The area of a square is expressed as the length of the side to the power of two, A = s^2. We were given the area of the enlarged photo which is 256 in^2. Also, it was stated that the length of the enlarged photo is the length of the original photo plus ten inches. So, from these statements we can make an equation to solve for x which represents the length of the original photo.
A = s^2
where s = (x+10)
A = (x+10)^2 = 256
Solving for x,
x= 6 in.
The dimensions of the original photo is 6 x 6.
Standard deviations of the four activities of the critical path are 1,2,4,2.
Standard deviation of this critical path = Sum of square root of variance of this corresponding critical path
Standard deviation of critical path 



Now we need to find the probability that the project will completed in 38 weeks given that its expected completion time is 40 weeks.
That is, we need to find P(X<38) :


Probability 
Thus the probability that the project will be completed in 38 weeks is 0.34.
Answer:
For x∈ {-∞,-3} y<0, below x-axis
x∈ {-3,-1} y>0, above x-axis
x∈ {-1,4} y<0, below x-axis
x∈ {4,∞} y>0, above x-axis
Step-by-step explanation:
f(x)=
=0



For x∈ {-∞,-3} y<0, below x-axis
x∈ {-3,-1} y>0, above x-axis
x∈ {-1,4} y<0, below x-axis
x∈ {4,∞} y>0, above x-axis
Answer:
The amount of heat required to raise the temperature of liquid water is 9605 kilo joule .
Step-by-step explanation:
Given as :
The mass of liquid water = 50 g
The initial temperature =
= 15°c
The final temperature =
= 100°c
The latent heat of vaporization of water = 2260.0 J/g
Let The amount of heat required to raise temperature = Q Joule
Now, From method
Heat = mass × latent heat × change in temperature
Or, Q = m × s × ΔT
or, Q = m × s × (
-
)
So, Q = 50 g × 2260.0 J/g × ( 100°c - 15°c )
Or, Q = 50 g × 2260.0 J/g × 85°c
∴ Q = 9,605,000 joule
Or, Q = 9,605 × 10³ joule
Or, Q = 9605 kilo joule
Hence The amount of heat required to raise the temperature of liquid water is 9605 kilo joule . Answer