Answer:
0.00597
Step-by-step explanation:
Given,
Total number of cards = 52,
In which flush cards = 20,
Also, the number of spade flush cards = 5,
Since,
![\text{Probability}=\frac{\text{Favourable outcomes}}{\text{Total outcomes}}](https://tex.z-dn.net/?f=%5Ctext%7BProbability%7D%3D%5Cfrac%7B%5Ctext%7BFavourable%20outcomes%7D%7D%7B%5Ctext%7BTotal%20outcomes%7D%7D)
Thus, the probability of a hand containing a spade flush, if each player has 5 cards
![=\frac{\text{Ways of selecting a spade flush card}}{\text{Total ways of selecting five cards}}](https://tex.z-dn.net/?f=%3D%5Cfrac%7B%5Ctext%7BWays%20of%20selecting%20a%20spade%20flush%20card%7D%7D%7B%5Ctext%7BTotal%20ways%20of%20selecting%20five%20cards%7D%7D)
![=\frac{^{20}C_5}{^{52}C_5}](https://tex.z-dn.net/?f=%3D%5Cfrac%7B%5E%7B20%7DC_5%7D%7B%5E%7B52%7DC_5%7D)
![=\frac{\frac{20!}{5!15!}}{\frac{52!}{5!47!}}](https://tex.z-dn.net/?f=%3D%5Cfrac%7B%5Cfrac%7B20%21%7D%7B5%2115%21%7D%7D%7B%5Cfrac%7B52%21%7D%7B5%2147%21%7D%7D)
![=\frac{15504}{2598960}](https://tex.z-dn.net/?f=%3D%5Cfrac%7B15504%7D%7B2598960%7D)
= 0.00597
The distance covered at time t is listed below:
Distance at 2 sec = d(2) = 64 feet
Distance at 4 sec = d(4) = 256 feet
Distance at 6 sec = d(6) = 576 feet
Distance at 8 sec = d(8) = 1024 feet
We are to find the average rate of change between 2 seconds and 6 seconds. The average rate of change will be:
![\frac{d(6)-d(2)}{6-2} \\ \\ = \frac{576-64}{4} \\ \\=128](https://tex.z-dn.net/?f=%20%5Cfrac%7Bd%286%29-d%282%29%7D%7B6-2%7D%20%5C%5C%20%5C%5C%20%3D%20%5Cfrac%7B576-64%7D%7B4%7D%20%5C%5C%20%5C%5C%3D128)
Therefore, the average rate of distance between 2 seconds and 6 seconds is 128 feet per second. This represents the speed of the object. So the speed of the object between 2 and 6 seconds was 128 feet per second.
Answer:
The value of the car after 3 years is R$58.320.
O valor do carro ao final desse período é R$58.320.
Step-by-step explanation:
The value of the car after t years is given by the following equation:
![V(t) = V(0)(1-r)^{t}](https://tex.z-dn.net/?f=V%28t%29%20%3D%20V%280%29%281-r%29%5E%7Bt%7D)
In which V(0) is the initial value and r is the yearly depreciation rate.
In this question, we have that:
![V(0) = 80000, r = 0.1](https://tex.z-dn.net/?f=V%280%29%20%3D%2080000%2C%20r%20%3D%200.1)
So
![V(t) = V(0)(1-r)^{t}](https://tex.z-dn.net/?f=V%28t%29%20%3D%20V%280%29%281-r%29%5E%7Bt%7D)
![V(t) = 80000(1-0.1)^{t}](https://tex.z-dn.net/?f=V%28t%29%20%3D%2080000%281-0.1%29%5E%7Bt%7D)
![V(t) = 80000(0.9)^{t}](https://tex.z-dn.net/?f=V%28t%29%20%3D%2080000%280.9%29%5E%7Bt%7D)
Value of the car at the end of 3 years:
![V(3) = 80000(0.9)^{3} = 58320](https://tex.z-dn.net/?f=V%283%29%20%3D%2080000%280.9%29%5E%7B3%7D%20%3D%2058320)
The value of the car after 3 years is R$58.320.
O valor do carro ao final desse período é R$58.320.
8cm because 15cm-7cm= 8cm
Answer:
b & d
Step-by-step explanation: