Let a,b,c,d and e are five numbers
mean = (a +b+ c+ d+e)/5 = 30
⇒ (a +b+ c+ d+e) = 150_______(1)
now, Let the number excluded be a
then, new mean = (b+ c+ d+e)/4 = 28
⇒ (b+ c+ d+e)= 112
putting this value in (1),
⇒a + 112 = 150
⇒a = 150 -112 = 38
excluded number = 38
Answer:
(2.83 , 1 , 4)
Step-by-step explanation:
Rewrite these equations in matrix form
![\left[\begin{array}{ccc}2&2&-1\\4&-2&-2\\3&3&-4\end{array}\right] \left[\begin{array}{ccc}x\\y\\z\end{array}\right]=\left[\begin{array}{ccc}4\\2\\-4\end{array}\right] \\](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D2%262%26-1%5C%5C4%26-2%26-2%5C%5C3%263%26-4%5Cend%7Barray%7D%5Cright%5D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Dx%5C%5Cy%5C%5Cz%5Cend%7Barray%7D%5Cright%5D%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D4%5C%5C2%5C%5C-4%5Cend%7Barray%7D%5Cright%5D%20%5C%5C)
we can write it like this,
so to solve it we need to take the inverse of the 3 x 3 matrix A then multiply it by B.
We get the inverse of matrix A,
![A^{-1}=\left[\begin{array}{ccc}7/15&1/6&-1/5\\1/3&-1/6&0\\3/5&0&-2/5\end{array}\right] \\](https://tex.z-dn.net/?f=A%5E%7B-1%7D%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D7%2F15%261%2F6%26-1%2F5%5C%5C1%2F3%26-1%2F6%260%5C%5C3%2F5%260%26-2%2F5%5Cend%7Barray%7D%5Cright%5D%20%20%5C%5C)
now multiply the matrix with B
![X=A^{-1}B\\\\\left[\begin{array}{ccc}x\\y\\z\end{array}\right] =\left[\begin{array}{ccc}7/15&1/6&-1/5\\1/3&-1/6&0\\3/5&0&-2/5\end{array}\right]\left[\begin{array}{ccc}4\\2\\-4\end{array}\right] \\\\\\\left[\begin{array}{ccc}x\\y\\z\end{array}\right] =\left[\begin{array}{ccc}2.83\\1\\4\end{array}\right] \\](https://tex.z-dn.net/?f=X%3DA%5E%7B-1%7DB%5C%5C%5C%5C%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Dx%5C%5Cy%5C%5Cz%5Cend%7Barray%7D%5Cright%5D%20%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D7%2F15%261%2F6%26-1%2F5%5C%5C1%2F3%26-1%2F6%260%5C%5C3%2F5%260%26-2%2F5%5Cend%7Barray%7D%5Cright%5D%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D4%5C%5C2%5C%5C-4%5Cend%7Barray%7D%5Cright%5D%20%5C%5C%5C%5C%5C%5C%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Dx%5C%5Cy%5C%5Cz%5Cend%7Barray%7D%5Cright%5D%20%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D2.83%5C%5C1%5C%5C4%5Cend%7Barray%7D%5Cright%5D%20%5C%5C)
Answer:
3
Step-by-step explanation:
lim(t→∞) [t ln(1 + 3/t) ]
If we evaluate the limit, we get:
∞ ln(1 + 3/∞)
∞ ln(1 + 0)
∞ 0
This is undetermined. To apply L'Hopital's rule, we need to rewrite this so the limit evaluates to ∞/∞ or 0/0.
lim(t→∞) [t ln(1 + 3/t) ]
lim(t→∞) [ln(1 + 3/t) / (1/t)]
This evaluates to 0/0. We can simplify a little with u substitution:
lim(u→0) [ln(1 + 3u) / u]
Applying L'Hopital's rule:
lim(u→0) [1/(1 + 3u) × 3 / 1]
lim(u→0) [3 / (1 + 3u)]
3 / (1 + 0)
3
