Question

) The product of the means in a proportion is 45. If one of the extremes is 15, find the other.​

Answer 1

Answer:

Heya.

This is your answer.

Let the other extreme be x.

As we know that product of mean and product of means is equal.

Given - 

Product of means = 1/45

Product of extremes = 1/5 X x

Now,   product of means = product of extremes

$</p><p>\begin{gathered} \frac{1}{45} = \frac{x}{5} \\ =\ \textgreater \ x = \frac{5}{45} \\ =\ \textgreater \ x = \frac{1}{9} \end{gathered}451=5x= \textgreater x=455= \textgreater x=91</p><p>Hence, the other extreme is 1/9.</p><p></p><p>$

Let us verify...

$\begin{gathered} \frac{1}{9} X \frac{1}{5} = \frac{1}{45} \\ =\ \textgreater \ \frac{1}{45} = \frac{1}{45} \end{gathered}91X51=451= \textgreater 451=451$

Hence, verified...

Hope it helps you   

Answer 2

Answer:

Answer:

Heya.

This is your answer.

Let the other extreme be x.

As we know that product of mean and product of means is equal.

Given -

Product of means = 1/45

Product of extremes = 1/5 X x

Now, product of means = product of extremes

\begin{gathered} < /p > < p > \begin{gathered} \frac{1}{45} = \frac{x}{5} \\ =\ \textgreater \ x = \frac{5}{45} \\ =\ \textgreater \ x = \frac{1}{9} \end{gathered}451=5x= \textgreater x=455= \textgreater x=91 < /p > < p > Hence, the other extreme is 1/9. < /p > < p > < /p > < p > \end{gathered}

</p><p>

45

1

=

5

x

= \textgreater x=

45

5

= \textgreater x=

9

1

451=5x= \textgreater x=455= \textgreater x=91</p><p>Hence,theotherextremeis1/9.</p><p></p><p>

Let us verify...

\begin{gathered}\begin{gathered} \frac{1}{9} X \frac{1}{5} = \frac{1}{45} \\ =\ \textgreater \ \frac{1}{45} = \frac{1}{45} \end{gathered}91X51=451= \textgreater 451=451\end{gathered}

9

1

X

5

1

=

45

1

= \textgreater

45

1

=

45

1

91X51=451= \textgreater 451=451

Hence, verified...

Hope it helps you