Answer:
Step-by-step explanation:
Given that a dairy scientist is testing a new feed additive. She chooses 13 cows at random from a large population of cows. She randomly assigns nold = 8 to get the old diet, and nnew = 5 to get the new diet including the additive.
From the data given we get the following
N Mean StDev SE Mean
Sample 1 8 43 5.1824 1.832
Sample 2 5 73 21.0832 9.429
df = 11
Std dev for difference = 13.3689
a) Yes the two are independent. The two sets of cows randomly chosen are definitely independent. Paired means equal number should be there and homogeneous conditions should be maintained.
b) Enclosed
c) Comparison of two means is the test recommended here. Because independent samples are used.\
d) Test statistic= -3.1233
(because of unequal variances we use that method)
95% confidence interval = ( -56.6676 , -3.3324 )
p value <0.05 our alpha
So reject null hypothesis.
The two means are statistically significantly different.
ok first you need to add e and 4 and get 95 out of haha
Answer:
Dieting (?) since proportions of the food can affect you're diet and it's also useful to stay on track.
Let
x------> the length side of the square base of the box
y-------> the height of the box
we know that
volume of the box=b²*h
b=x
h=y
volume=256 cm³
so
256=x²*y------>y=256/x²--------> equation 1
<span>The amount of material used is directly proportional to the surface area, so we will minimize the amount of material by minimizing the surface area.</span>
surface area of the cardboard=area of the base+perimeter of base*height
area of the base=x²
perimeter of the base=4*x
height=y
surface area=x²+4x*y-----> equation 2
substitute equation 1 in equation 2
SA=x²+4x*[256/x²]-----> SA=x²+1024/x
step 1
find the first derivative of SA and equate to zero
2x+1024*(-1)/x²=0------> 2x=1024/x²----> x³=512--------> x=8 cm
y=256/x²------> y=256/8²-----> y=4 cm
the answer is
the length side of the square base of the box is 8 cm
the height of the box is 4 cm
Answer:
(2.83 , 1 , 4)
Step-by-step explanation:
Rewrite these equations in matrix form
![\left[\begin{array}{ccc}2&2&-1\\4&-2&-2\\3&3&-4\end{array}\right] \left[\begin{array}{ccc}x\\y\\z\end{array}\right]=\left[\begin{array}{ccc}4\\2\\-4\end{array}\right] \\](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D2%262%26-1%5C%5C4%26-2%26-2%5C%5C3%263%26-4%5Cend%7Barray%7D%5Cright%5D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Dx%5C%5Cy%5C%5Cz%5Cend%7Barray%7D%5Cright%5D%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D4%5C%5C2%5C%5C-4%5Cend%7Barray%7D%5Cright%5D%20%5C%5C)
we can write it like this,
so to solve it we need to take the inverse of the 3 x 3 matrix A then multiply it by B.
We get the inverse of matrix A,
![A^{-1}=\left[\begin{array}{ccc}7/15&1/6&-1/5\\1/3&-1/6&0\\3/5&0&-2/5\end{array}\right] \\](https://tex.z-dn.net/?f=A%5E%7B-1%7D%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D7%2F15%261%2F6%26-1%2F5%5C%5C1%2F3%26-1%2F6%260%5C%5C3%2F5%260%26-2%2F5%5Cend%7Barray%7D%5Cright%5D%20%20%5C%5C)
now multiply the matrix with B
![X=A^{-1}B\\\\\left[\begin{array}{ccc}x\\y\\z\end{array}\right] =\left[\begin{array}{ccc}7/15&1/6&-1/5\\1/3&-1/6&0\\3/5&0&-2/5\end{array}\right]\left[\begin{array}{ccc}4\\2\\-4\end{array}\right] \\\\\\\left[\begin{array}{ccc}x\\y\\z\end{array}\right] =\left[\begin{array}{ccc}2.83\\1\\4\end{array}\right] \\](https://tex.z-dn.net/?f=X%3DA%5E%7B-1%7DB%5C%5C%5C%5C%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Dx%5C%5Cy%5C%5Cz%5Cend%7Barray%7D%5Cright%5D%20%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D7%2F15%261%2F6%26-1%2F5%5C%5C1%2F3%26-1%2F6%260%5C%5C3%2F5%260%26-2%2F5%5Cend%7Barray%7D%5Cright%5D%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D4%5C%5C2%5C%5C-4%5Cend%7Barray%7D%5Cright%5D%20%5C%5C%5C%5C%5C%5C%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Dx%5C%5Cy%5C%5Cz%5Cend%7Barray%7D%5Cright%5D%20%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D2.83%5C%5C1%5C%5C4%5Cend%7Barray%7D%5Cright%5D%20%5C%5C)
