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Nata [24]
2 years ago
8

If w = 9, x = 10, and y = 3, what does 6x over 5y equal?

Mathematics
2 answers:
zlopas [31]2 years ago
7 0

Answer:

<em>4</em>

Step-by-step explanation:

To solve this, we first need to apply what we know (the value of the variables that have been given to us, for example, x = 10) to the equation. *Take note that there is no <em>w </em>variable in the problem given to us, so that info is only there to confuse us*

It'll look something like this:

\frac{6x}{5y} = ?

\frac{6 (10)}{5 (3)} = ?

Now that we've applied the values we know, we can just solve as is (which is going to be multiplying the top and bottom values together). In other words, it'll be like this now:

\frac{60}{15}

Since this is an improper fraction (meaning the numerator is larger than the denominator) we need to simply, which can be done by dividing (since fractions basically mean division). So 60 ÷ 15 is the last step.

\frac{60}{15} = 4

Hope that wasn't too complicated. Have a great day <3

nirvana33 [79]2 years ago
3 0

Answer:

6x = 6 (10) => 60

5y = 5 (3) => 15

\frac{6x}{5y}  =  >  \frac{60}{15}  =  > 4

hope that helps uhh!!

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Find the probability that the person is frequently or occasionally involved in charity work.
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Given the table below which shows the result of a survey that asked 2,881 people whether they are involved in any type of charity work.

\begin{tabular}&#10;{|c|c|c|c|c|c|}&#10; &Frequently&Occassionally&Not at all&Total\\[1ex]&#10;Male&227&454&798&1,479\\&#10;Female &205&450&747&1,402\\&#10;Total&432&904&1,545&2,881&#10;\end{tabular}

Part A:

If a person is chosen at random, the probability that the person is frequently or occassinally involved in charity work is given by

P(being \ frequently \ involved \ or \ being \ occassionally \ involved)\\ \\= \frac{432}{2881} + \frac{904}{2881} = \frac{1336}{2881}=\bold{0.464}



Part B:

If a person is chosen at random, the probability that the person is female or not involved in charity work at all is given by

P(being&#10; \ female \ or \ not \ being \ involved)\\ \\= &#10;\frac{1402}{2881} + \frac{1545}{2881}-\frac{747}{2881} = &#10;\frac{2200}{2881}=\bold{0.764}



Part C:

If a person is chosen at random, the probability that the person is male or frequently involved in charity work is given by

P(being&#10; \ male \ or \ being \ frequently \ involved)\\ \\= &#10;\frac{1479}{2881} + \frac{432}{2881}-\frac{227}{2881} = &#10;\frac{1684}{2881}=\bold{0.585}



Part D:

If a person is chosen at random, the probability that the person is female or not frequently involved in charity work is given by

P(being&#10; \ female \ or \ not \ being \ frequently \ involved)\\ \\= &#10;\frac{1402}{2881} + \frac{904}{2881} + \frac{1545}{2881}-\frac{450}{2881}-\frac{747}{2881} = &#10;\frac{2654}{2881}=\bold{0.921}



Part E:

The events "being female" and "being frequently involved in charity work" are not mutually exclusive because being a female does not prevent a person from being frequently involved in charity work.

Indeed from the table, there are 205 females who are frequently involved in charity work.

Therefore, the answer to the question is "No, because 205 females are frequently involved charity work".
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