Answer:
The minimum sample size required to create the specified confidence interval is 2229.
Step-by-step explanation:
We have that to find our 
level, that is the subtraction of 1 by the confidence interval divided by 2. So:
Now, we have to find z in the Ztable as such z has a pvalue of 
.
So it is z with a pvalue of 
, so 
Now, find the margin of error M as such
In which 
is the standard deviation of the population and n is the size of the sample.
What is the minimum sample size required to create the specified confidence interval
This is n when 
Rounding up
The minimum sample size required to create the specified confidence interval is 2229.
Answer:
Step-by-step explanation:
You know the lengths of two sides and the angle between them, so use the Law of Cosines to find the third side.
x^2 = 10.5^2+ 5.4^2 -2(10.5)(5.4)cos20°
x ≈ 5.73 units
Answer:
Step-by-step explanation:
The picture of the question in the attached figure
step 1
Let
r ---> the radius of the sector
s ---> the arc length of sector
Find the radius r
we know that
solve for r
step 2
Find the value of s
substitute the value of r
step 3
we know that
The area of complete circle is equal to
The complete circle subtends a central angle of 2π radians
so
using proportion find the area of the sector by a central angle of angle theta
Let
A ---> the area of sector with central angle theta
substitute the value of r
Convert to function notation
60 = a * (-30)^2
a = 1/15
So y = (1/15)x^2
abc)
The derivative of this function is 2x/15. This is the slope of a tangent at that point.
If Linda lets go at some point along the parabola with coordinates (t, t^2 / 15), then she will travel along a line that was TANGENT to the parabola at that point.
Since that line has slope 2t/15, we can determine equation of line using point-slope formula:
y = m(x-x0) + y0
y = 2t/15 * (x - t) + (1/15)t^2
Plug in the x-coordinate "t" that was given for any point.
d)
We are looking for some x-coordinate "t" of a point on the parabola that holds the tangent line that passes through the dock at point (30, 30).
So, use our equation for a general tangent picked at point (t, t^2 / 15):
y = 2t/15 * (x - t) + (1/15)t^2
And plug in the condition that it must satisfy x=30, y=30.
30 = 2t/15 * (30 - t) + (1/15)t^2
t = 30 ± 2√15 = 8.79 or 51.21
The larger solution does in fact work for a tangent that passes through the dock, but it's not important for us because she would have to travel in reverse to get to the dock from that point.
So the only solution is she needs to let go x = 8.79 m east and y = 5.15 m north of the vertex.
24 I think is the answer but I'm not sure. Hope I helped:)
