Answer:
the same
Step-by-step explanation:
Not as histograms where space must equal 0 as calculating area is needed.
Answer:
n=40 boys
Step-by-step explanation:
-Margin of error is given by the formula:

-The z score for a probability of 92% is 1.75
#We make n the subject of the formula and substitute at a 92% confidence:
![E=z\frac{\sigma}{\sqrt{n}}\\\\\sqrt{n}=\frac{z\sigma}{E}\\\\n=[\frac{z\sigma}{E}]^2\\\\\therefore n=[\frac{1.75\times 1.8}{0.5}]^2\\\\=39.69\approx 40](https://tex.z-dn.net/?f=E%3Dz%5Cfrac%7B%5Csigma%7D%7B%5Csqrt%7Bn%7D%7D%5C%5C%5C%5C%5Csqrt%7Bn%7D%3D%5Cfrac%7Bz%5Csigma%7D%7BE%7D%5C%5C%5C%5Cn%3D%5B%5Cfrac%7Bz%5Csigma%7D%7BE%7D%5D%5E2%5C%5C%5C%5C%5Ctherefore%20n%3D%5B%5Cfrac%7B1.75%5Ctimes%201.8%7D%7B0.5%7D%5D%5E2%5C%5C%5C%5C%3D39.69%5Capprox%2040)
Hence, he needs approximately 40 boys for the study
Cot(t)= cos(t) / sin(t)
Cos(t) = sqrt(1 - sin^2(t) ) But this is in quad 4, so sin(t) <0
cot(t) = sqrt(1 -sin^2(t)) / - sint(t)
cot(t) = - sqrt( 1 - sin^2(t) ) / sin(t)
Their are 6 equilateral triangles in a regular hexagon:
The area of one triangle = (x²√3)/4, then:
1st answer: The area of the hexagon base: is 6 TIMES the area of one equilateral triangle [or, needed for 2nd, question, TOT AREA:(6x²√3)/4] = (3x²√3)/2] unit²
2nd answer: :
Volume of pyramid: (base area)(height)/3
Volume of pyramid: (3x²√3).(3x)/3 (because height = 3x, given)
Then Volume of pyramid= 3x³√3 unit³.
3 is the answer for the second question
and 6 for the 1st