Basically we need to add and subtract 1/2 from 1/6:
(1) 1/6 - 1/2 = 1/6 - 3/6
= -2/6
= -1/3
(2) 1/6 + 1/2 = 1/6 + 3/6
= 4/6
= 2/3
Therefor the two numbers that are located 1/2 unit from 1/6 are -1/3 and 2/3
Yes this is the correct answer. The slope field shown is the differential equation dy/dx = x+y
All points along the line x+y = k, for some constant k, will have the same tangent slope k. For example, the line x+y = 0 will have every point with tangent slope 0 for the solution y(x). It may not be 100% clear, but this graph has horizontal tickmarks on the line x+y = 0, or the tickmarks are close to being horizontal. Using geogebra, I checked the others and they produced completely different graphs, which allowed me to rule them out.
Answer:
P ( 5 < X < 10 ) = 1
Step-by-step explanation:
Given:-
- Sample size n = 49
- The sample mean u = 8.0 mins
- The sample standard deviation s = 1.3 mins
Find:-
Find the probability that the average time waiting in line for these customers is between 5 and 10 minutes.
Solution:-
- We will assume that the random variable follows a normal distribution with, then its given that the sample also exhibits normality. The population distribution can be expressed as:
X ~ N ( u , s /√n )
Where
s /√n = 1.3 / √49 = 0.2143
- The required probability is P ( 5 < X < 10 ) minutes. The standardized values are:
P ( 5 < X < 10 ) = P ( (5 - 8) / 0.2143 < Z < (10-8) / 0.2143 )
= P ( -14.93 < Z < 8.4 )
- Using standard Z-table we have:
P ( 5 < X < 10 ) = P ( -14.93 < Z < 8.4 ) = 1
Answer:
3. 160
4. 105
5. 95
6. 30
7. 86
8. 130
All triangles equal 180 so when you add the other 2 degrees in the triangle then subtract that by 180 then you get the third angle. The a straight line also equals 180 degrees so when you get the third angle subtract it by 180 then you get the answer to the "?"
That's how I was taught so, hope I got it all right! : )
If you are looking to write an inequality equation that models this situation, then the answer would be: 
. The "at most" signals to use the less than or equal to sign.
