How do i solve this

Mathematics

2 answers:

galina1969 [7]2 years ago

6 0

Here, hope you can understand it lol

REY [17]2 years ago

3 0

Multiply -2 with 2 so the 2 is out of the equation. Then you should get 4 - m = 10. Subtract -4 from 4 and cross our 4 bevause it cancels 4 out and subtract -4 from 10. When you subtract 10 and 4 you get 6. M= 6 is the answer.

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7/11 converted to a decimal rounded up 2 places

astraxan [27]

Answer:

0.636

Step-by-step explanation:

First off, we'll interpret the fraction bar to mean "divided by." This means that 7/11 is the same as 7 divided by 11.

Second, you have requested to see 2 decimal places in your answer, so we'll write the 7 as 7.00 instead (a 7 with 2 zeros after the decimal point).

Now, we'll just do what the fraction bar says: divide 7.00 by 11:

And that's about it! 7/11 written as a decimal to 3 decimal places is 0.636.

3 0

3 years ago

Read 2 more answers

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 =

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