The number of possible batting lineups is 720
<u><em>Explanation</em></u>
Total number of players in the team = 9
The center fielder must bat fourth, the second baseman must bat third and the pitcher last. That means, <u>3 slots in the batting line-up are already assigned to 3 players</u>.
So, the remaining (9 - 3) = 6 players will be assigned to remaining 6 slots.
Thus, the possible number of ways so that 6 players can be assigned to 6 slots will be: 
<span>6 servings ... 1/2 cup of butter
8 servings .... x cup of butter
We would like to know what is the value of the x.
</span>6 * x = 8 * 1/2<span>
6 * x = 4 /6
x = 4/6
x = 2/3
Result: Nancy needs</span> 2/3 cup of butter for 8 servings.
Step-by-step explanation:
1)......mean= sum of observations/ total observations
= 5.2+8.4+4.3+6.7+5.8/5= 6.08
median= n=5(odd)=n+1/2=6/2=3rdterm= 5.8( when arranged in ascending order).....
mode= 3median- 2meam= 3×5.8-2×6.08= 5.24
range=8.4-4.3=4.1(when done A.O)
2)......mean= -2-13-13-5-7/5= -8
median= 3rd term= -5.....(n=5..odd)(when done A.o)
mode=largest frequency= -13
range= -13-(-2) = -13+2= -11(A.O)
3).....mean= sum/no.= 8+6+13/2+17/2+15/2+7+8+11/2+7+8//10=7.2
median=n=10(even)
=
mode= 8
range=8.5-5.5=3
median:==
(10/2)+(10/2+1)//2=5 term + 6term/2=7+7.5=14.5/2=7.25
4)....
5)...remove 62...then 73will be having highest frequency
(64/56)^2 x (1/343) x 7^-9
=9.44x10^-11
