Answer:
0.665
Step-by-step explanation:
Given: 100 people are split into two groups 70 and 30. I group is given cough syrup treatment but second group did not.
Prob for a person to be in the cough medication group = 0.70
Out of people who received medication, 34% did not have cough
Prob for a person to be in cough medication and did not have cough
=
Prob for a person to be not in cough medication and did not have cough
=
Probability for a person not to have cough
= P(M1C')+P(M2C')
where M1 = event of having medication and M2 = not having medication and C' not having cough
This is because M1 and M2 are mutually exclusive and exhaustive
SO P(C') = 0.397+0.2=0.597
Hence required prob =P(M1/C') = 
Basically you’re multiplying them both
(x^3 + 2x - 1)(x^4 - x^3 + 3)
so you need to make sure you multiply each one, if you do it right, you should end up with
x^7 - x^6 + 3x^3 + 2x^5 - 2x^4 + 6x -x^4 +x^3 -3
simplify by adding like terms
x^7 - x^6 + 2x^5 - 3x^4 + 4x^3 + 6x - 3
your answer would be the third option
It would be 22 x 3.14 = 69.08 i’m pretty sure lolol sorry if i’m wrong
Answer: m1+m3
Step-by-step explanation: 1+3=4 and m1+m3 has the same varibles in the expression.
The answer is
<span>The two shipments have the same range of weights but a different mean weight.
Hope i helped </span>