Answer:
![4^3 =64\\\\(-3)^4 = 81](https://tex.z-dn.net/?f=4%5E3%20%3D64%5C%5C%5C%5C%28-3%29%5E4%20%3D%2081)
Step-by-step explanation:
![4^3 = 4\times 4 \times 4 = 64\\\\(-3)^4 = (-1)^4\cdot 3^4 = 81](https://tex.z-dn.net/?f=4%5E3%20%3D%204%5Ctimes%204%20%5Ctimes%204%20%3D%2064%5C%5C%5C%5C%28-3%29%5E4%20%3D%20%28-1%29%5E4%5Ccdot%203%5E4%20%3D%2081)
Solve these equations the same as solving multistep equations,
for example
#6.
Add six to both sides
now the inequality says
f < 5
now solve the second equation
F-4 Is equal to, or greater than 2
Add four to both sides
now the equation says:
F is equal to, or greater than 6
so your two answers are
f < 5 and F is equal to, or greater than 6
Answer:
8(7+4) (C.)
Step-by-step explanation:
First find the biggest number thta can go into 56 and 32:
8!
Since C. is the only one that has an 8 out in the front, let's try that one:
![8*7=56](https://tex.z-dn.net/?f=8%2A7%3D56)
![8*4=32](https://tex.z-dn.net/?f=8%2A4%3D32)
So C. Is correct!
We are dealing here with a uniform distribution ranging from 0 to 30 minutes. We need to calculate the probability that the unreliable bus will arrive before the reliable one. This probability is the area under the uniform distribution "curve" from 0 to 10 minutes. This constitutes 1/3 of the entire unform distr. curve. So the probability that the unreliable bus will arrive before the reliable one is 1/3, or 0.33. The probability that it will arrive AFTER the reliable bus is 2/3, or 0.67.
Answer:
14 :)
Step-by-step explanation:
all you need to do is add the ones