A home has a living room that is 14 ft. wide, 22 ft. long, and a height of 9ft. Bob needs to paint

Mathematics

1 answer:

aleksandrvk [35]3 years ago

3 0

Answer:

Step-by-step explanation:

Let's find the volume of the room in ft³ and that will give

14×22×9=2772ft³

There are deductions. Hence we subtract them from this value

3×5=15, but there are two, hence 15×2=30ft²

And 4×7=28ft²

Hence 2772-(30+28)=2714

Dividing this value by 320 gives

8.4(to 2s.f)

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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 =

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