To find : outcomes for choosing 2 bottles (without replacement) from 3 various soft drink bottles in different numbers - coke , fanta , sprite
There can be total 2 x 3 , ie = 6 outcomes in this case.
Like two alphabets from A,B,C can be chosen without replacement in following 6 cases - A B & B A , A C & C A , B C & C B
Similarly for Coke , Fanta , Sprite case ; total <u>outcomes</u><u> </u>are :-
- Coke, Fanta
- Fanta, Coke
- Coke, Sprite
- Sprite , Coke
- Fanta , Sprite
- Sprite , Fanta
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Answer:
You already answered it.
Step-by-step explanation:
Answer:
The variation needed for the daily buget to follow the increase in production for the first year is 12.38 $/year.
This value of Δy is not constant for a constant increase in production.
Step-by-step explanation:
We know that the production function is
, and in the current situation
and
.
With this information we can calculate the actual budget level:
![p_0 = 10x^{0.2} y^{0.8}\\\\1200=10*130^{0.2} y^{0.8}\\\\1200=26.47*y^{0.8}\\\\y=(1200/26.47)^{1/0.8}=45.33^{1.25}=117.62](https://tex.z-dn.net/?f=p_0%20%3D%2010x%5E%7B0.2%7D%20y%5E%7B0.8%7D%5C%5C%5C%5C1200%3D10%2A130%5E%7B0.2%7D%20y%5E%7B0.8%7D%5C%5C%5C%5C1200%3D26.47%2Ay%5E%7B0.8%7D%5C%5C%5C%5Cy%3D%281200%2F26.47%29%5E%7B1%2F0.8%7D%3D45.33%5E%7B1.25%7D%3D117.62)
The next year, with an increase in demand of 100 more automobiles, the production will be
.
If we calculate y for this new situation, we have:
![y_1=(\frac{p_1}{10x^{0.2}} )^{1.25}=(\frac{1300}{26.47} )^{1.25}=49.10^{1.25}=130](https://tex.z-dn.net/?f=y_1%3D%28%5Cfrac%7Bp_1%7D%7B10x%5E%7B0.2%7D%7D%20%29%5E%7B1.25%7D%3D%28%5Cfrac%7B1300%7D%7B26.47%7D%20%29%5E%7B1.25%7D%3D49.10%5E%7B1.25%7D%3D130)
The budget for the following year is 130.
The variation needed for the daily buget to follow the increase in production for the first year is 12.38 $/year.
![\Delta y=y_1-y_0=130.00-117.62=12.38](https://tex.z-dn.net/?f=%5CDelta%20y%3Dy_1-y_0%3D130.00-117.62%3D12.38)
This value of Δy is not constant for a constant increase in production.
Answer:
P(n,4)÷P(n,2)
= (n-2)(n-3)
20 = 4 × 5 (two consecutive numbers)
(n-2)(n-3)=5×4
n-2=5
n=7