Answer:
A sample size of 345 is needed so that the confidence interval will have a margin of error of 0.07
Step-by-step explanation:
In a sample with a number n of people surveyed with a probability of a success of 
, and a confidence level of 
, we have the following confidence interval of proportions.
In which
z is the zscore that has a pvalue of 
.
The margin of error of the interval is given by:
In this problem, we have that:
99.5% confidence level
So 
, z is the value of Z that has a pvalue of 
, so 
.
Using this estimate, what sample size is needed so that the confidence interval will have a margin of error of 0.07?
This is n when M = 0.07. So
A sample size of 345 is needed so that the confidence interval will have a margin of error of 0.07
Answer:
76017.69 mm²
631000 mm³
295045.04 mm³
Step-by-step explanation:
Surface area = bh + 2ls + lb
Since the sides of the triangular prism are equal (equilateral triangle) = 60 mm each
Height = 405 mm
The surface area of triangular = 76017.69 mm
The volume of triangular prism :
1/2 * base * height * length
Volume = 631000 mm³
If amount of chocolate inside = 335954.96 mm3
Empty space =
Volume of triangular prism - amount of chocolate inside
631000 - 335954.96
= 295045.04 mm³
Answer:
maybe
Step-by-step explanation:
Dora is apparently assuming the dimensions are integers. In that case she is correct.
If the dimensions are unconstrained, the perimeter will be largest when a pair of opposite sides will be the smallest measure allowed.
For some perimeter P and side length x, the area is ...
A = x(P/2 -x)
Conversely, the perimeter for a given area is ...
P = 2(A/x +x)
This gets very large when x gets very small, so Dora is correct in saying that the side lengths that are as small as they can be will result in the largest perimeter. We have no way of telling if her assumption of integer side lengths is appropriate. If it is not, her statement makes no sense.
Answer:
A
Step-by-step explanation:
an=2(1)=2
an=2(3)=6
Answer:
Step-by-step explanation:
Domain:
![\dfrac{x^2+9y^2}{x-3y}+\dfrac{6xy}{3y-x}=\dfrac{x^2+9y^2}{x-3y}+\dfrac{6xy}{-(x-3y)}\\\\=\dfrac{x^2+9y^2}{x-3y}-\dfrac{6xy}{x-3y}=\dfrac{x^2+9y^2-6xy}{x-3y}\\\\=\dfrac{x^2-2(x)(3y)+(3y)^2}{3y-x}=\dfrac{(x-3y)^2}{3y-x}\\\\=\dfrac{\bigg[-1(3y-x)\bigg]^2}{3y-x}=\dfrac{(-1)^2(3y-x)^2}{3y-x}\\\\=\dfrac{1(x-3y)(x-3y)}{x-3y}=x-3y](https://tex.z-dn.net/?f=%5Cdfrac%7Bx%5E2%2B9y%5E2%7D%7Bx-3y%7D%2B%5Cdfrac%7B6xy%7D%7B3y-x%7D%3D%5Cdfrac%7Bx%5E2%2B9y%5E2%7D%7Bx-3y%7D%2B%5Cdfrac%7B6xy%7D%7B-%28x-3y%29%7D%5C%5C%5C%5C%3D%5Cdfrac%7Bx%5E2%2B9y%5E2%7D%7Bx-3y%7D-%5Cdfrac%7B6xy%7D%7Bx-3y%7D%3D%5Cdfrac%7Bx%5E2%2B9y%5E2-6xy%7D%7Bx-3y%7D%5C%5C%5C%5C%3D%5Cdfrac%7Bx%5E2-2%28x%29%283y%29%2B%283y%29%5E2%7D%7B3y-x%7D%3D%5Cdfrac%7B%28x-3y%29%5E2%7D%7B3y-x%7D%5C%5C%5C%5C%3D%5Cdfrac%7B%5Cbigg%5B-1%283y-x%29%5Cbigg%5D%5E2%7D%7B3y-x%7D%3D%5Cdfrac%7B%28-1%29%5E2%283y-x%29%5E2%7D%7B3y-x%7D%5C%5C%5C%5C%3D%5Cdfrac%7B1%28x-3y%29%28x-3y%29%7D%7Bx-3y%7D%3Dx-3y)
Used:
The distributive property: a(b + c) = ab + ac
(a - b)² = a² - 2ab + b²
