Answer:
dx/dt = 0,04 m/sec
Step-by-step explanation:
Area of the circle is:
A(c) =π*x² where x is a radius of the circle
Applying differentiation in relation to time we get:
dA(c)/dt = π*2*x* dx/dt
In this equation we know:
dA(c)/dt = 0,5 m²/sec
And are looking for dx/dt then
0,5 = 2*π*x*dx/dt when the area of the sheet is 12 m² (1)
When A(c) = 12 m² x = ??
A(c) = 12 = π*x² ⇒ 12 = 3.14* x² ⇒ 12/3.14 = x²
x² = 3,82 ⇒ x = √3,82 ⇒ x = 1,954 m
Finally plugging ths value in equation (1)
0,5 = 6,28*1,954*dx/dt
dx/dt = 0,5 /12.28
dx/dt = 0,04 m/sec
Answer:
Boat traveled 553.24 feet towards the lighthouse.
Step-by-step explanation:
In the figure attached AB is the light house of height 200 feet.
Angle of depression of the boat from the top of a lighthouse = angle of elevation of the lighthouse from the boat = 14°52'
so 1' =
degree
so angle of elevation at point C = 14 + 
So angle of elevation from C = (14 + 0.87) = 14.87°
Similarly, when boat arrives at point D angle of elevation = 45°10' = 45 +
= 45.17°
Now we have to calculate the distance CD, traveled by the boat.
In ΔABC
tan14.87 = 
0.2655 = 
BC = 
BC = 753.239 feet
Similarly in ΔABD
tan45.17 = 
1 = 
BD = 200 feet
So distance CD = BC - BD
CD = 753.239 - 200
= 553.24 feet
Therefore, Boat traveled 553.24 feet towards the lighthouse.
Let s and a represent the position of Sal and Amir respectively.
s=.75+4.5t and a=-.25+6.7t
When Sal catches Amir, s=a so we can say.
-.25+6.7t=.75+4.5t subtract 4.5t from both sides
-.25+2.2t=.75 add .25 to both sides
2.2t=1 divide both sides by 2.2
t=5/11 hours
t≈0.45 hours (to nearest hundredth of an hour)
They are called invertebrats
Answer:


Step-by-step explanation:
Previous concepts
Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".
The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".
The central limit theorem states that "if we have a population with mean μ and standard deviation σ and take sufficiently large random samples from the population with replacement, then the distribution of the sample means will be approximately normally distributed. This will hold true regardless of whether the source population is normal or skewed, provided the sample size is sufficiently large".
Solution to the problem
For this case we select a sample of n =100
From the central limit theorem we know that the distribution for the sample mean
is given by:
So then the sample mean would be:

And the standard deviation would be:
