6.75 in by 5 in. the area is 33.75
Answer:
So the answer for this case would be n=22547 rounded up to the nearest integer
Step-by-step explanation:
Let's define some notation
represent the sample mean
population mean (variable of interest)
represent the population standard deviation
n represent the sample size
represent the margin of error desire
The margin of error is given by this formula:
(a)
And on this case we have that ME =+1 and we are interested in order to find the value of n, if we solve n from equation (a) we got:
(b)
The critical value for 99% of confidence interval now can be founded using the normal distribution. The significance would be 
and the critical value 
, replacing into formula (b) we got:
So the answer for this case would be n=22547 rounded up to the nearest integer
1/2sqrt((x_1^2+y_1^2)(x_2^2+y_2^2))
The area of a triangle is equal to 1/2bh (one half base times height). Since this is a right triangle, the base and height are the two legs connected to the 90* angle. To find the values of these sides, we will use Pythagorean Theorem, root a squared plus b squared.
Short leg: <x(1),y(1)>
This leg can be seen as the hypotenuse of an invisible right triangle. The x value, x(1), is how far over the x value has gone from the origon at x=0. Imagine a leg alone the x-axis, going from (0,0) to (x(1),0). The y value of the point, y(1), works the same way. This leg will go from our previous mark at (x(1),0) to the point (x(1),y(1)). This shows that the short leg of the main triangle is the hypotenuse, with a height of y(1) and base of x(1). Pythagoreum Theorem shows that the length of this leg is equal to sqrt(x_1^2+y_1^2).
Long leg: <x(2), y(2)>
The same process works here, giving us sqrt(x_2^2+y_2^2).
Now for the area, we have the b and h values. Our equation reads 1/2sqrt(x_1^2+y_1^2)sqrt(x_2^2+y_2^2).
But we can simplify this (yay). The two square roots can be written together as sqrt((x_1^2+y_1^2)(x_2^2+y_2^2))
So the correct answer is 1/2sqrt((x_1^2+y_1^2)(x_2^2+y_2^2))
We don’t have any of the “images” to answer this question.
