Answer:
The athlete will run 200 kilometers.
A atleta percorrerá 200 quilômetros no total.
Step-by-step explanation:
She will run x kilometers.
She has already run 5/8 of the distance(5/8 of x), and still needs to run 75 kilometers.
So
The athlete will run 200 kilometers.
A atleta percorrerá 200 quilômetros no total.
Answer:
D. a is equal to
Step-by-step explanation:
Let us assume that segment PQ is 8 units, so that its midpoint M is at 4 units to both P and Q.
Given that X is the midpoint of PM, this implies that X is at 2 units to both P and M.
Then;
PX = aPQ
⇒ = a x 8
= 8a
Therefore, a must be equal to , so that;
PX = 8a
= 8 x
= 2 units
Thus, a is equal to is the correct option.
Area of a triangle = (1/2)*base*height
For both of the triangles, you have the base (8.8 for the triangle on the left, 7.6 for the triangle on the right) and the side lengths, but not the height. But since both are isosceles triangles, you can find the height using the pythagorean theorem.
5.
First divide the triangle vertically into two triangles (see attached picture). Now you have two right triangles, you can apply the pythagorean theorem on either one of them to find the height. The pythagorean theorem says that for a right triangle, , where c is the hypotenuse and a and b are the sides of the triangle.
Substituting the given values and rounding to nearest tenth:
Now that you have the height, you can find the area of the entire triangle.
A = (1/2)*base*height
A = (1/2)*8.8*9.0 = 39.6
6.
Same procedure.
A = (1/2)*base*height
A = (1/2)*7.6*9.2 = 35.0
Answer:
4 days
Step-by-step explanation:
We need to use the probability functions of each of the intervals to know the Probability number and then use it in the expected value.
P(x>3 cargo ships) = 1-P(x<=3)
P(x>3) = 1-P(x<=3)
P(x>3) = 1-[P(x=0)+P(x=1)+P(x=2)+P(x=3)]
P(x>3) = 1- [0.1353(1+2+2+1.33)]
P(x>3) = 1-0.856
P(x>3) = 0.1431
Als n=30, expected number is
E(x) =30*P(x>3)
E(x) = 30*0.1431
E(x) = 4.293
I expect 4.293 or 4 days per month the block being unable to hanlde all arriving ships