-1(x-1)(4x-9) +(x-1)(x-5)
(x-1)(-1(4x-9) +x-5)
(x - 1)(-4x + 9 + x - 5)
(x - 1)(-3x + 4)
Answer:
<x = 20°
Step-by-step explanation:
<x = 1/2(arc UT - arc SV)
<x = 1/2(65° - 25°)
<x = 1/2 (40°)
<x = 20°
4 is 18 because 6x3 is 18
Answer:
Step-by-step explanation:
The logistic differential equation is as follows:
In this problem, we have that:
, which is the carring capacity of the population, that is, the maximum number of people allowed on the beach.
At 10 A.M., the number of people on the beach is 200 and is increasing at the rate of 400 per hour.
This means that 
when 
. With this, we can find r, that is, the growth rate,
So
So the differential equation is:
Answer:
Step-by-step explanation:
Let us denote probability of spoilage as follows
Transformer spoilage = P( T ) ; line spoilage P ( L )
Both P ( T ∩ L ) .
Given
P( T ) = .05
P ( L ) = .08
P ( T ∩ L ) = .03
a )
For independent events
P ( T ∩ L ) = P( T ) x P ( L )
But .03 ≠ .05 x .08
So they are not independent of each other .
b )
i )
Probability of line spoilage given that there is transformer spoilage
P L/ T ) = P ( T ∩ L ) / P( T )
= .03 / .05
= 3 / 5 .
ii )
Probability of transformer spoilage but not line spoilage.
P( T ) - P ( T ∩ L )
.05 - .03
= .02
iii )Probability of transformer spoilage given that there is no line spoilage
[ P( T ) - P ( T ∩ L ) ] / 1 - P ( L )
= .02 / 1 - .08
= .02 / .92
= 1 / 49.
i v )
Neither transformer spoilage nor there is no line spoilage
= 1 - P ( T ∪ L )
1 - [ P( T ) + P ( L ) - P ( T ∩ L ]
= 1 - ( .05 + .08 - .03 )
= 0 .9
