Answer:
The probability of selecting a black card or a 6 = 7/13
Step-by-step explanation:
In this question we have given two events. When two events can not occur at the same time,it is known as mutually exclusive event.
According to the question we need to find out the probability of black card or 6. So we can write it as:
P(black card or 6):
The probability of selecting a black card = 26/52
The probability of selecting a 6 = 4/52
And the probability of selecting both = 2/52.
So we will apply the formula of compound probability:
P(black card or 6)=P(black card)+P(6)-P(black card and 6)
Now substitute the values:
P(black card or 6)= 26/52+4/52-2/52
P(black card or 6)=26+4-2/52
P(black card or 6)=30-2/52
P(black card or 6)=28/52
P(black card or 6)=7/13.
Hence the probability of selecting a black card or a 6 = 7/13 ....
Answer:
90 red pens
Step-by-step explanation:
we need to find the least common multiple of the numbers 3 , 5 , 6 , 9
6 = 2 × 3
9 = 3²
then
LCM(3,5,6,9) = 2×5×3² = 2×5×9 = 90
The number 90 can be divided into 3 ,5 ,6 ,9 with no remainder
Therefore the pack must have 90 pens or a multiple of 90
For example 180 , 270 ...
The answer is n=10 I already did this
I hope this helps you
answer is D
The value is J = 1.71*10^-6 kmol/m^2.s
According to the given conditions we get,
The length the tube l= 0.20m
The diameter of the tube d= 0.01m
The total pressure inside the tube P= 101.32kPa
The partial pressure of CO2 at the first end is
P1= 456mm Hg = 60794.832 Pa
The partial pressure of CO2 at the other end is
P2= 76mm Hg = 10132.472 Pa
The temperature is T = 298 K
The diffusion coefficient D= 1.67* 10^-5 m^2/s
Generally the molar flux of CO2 is mathematically represented as
J= D{p1-p2} / R.T
Here R is the gas constant with value R= 8.314 j/kmol
So we get J= 1.71 * 10^-3 mol/m^2.sec
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