A. 
To find greater than or smaller than relation, we multiply the terms like (numerator of L.H.S with denominator of R.H.S and put the value on the left side. Then multiply the denominator of L.H.S with numerator of R.H.S and put the value on right side. Now compare the digits.)
So, solving A, we get 810<209 ... This is false
B. 
= 238>589 ..... This is false
C. 
= 496>780 .... This is false
D. 
= 420<660 ..... This is true
Hence, option D is true.
Answer:
Let x = the third side
In a triangle, the sum of any 2 sides must be larger than the third side.
I believe this is called the triangle inequality theorem.
We can construct 3 inequalities to obtain the range of values for the third side.
(Inequality #1) 12 + 4 > x
16 > x
(Inequality#2) 12 + x > 4
x > -8 (we can discard this ... we know all sides will be positive)
(Inequality #3) 4 + x > 12
x > 8
So when we combine these together,
8 < x < 16
X (the third side) must be a number between 8 and 16. but not including 8 and 16
Answer:
a.) f(x) = 
where 90 < x < 120
b.) 
c.)
d.) 
Step-by-step explanation:
Let
X be a uniform random variable that denotes the actual charging time of battery.
Given that, the actual recharging time required is uniformly distributed between 90 and 120 minutes.
⇒X ≈ ∪ ( 90, 120 )
a.)
Probability density function , f (x) = 
where 90 < x < 120
b.)
P(x < 110) = 
= ![\frac{1}{30}[x]\limits^{110}_{90} = \frac{1}{30} [ 110 - 90 ] = \frac{1}{30} [ 20] = \frac{2}{3}](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B30%7D%5Bx%5D%5Climits%5E%7B110%7D_%7B90%7D%20%20%3D%20%5Cfrac%7B1%7D%7B30%7D%20%5B%20110%20-%2090%20%5D%20%3D%20%5Cfrac%7B1%7D%7B30%7D%20%5B%2020%5D%20%3D%20%5Cfrac%7B2%7D%7B3%7D)
c.)
P(x > 100 ) = 
= ![\frac{1}{30}[x]\limits^{120}_{100} = \frac{1}{30} [ 120 - 100 ] = \frac{1}{30} [ 20] = \frac{2}{3}](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B30%7D%5Bx%5D%5Climits%5E%7B120%7D_%7B100%7D%20%20%3D%20%5Cfrac%7B1%7D%7B30%7D%20%5B%20120%20-%20100%20%5D%20%3D%20%5Cfrac%7B1%7D%7B30%7D%20%5B%2020%5D%20%3D%20%5Cfrac%7B2%7D%7B3%7D)
d.)
P(95 < x< 110) = 
= ![\frac{1}{30}[x]\limits^{110}_{95} = \frac{1}{30} [ 110 - 95 ] = \frac{1}{30} [ 15] = \frac{1}{2}](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B30%7D%5Bx%5D%5Climits%5E%7B110%7D_%7B95%7D%20%20%3D%20%5Cfrac%7B1%7D%7B30%7D%20%5B%20110%20-%2095%20%5D%20%3D%20%5Cfrac%7B1%7D%7B30%7D%20%5B%2015%5D%20%3D%20%5Cfrac%7B1%7D%7B2%7D)
Answer:
lower than Amanda: 816 students
Step-by-step explanation:
An equivalent way in which to state this problem is: Find the area under the standard normal curve to the left (below) 940.
Most modern calculators have built in distribution functions.
In this case I entered the single command normalcdf(-1000,940, 850, 100)
and obtained 0.816.
In this particular situation, this means that 0.816(1000 students) scored lower than Amanda: 816 students.
