4 - t = 3(t - 1) - 5
= 4 - t = (3t - 3) - 5
= 4 - t = 3t - 8
Add 8 to both sides:
12 - t = 3t
Add t on both sides:
12 = 4t
Divide both sides by 4:
t = 3
Hope I helped :)
EDIT: Yeah I see where I went wrong - answer is 3
Answer:
PR5T represents <em>6754.</em>
P = 6
R = 7
T = 4
Y = 5
Step-by-step explanation:
Let us try to understand the subtraction starting from the unit's digit.
P R 5 T
- 4 7 Y 6
-----------------
1 9 9 8
To get 8, 6 should be subtracted from 14
14 - 6 = 8
So, T should be equal unit digit of 14 <em>i.e.</em> T = 4 and 1 should be carried over from the ten's digit of PR5T.
Now, let us move to ten's digit:
P R 5 4
- 4 7 Y 6
-----------------
1 9 9 8
To get 9, 6 should be subtracted from 15. But 1 is already carried to unit's place So, we have actually 14 here
14 - Y = 9
So, Y = 5
Moving to hundred's place:
R - carry - 7 = 9
Carry is 1, so R = unit's place of 9 + 8 i.e. R = 7
Moving to thousand's place:
P - carry - 4 = 1
Here carry is 1
So, P = 1 + 4 + 1 = 6
So, the number PR5T is <em>6754.</em>
Answer:
20% I think this is the answer
Step-by-step explanation:
![f_X(x)=\begin{cases}ke^{kx}&\text{for }x\in[0,2]\\0&\text{otherwise}\end{cases}](https://tex.z-dn.net/?f=f_X%28x%29%3D%5Cbegin%7Bcases%7Dke%5E%7Bkx%7D%26%5Ctext%7Bfor%20%7Dx%5Cin%5B0%2C2%5D%5C%5C0%26%5Ctext%7Botherwise%7D%5Cend%7Bcases%7D)
to be a valid PDF, the integral over its support must equal 1:
, so that 
and the integral becomes
