Answer : Remaining two observation becomes 97 and 107.
Explanation :
Since we have given that
Mean = 100
Modal value = 98
Range = 10
As we know that ,
Range = Highest-Lowest
Let highest observation be x
Let lowest observation be y
So equation becomes x-y=10 ----equation 1
So, observation becomes
x,98,98,y
Now, we use the formula of mean i.e.
Mean = 
So, mean =
So our 2nd equation becomes
x+y=204
On using elimination method of system of linear equation on these two equation we get,
x=97
and
Hence , remaining two observation becomes 97 and 107.
Answer:
5% I think
Step-by-step explanation:
Its mostly just a guess but I think its 5%
So the given series is "16, 06, 68, 88, __"
Count all the cyclical opening in each of these numbers. For example in 16, there is a one cyclical loop present in it(the one in 6), similarly in 06 it is two(one in zero and one in 6), going ahead, in 68 it is 3(one in 6 and two in 8).
From here on things become simple: hence, the cyclical figures in these equations written down becomes 1,2,3,4,_,3.
Let's now try solving the above sequence, going by the logical reasoning the only number that can fill in the gap should be 4.
Answer:
Z = 8 + 2x2 + 2y2
Convert to polar coordinates
Z = 8 + 2r2
Now theta will go from 0 to pi/2 because it's in the first quadrant.
R will go from 0 to the radius of the circle formed at the intersection of the plane and the paraboloid.
14 = 8 + 2r2
r = sqrt(3)
So r goes from 0 to sqrt(3).
You integrate 14-z where 0<r<sqrt(3) and 0<theta<pi/2.
It is 14-z and not z because just z would give the volume under the paraboloid.
Step-by-step explanation: please go answer my recent question
Step-by-step explanation:
Maths
Bookmark
if 2
2008
−2
2007
−2
2006
+2
2005
=k ⋅2
2005
then the value of k is equal to
Answer
Correct option is
B
3
2
2008
−2
2007
−2
2006
+2
2005
=k⋅2
2005
⇒[2
2008
−2
2006
]−[2
2007
−2
2005
]=k⋅2
2005
⇒2
2006
(2
2
−1)−2
2005
(2
2
−1)=k⋅2
2005
⇒3(2
2006
−2
2005
)=k⋅2
2005
⇒3[2
2005
(2−1)]=k⋅2
2005
⇒3(2
2005
)=k⋅2
2005
On comparing, we get
k=3
Hence, Option B is correct.
