Answer:

Step-by-step explanation:
6 symbols and 4-symbol code so number of combinations wich doesn't contain the same symbol twice would be:
6•5•4•3 = 360
If code starts with a star followed by triangle that means only one posibility for first two symbols and code doesn't containing the same symbol twice means that left 4 symbols for next two.
So number of combinations of code starting with a star followed by triangle and doesn't containing the same symbol twice will be:
1•1•4•3 = 12
The probability that a randomly chosen code starts with a star followed by triangle and doesn't contain the same symbol twice:

Answer: -5100
<u>Step-by-step explanation:</u>
![\sum^4_1[100(-4)^{n-1}]\qquad \rightarrow \qquad a_1=100\ \text{and r = -4}\\\\\\S_n=\dfrac{a_1(1-r^n)}{1-r}\\\\\\\\S_4=\dfrac{100(1-(-4)^4)}{1-(-4)}\\\\\\.\quad=\dfrac{100(1-256)}{1+4}\\\\\\.\quad=\dfrac{100(-255)}{5}\\\\.\quad=20(-255)\\\\.\quad=-5100\\](https://tex.z-dn.net/?f=%5Csum%5E4_1%5B100%28-4%29%5E%7Bn-1%7D%5D%5Cqquad%20%5Crightarrow%20%5Cqquad%20a_1%3D100%5C%20%5Ctext%7Band%20r%20%3D%20-4%7D%5C%5C%5C%5C%5C%5CS_n%3D%5Cdfrac%7Ba_1%281-r%5En%29%7D%7B1-r%7D%5C%5C%5C%5C%5C%5C%5C%5CS_4%3D%5Cdfrac%7B100%281-%28-4%29%5E4%29%7D%7B1-%28-4%29%7D%5C%5C%5C%5C%5C%5C.%5Cquad%3D%5Cdfrac%7B100%281-256%29%7D%7B1%2B4%7D%5C%5C%5C%5C%5C%5C.%5Cquad%3D%5Cdfrac%7B100%28-255%29%7D%7B5%7D%5C%5C%5C%5C.%5Cquad%3D20%28-255%29%5C%5C%5C%5C.%5Cquad%3D-5100%5C%5C)
Answer:
3. do rise over run to get 3/1 and that simplifies to 3!
Step-by-step explanation:
Answer:
8
Step-by-step explanation:
because if you do it then the anwser would be 8.5 so it could be 8 or 9
Remember to do PEMDAS so your answer is not correct, since you multiplied before doing the parentheses/exponents.
Explanation:
3(3)^2- 5(4)
= 3(9) - 5(4)
= 27 - 20 = 7