Answer: 
Step-by-step explanation:
We know that mean and standard deviation of sampling distribution is given by :-
, where 
= population mean
=Population standard deviation.
n= sample size .
In the given situation, we have
n= 2
Then, the expected mean and the standard deviation of the sampling distribution will be :_
[Rounded to the nearest whole number]
Hence, the the expected mean and the standard deviation of the sampling distribution :
Okay, so Sang is standing 20 yards away from one corner, and Jazmin is standing 99 yards away from the same corner. If this is a rectangle (I like visuals, so I'll use them to explain), then:
99ft
A ------------------------- B
| |
20 ft | |
| |
C -------------------------- D
The question is asking you to solve for the diagonal line between points C and B. If you imagine a line there, you actually have the rectangle split into two triangles. So if you have triangle ABC, side CB would be the longest line, or the hypotenuse. That means you can use the Pythagorean Theorem to solve the problem.
A^2 + B^2 = C^2
99^2 + 20^2 = C^2
9,801 + 400 = C^2
10,201 = C^2
Now you solve for the square root of 10,201 to get C.
sqr (10,201) = C
C = 101 yards
Answer:
B, 5 degrees
Step-by-step explanation:
-20 -5 = -25
Answer:
Noah
Step-by-step explanation:
75%*10=7.5
25%*30=7.5
The answers are both 7.5, which are equal
In the general case in Cartesian coordinates, you would use the definition of a parabola as the locus of points equidistant from the focus and directrix. The equation would equate the square of the distance from a general point (x, y) to the focus with the square of the distance from that point to the directrix line.
Suppose the focus is located at (h, k) and the equation of the directrix is ax+by+c=0. The expression for the square of the distance from (x, y) to the point (h, k) is ...
(d₁)² = (x-h)²+(y-k)²
The expression for the square of the distance from (x, y) to the directrix line is
(d₂)² = (ax+by+c)²/(a²+b²)
Equating these expressions gives the equation of the parabola.
(x-h)²+(y-k)² = (ax+by+c)²/(a²+b²)
When the directrix is parallel with one of the axes, one of the coefficents "a" or "b" is zero and the equation becomes much simpler. Often, it would be easier to make use of the formula (for a directrix parallel to the x-axis):
y = 1/(4p)*(x -h)² +k
where the (h, k) here is the vertex, the point halfway between the focus and directrix, and "p" is the (signed) distance from the focus to the vertex. (p is positive when the focus is above or to the right of the vertex.)
