Oldest = 2 times Youngest -> O = 2*Y
Middle = Youngest + 5 -> M = Y+5
All of them together is 57 -> O + M + Y = 57

So you have these three equations:
(1) O = 2*Y
(2) M = Y+5
(3) O + M + Y = 57

Now you want to reduce the number of variables. You can change the second equation to be Y = M-5 and then plug in "M-5" wherever there is currently a Y:

(4) O = 2*(M-5) = 2*M - 10
(5) O + M + (M-5) = 57
which becomes O + 2M = 62

Then you plug in the "O" equation (4) into (5) which gives you
(2M-10) + 2M = 62 which reduces to 4M = 72.

So now I know M is 18.
I can now plug that into my other equations:
(4) O = 2*18 - 10 which means O = 26.
Now I plug that into (1) from the top:
26 = 2*Y which becomes 13 = Y

So now I have O, Y, and M
Oldest is 26
Middle is 18
Youngest is 13

Reading the sentence again, you can see that this makes sense.

4 0

3 years ago

Read 2 more answers

May someone help me with this question please anyone please

djverab [1.8K]

$5;\ 5;\ 5;\ 5;\ 7;\ 9\\\\\overline{x}=\dfrac{5+5+5+5+7+9}{6}=\dfrac{36}{6}=6\\\\\delta^2=\dfrac{(x_1-\overline{x})^2+(x_2-\overline{x})^2+(x_3-\overline{x})^2+...+(x_n-\overline{x})^2}{n}\\\\\delta^2=\dfrac{(5-6)^2+(5-6)^2+(5-6)^2+(5-6)^2+}{6}$
$\dfrac{+(7-6)^2+(9-6)^2}{6}\\\\\delta^2=\dfrac{(-1)^2+(-1)^2+(-1)^2+(-1)^2+1^2+3^2}{6}=\dfrac{14}{6}=\dfrac{7}{3}$

$standard\ deviation:\\\\\sqrt{\delta^2}=\sqrt{\dfrac{7}{3}}=\dfrac{\sqrt7}{\sqrt3}=\dfrac{\sqrt7\cdot\sqrt3}{\sqrt3\cdot\sqrt3}=\boxed{\dfrac{\sqrt{21}}{3}}\approx\boxed{1.53}$

4 0

4 years ago