Answer:
Step-by-step explanation:
We assume that there are 100 sour candies, Thus-
26 % of candy are grape implies that 26% of 100 candies are grape that is equal to 26
Now remaing candies that are not grape are 100-26 = 74
Based on the rule of multiplication:
P(A ∩ B) = P(A)/ P(B|A)
In the beginning, there are 26grape candies, probability of choosing first grape candy = 26C1 = 26
After the first selection, we replace the selected grape candy so there are still 100 candies in the bag P(B|A) = 100C3 = 100 x 99 x 98 x 97!/3! X 97!
= 50 x 33 x 98
So probability =1/ 50 x 33 x 98 x 26
= 1/4204200
For the answer to the question above, are your diagram shows that the horizontal lines and is a vertical line segment where as FB: 3 , If C is equaled to 3, so the coordinate of the point D would be (-6,-4).I hope my answer helped you.
X = <span>weight of the baby.
y = </span>weight of the doctor.
z = weight of the nurse.
x + y = 78 so y = 78 - x
x + z = 69 so z = 69 - x
x + y + z = 142
substitute y = 78 - x and z = 69 - x into x + y + z = 142
x + y + z = 142
x +78 - x + 69 - x = 142
-x + 147 = 142
-x = - 5
x = 5
answer
<span>the weight of the baby was 5 kg</span>
Class B has the most consistant sleep because there is less of a difference between 6.87 and 3.65 than the other classes.
Answer:
A and C
Step-by-step explanation:
To determine which events are equal, we explicitly define the elements in each set builder.
For event A
A={1.3}
for event B
B={x|x is a number on a die}
The possible numbers on a die are 1,2,3,4,5 and 6. Hence event B is computed as
B={1,2,3,4,5,6}
for event C
![C=[x|x^{2}-4x+3]\\solving x^{2}-4x+3\\x^{2}-4x+3=0\\x^{2}-3x-x+3=0\\x(x-3)-1(x-3)=0\\x=3 or x=1](https://tex.z-dn.net/?f=C%3D%5Bx%7Cx%5E%7B2%7D-4x%2B3%5D%5C%5Csolving%20%20x%5E%7B2%7D-4x%2B3%5C%5Cx%5E%7B2%7D-4x%2B3%3D0%5C%5Cx%5E%7B2%7D-3x-x%2B3%3D0%5C%5Cx%28x-3%29-1%28x-3%29%3D0%5C%5Cx%3D3%20or%20x%3D1)
Hence the set c is C={1,3}
and for the set D {x| x is the number of heads when six coins re tossed }
In the tossing a six coins it is possible not to have any head and it is possible to have head ranging from 1 to 6
Hence the set D can be expressed as
D={0,1,2,3,4,5,6}
In conclusion, when all the set are compared only set A and set C are equal