Answer: Yes , it is unusual for a boiler to weigh more than 1550 grams .
Step-by-step explanation:
Given : Big chickens: According to a poultry industry news website, the weights of broilers (commercially raised chickens) are approximately normally distributed with mean
1358 grams and standard deviation
161 grams.
When the probability that broiler weigh more than 1550 grams < 0.5 , then is unusual otherwise not.
Let x denotes the weight of broiler, then the probability that broiler weigh more than 1550 grams :-
![P(x>1550)=1-P(x\leq1550)\\\\=1-P(\dfrac{x-\mu}{\sigma}\leq\dfrac{1550-1358}{161})\\\\\approx 1-P(z\leq1.19)\ \ [\because\ z=\dfrac{x-\mu}{\sigma}]\\\\=\\\\=1-0.8830\ \ [\text{By using z-table}]\\\\=0.117](https://tex.z-dn.net/?f=P%28x%3E1550%29%3D1-P%28x%5Cleq1550%29%5C%5C%5C%5C%3D1-P%28%5Cdfrac%7Bx-%5Cmu%7D%7B%5Csigma%7D%5Cleq%5Cdfrac%7B1550-1358%7D%7B161%7D%29%5C%5C%5C%5C%5Capprox%201-P%28z%5Cleq1.19%29%5C%20%5C%20%5B%5Cbecause%5C%20z%3D%5Cdfrac%7Bx-%5Cmu%7D%7B%5Csigma%7D%5D%5C%5C%5C%5C%3D%5C%5C%5C%5C%3D1-0.8830%5C%20%5C%20%5B%5Ctext%7BBy%20using%20z-table%7D%5D%5C%5C%5C%5C%3D0.117)
Since 0.117<0.5
Therefore, it is unusual for a boiler to weigh more than 1550 grams .
X would be: 0.5, 1.5, 4.5
Y would be: 3, 9, 27
When you divide 3 by 0.5 you get six so I basically just did trial and error with all the number
Answer:
c
Step-by-step explanation:
let the number be n then multiplied by itself gives n², thus
n² + 7 = 16 ( subtract 7 from both sides )
n² = 9 ( take the square root of both sides )
n = 
= 3
If you look at the pyramid, the pyramid is really just a hexagon with 6 triangles stacked up to each other at the edges, with the hexagon at the bottom.
now, the perpendicular distance from the center of the hexagon to a side, namely the
apothem, is 6√3, and each side is 12 units long, since there are 6 of them that'd be 72 for all, namely the
perimeter.
each of the triangular faces have a base of 12 and an altitude or height of 11, recall that area of a triangle is (1/2)bh, and area of a regular polygon is (1/2)(apothem)(perimeter).
so if we just get the area of the hexagon at the bottom, and the triangles, sum them up, that's the surface area of the pyramid.
![\bf \stackrel{\textit{area of the hexagon}}{\left[\cfrac{1}{2}(6\sqrt{3})(72) \right]}~~~~+~~~~\stackrel{\textit{area of the 6 triangles}}{6\left[\cfrac{1}{2}(12)(11) \right]}](https://tex.z-dn.net/?f=%5Cbf%20%5Cstackrel%7B%5Ctextit%7Barea%20of%20the%20hexagon%7D%7D%7B%5Cleft%5B%5Ccfrac%7B1%7D%7B2%7D%286%5Csqrt%7B3%7D%29%2872%29%20%20%5Cright%5D%7D~~~~%2B~~~~%5Cstackrel%7B%5Ctextit%7Barea%20of%20the%206%20triangles%7D%7D%7B6%5Cleft%5B%5Ccfrac%7B1%7D%7B2%7D%2812%29%2811%29%20%20%5Cright%5D%7D)
Answer:
i cant see the picture
Step-by-step explanation:
