay a fee based on the nu

mber of potential customers who click on their ad. Unfortunately, click fraud—the practice of someone clicking on an ad solely for the purpose of driving up advertising revenue—has become a problem. According to BusinessWeek, 33% of advertisers claim they have been a victim of click fraud. Suppose a simple random sample of 330 advertisers will be taken to learn more about how they are affected by this practice.
a. What is the probability that the sample proportion will be within ±0.02 of the population proportion experiencing click fraud?

b. What is the probability that the sample proportion will be greater than 0.36?
Mathematics
1 answer:
larisa [96]3 years ago
7 0
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You might be interested in
Need help ASAP please​
victus00 [196]

Answer:

x = 5

Step-by-step explanation:

im too tired to explain just trust me

(5) + 2 = 7

2(5) + 3 = 13

13 * 7 = 91

4 0
3 years ago
Read 2 more answers
Suppose that a recent poll of American households about car ownership found that for households with a car, 39% owned a sedan, 3
Rama09 [41]

Answer:

The probability that of the 3 households randomly selected at least 1 owns a sports car is 0.1956.

Step-by-step explanation:

Let <em>X</em> = number of household owns a sports car.

The probability of <em>X</em> is, P (X) = p = 0.07.

Then the random variable <em>X</em> follows a Binomial distribution with <em>n</em> = 3 and <em>p</em> = 0.07.

The probability function of a binomial distribution is:

P(X=x) = {n\choose x}p^{x}[1-p]^{n-x}\\

Compute the probability that of the 3 households randomly selected at least 1 owns a sports car:

P(X\geq 1)=1-P(X

Thus, the probability that of the 3 households randomly selected at least 1 owns a sports car is 0.1956.

4 0
3 years ago
PLEASE HELP
Tems11 [23]

QUESTION A

The given multiplication problem is

\frac{39}{64} \times \frac{8}{13}

Factor each term to obtain;

\frac{13\times 3}{8\times8} \times \frac{8}{13}

Cancel out the common factors to obtain;

\frac{1\times 3}{8\times1} \times \frac{1}{1}

Simplify to get;

\frac{3}{8}

QUESTION B

The given multiplication problem is

\frac{2}{3}\times \frac{1}{5}\times \frac{4}{7}

This the same as

\frac{2\times 1\times 4}{3\times 5\times 7}

This simplifies to;

\frac{8}{105}

QUESTION C

The given problem is

\frac{3}{5}\times \frac{10}{12} \times \frac{1}{2}

This is the same as

\frac{3}{5}\times \frac{5}{6} \times \frac{1}{2}

=\frac{1}{1}\times \frac{1}{2} \times \frac{1}{2}

This simplifies to

=\frac{1}{4}

QUESTION D.

The given expression is

\frac{4}{9}\times 54

Factor the 54 to obtain;

\frac{4}{9}\times 9\times 6

Cancel the common factors to get;

\frac{4}{1}\times 1\times 6

This simplifies to;

=24

QUESTION E

The given problem is

20\times 3\frac{1}{5}

Convert the mixed numbers to improper fraction to obtain;

=20\times \frac{16}{5}

=4\times5 \times \frac{16}{5}

Cancel the common factors to get;

=4\times1 \times \frac{16}{1}

=64

QUESTION F

The multiplication problem is

11 \times 2 \frac{7}{11}

Convert the mixed numbers to improper fractions to obtain;

11 \times \frac{29}{11}

Cancel out the common factors to get;

=1 \times \frac{29}{1}

Simplify;

=29

QUESTION G

The given problem is

5\frac{1}{3}\times 5\frac{1}{8}

Convert to improper fractions;

=\frac{16}{3}\times \frac{41}{8}

Cancel out the common factors to get;

=\frac{2}{3}\times \frac{41}{1}

=\frac{82}{3}

Convert back to mixed numbers

=27\frac{1}{3}

QUESTION H

The given expression is

10\frac{2}{3} \times 1\frac{3}{8}

Convert to improper fraction to get;

\frac{32}{3} \times \frac{11}{8}

Cancel common factors to get;

=\frac{4}{3} \times \frac{11}{1}

Simplify

=\frac{44}{3}

Convert back to mixed numbers;

=14\frac{2}{3}

7 0
3 years ago
The high school took two field trips to the Newport Aquarium many many years ago. They used school buses and vans to transport a
il63 [147K]

Answer:

bus = 43

van 12

Step-by-step explanation:

This can be solved using simultaneous equations

Let v represent the number of students that a van carries

Let b represent the number of students that a bus carries

the following equations can be derived from the question

3v + 2b = 122 eqn 1

5v + 3b = 189  eqn 2

Multiply eqn 1 by 5 and eqn 2 by 3

15v + 10b = 610  eqn 3

15v + 9b = 567 eqn 4

Subtract equation 4 from 3

b = 43

Substitute for b in equation 1

3v + 2(43) = 122

solve for v

v = 12

5 0
3 years ago
The cost to rent a bike is $8 per hour plus a 24$ flat fee if your final bill was $48 how many hours did you rent a scooter for?
Luda [366]
The scooter was rented for 3 hours
8 0
3 years ago
Read 2 more answers
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