Answer:
The right answer is "1.818".
Step-by-step explanation:
The given values are:
a = 0.1
![\bar{x} = 94](https://tex.z-dn.net/?f=%5Cbar%7Bx%7D%20%3D%2094)
![\mu = 90](https://tex.z-dn.net/?f=%5Cmu%20%3D%2090)
Now,
The test statistic will be:
⇒ ![t=\frac{\bar{x}-\mu}{\frac{s}{vn} }](https://tex.z-dn.net/?f=t%3D%5Cfrac%7B%5Cbar%7Bx%7D-%5Cmu%7D%7B%5Cfrac%7Bs%7D%7Bvn%7D%20%7D)
On substituting the given values, we get
⇒ ![=\frac{94-90}{\frac{22}{10} }](https://tex.z-dn.net/?f=%3D%5Cfrac%7B94-90%7D%7B%5Cfrac%7B22%7D%7B10%7D%20%7D)
⇒ ![=\frac{4}{2.2}](https://tex.z-dn.net/?f=%3D%5Cfrac%7B4%7D%7B2.2%7D)
⇒ ![=1.818](https://tex.z-dn.net/?f=%3D1.818)
To find the number or rows, divide the total number of students by the number in each row.
63 / 7 = 9 rows.
Answer:
<h2>Volume!</h2>
Step-by-step explanation:
it the amount of space that can be occupied
X²(x - 4) +4 (x - 4)
(x² + 4) (x - 4)
First find the common terms that can enter into both x³ and 4x² then write its down in this case it’s x² that can enter x³ leaving only x _since x³/x² = subtract of the indices. x² will also enter 4x² leaving only four hence you having x² (x - 4)
then do the same for the next pair of terms giving you 4 that can enter into both 4 and 16
Leaving you with +4 (x - 4)
Now you can put the common terms together like so (x² + 4) and choose get one of the other two which are the same= (x - 4)
= (x² + 4) (x - 4)