Ok what is the question lol
Let, snake catches the mouse in t hours. Till then mouse is travelling from
( t + 1.5 ) hours.
Distance, travelled by mouse in t hours is,
.
To catch the mouse snake had to travel the same distance d in t hours.
So,

Therefore, snake will catch mouse in 1.5 hours.
Answer:
A) 68.33%
B) (234, 298)
Step-by-step explanation:
We have that the mean is 266 days (m) and the standard deviation is 16 days (sd), so we are asked:
A. P (250 x < 282)
P ((x1 - m) / sd < x < (x2 - m) / sd)
P ((250 - 266) / 16 < x < (282 - 266) / 16)
P (- 1 < z < 1)
P (z < 1) - P (-1 < z)
If we look in the normal distribution table we have to:
P (-1 < z) = 0.1587
P (z < 1) = 0.8413
replacing
0.8413 - 0.1587 = 0.6833
The percentage of pregnancies last between 250 and 282 days is 68.33%
B. We apply the experimental formula of 68-95-99.7
For middle 95% it is:
(m - 2 * sd, m + 2 * sd)
Thus,
m - 2 * sd <x <m + 2 * sd
we replace
266 - 2 * 16 <x <266 + 2 * 16
234 <x <298
That is, the interval would be (234, 298)
Answer:
A ) Not orthogonal to each other
B) 50i + 40j + 105k
C) The tensor product is attached below
D ) The value of X = F.X is attached below
Step-by-step explanation:
attached below is the detailed solution of the above problem
A) for the vectors ( u ) and ( v ) to be orthogonal to each other [ U.V has to be = 0 ] but in this scenario U.V = 4 hence they are not orthogonal to each other
b) The vector normal to plane is gotten by : U x V
= 50i + 40j + 105k
Answer:
787.50
Step-by-step explanation:
interest=principle*rate*time/100
interest=3500*7.5*3/100
interest=35*7.5*3
interest=787.50