Chris answered 61 out of 66 questions correctly on a test. What is his test average to the nearest thousandth

1 answer:

4 0

If you divide 61 by 66, you will get .9242424. The thousandths place is the third number, so you base it off of the four number. If the fourth number is under 5, the third number stays the same and all numbers after it disappear. If the number is above 5, the third number rounds up 1 and the rest disappear. Because the 3rd number is 4 and the 4th number is 2, you will get the answer of .924

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A city has a population of 210,000 people. Suppose that each year the population grows by 3.25% . What will the population be af

serg [7]

Answer: the population be after 10 years is 289145

Step-by-step explanation:

The growth rate is exponential. We would apply the formula for exponential growth which is expressed as

y = b(1 + r)^t

Where

y represents the population after t years.

t represents the number of years.

b represents the initial population.

r represents rate of growth.

From the information given,

b = 210000

r = 3.25% = 3.25/100 = 0.0325

t = 10 years

Therefore,

y = 210000(1 + 0.0325)^10

y = 210000(1.0325)^10

y = 289145

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2 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 =

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