The probability of flowering plants bearing flowers in January is given in the table.
2 answers:
Solution:
As, we know Probability of an event =
In the table given,probability of flowering plants bearing flowers in January is given.
It means out of 100 tulip 68 tulip flowers in January.
There are five kinds of flowers.
We have to select tulip and find it's Probability.
As all flowers are independent, there flowering process or Phenomenon does not depend on each other.
So,Probability of a flowering plant bearing flowers in January if it is a tulip= 68 %
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Answer:
The value of y is 5 , and The value of x is - 5
Step-by-step explanation:
Given as :
The line passes through point ( - 2, 5)
The slope of line is m =
Equation of line is written as
y - = m × (x -
)
Since line passes through line ( - 2 , 5 ) and have slope
So, y - 5 = × (x + 2)
Or, 3 × ( y - 5 ) = 2 × ( x + 2 )
Or, 3 y - 15 = 2 x + 4
Or, 3 y - 2 x - 19 = 0 is the equation of line
Now points ( x , 3 ) lies on the line 3 y - 2 x - 19 = 0
So, 3 × 3 - 2 x - 19 = 0
or, 9 - 2 x - 19 = 0
or, 2 x = - 10
∴ x = -
I.e x = - 5
Again points ( - 2 , y ) lies on the line 3 y - 2 x - 19 = 0
So, 3 × y + 2 × 2 - 19 = 0
Or, 3 × y + 4 -19 = 0
or, 3 y = 15
∴ y =
I.e y = 5
Hence The value of y is 5 , and The value of x is - 5 . Answer