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2 years ago

2a + 3b = 5 b = a - 5

Mathematics

1 answer:

Lunna [17]2 years ago

5 0

<h2>1st Equation:</h2>

Answer for A:

$a = \frac{5 - 3b}{2}$

Answer for B:

$b = \frac{5 - 2b}{3}$

<h2>2nd Equation:</h2>

Answer:

$a = b + 5$

<u>hope it helps</u><u>.</u>

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Jane has 18 juice boxes, 24 bags of peanuts and 36 apples. She wants to make snack bags with same number of each thing in each b

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Answer: 6 snack bags.

Step-by-step explanation:

For this exercise it is important to remember the Greatest Common Divisor (GCD). This is defined as the largest positive intener by which you can divided evenly "n" numbers.

In this case, you know that Jane has 18 juice boxes, 24 bags of peanuts and 36 apples. Therefore, the steps are the following:

1. You need to descompose 18, 24 and 36 into their prime factors, as below:

$18=2*3*3=2*3^2\\\\24=2*2*2*3=2^3*3\\\\36=2*2*3*3=2^2*3^2$

2. Now you must choose the common ones with the smallest exponent and multiply them.

Then, you get:

$GCD=3*2\\\\GCD=6$

Therefore, the greatest number of snack bags that Jane can make is: 6 snack bags.

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3 years ago

There are three monomials such that the greatest common factor of the first and second monomials is 2xy, and the greatest common

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You just obtained a credit card. You immediately purchase a stereo system for $200. your credit limit is $1000. Let's assume tha

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Step-by-step explanation:

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2 years ago

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Answer:

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Step-by-step explanation:

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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.)

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