Answer:
5:7.
...................................
The area of the shaded region is 15.453 square units.
Step-by-step explanation:
Step 1:
The given shape consists of a rectangle and two circles.
The length of the rectangle is given as 12 cm. Both the diameters of the circles equal 12 cm. So each circle has a diameter of 6 cm and thus a radius of 3 cm.
The diameter of the circle is equal to the width of the rectangle.
So the rectangle has a length of 12 cm and a width of 6 cm. The circles have radii of 3 cm each.
Step 2:
The area of the rectangle
square cm.
The area of each circle
square cm.
The area of both circles
square cm.
The area of the shaded region is the difference between the area of the rectangle and the area of both circles.
The area of the shaded region
square units.
The area of the shaded region is 15.453 square units.
Answer:
x=2 is incorrect
Step-by-step explanation:
Y(2)=(3/4)*x^2=(3/4)*4=3
Assume that a step to the north is in the positive direction and a step in the south is in the negative direction.
Then,

and

After 1 hour of independent steps,

and

Therefore, by Central Limit Theorem, the distribution of the random walk is normal with a mean of 0 cm and a standard deviation of 17.32 cm. He is expected to be somewhere around his starting point after 1 hour.