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FrozenT [24]
3 years ago
11

A health inspector wants to check compliance with a new city ordinance on meat storage. Since he can only inspect 10 of the 33 s

tores, he chooses to do a stratified random sample that consists of all 3 of the large chain stores, 4 of the 10 smaller chain stores, and 3 of the 20 locally-owned stores. From his inspection he finds that 2 of the 3 large chain stores are compliant, 1 of the 4 smaller chain stores is compliant, and 1 of the 3 locally-owned chain stores is compliant. Using the sample results, estimate the proportion of the entire population of stores that is compliant with the ordinance.
A) 0.121
B) 0.4
C) 0.6
D) 0.879
Mathematics
1 answer:
cestrela7 [59]3 years ago
3 0
The answer is B)0.4 because four of the ten stores that were surveyed were compliant so the proportion of the entire population of stores that would be compliant is <span>410</span> or 0.4<span> </span>
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2 years ago
A rectangular exercise mat has a perimeter of 36 feet . The length mat is twice the width. Write and solve an equation to determ
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72 square feet

EXPLANATION

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P= 2l+2w

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4 0
3 years ago
A consulting firm has received 2 Super Bowl playoff tickets from one of its clients. To be fair, the firm is randomly selecting
Soloha48 [4]

Answer:

Therefore the only statement that is not true is b.)

Step-by-step explanation:

There employees are 6 secretaries, 5 consultants and 4 partners in the firm.

a.) The probability that a secretary wins in the first draw

= \frac{number \hspace{0.1cm} of \hspace{0.1cm} secreataries}{total \hspace{0.1cm} number \hspace{0.1cm} of \hspace{0.1cm} employees}  = \frac{6}{15}

b.) The probability that a secretary wins a ticket on second draw.  It has been given that a ticket was won on the first draw by a consultant.

p(secretary wins on second draw | consultant  wins on first draw)

=\frac{p((consultant \hspace{0.1cm} wins \hspace{0.1cm} on \hspace{0.1cm} first \hspace{0.1cm}draw)\cap( secretary\hspace{0.1cm} wins\hspace{0.1cm} on second \hspace{0.1cm}draw))}{p(consultant \hspace{0.1cm} wins \hspace{0.1cm} on \hspace{0.1cm} first \hspace{0.1cm}draw)}

= \frac{\frac{5}{15}  \times \frac{6}{14}}{\frac{5}{15} }  = \frac{6}{14}  .

The probability that  a ticket was won on the first draw by a consultant a secretary wins a ticket on second draw  = \frac{6}{15} is not true.

The probability that a secretary wins on the second draw  = \frac{number \hspace{0.1cm} of  \hspace{0.1cm} secretaries  \hspace{0.1cm} remaining } { number  \hspace{0.1cm} of  \hspace{0.1cm} employees  \hspace{0.1cm} remaining}  = \frac{6 - 1}{15 - 1}  = \frac{5}{14}

c.) The probability that a consultant wins on the first draw  =

\frac{number \hspace{0.1cm} of  \hspace{0.1cm} consultants  \hspace{0.1cm}  } { number  \hspace{0.1cm} of  \hspace{0.1cm} employees  \hspace{0.1cm} }  = \frac{5 }{15}  = \frac{1}{3}

d.) The probability of two secretaries winning both tickets

= (probability of a secretary winning in the first draw) × (The probability that a secretary wins on the second draw)

= \frac{6}{15}  \times \frac{5}{14}  = \frac{1}{7}

Therefore the only statement that is not true is b.)

5 0
3 years ago
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