Aisha runs a tutoring business.with plan 1 students may choose to pay $15 per hour
1 answer:
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Answer:
All points between z = 0 and z = 7 along the z-azis in R3 x-y-z plane
Step-by-step explanation:
The inequality 1 z
7 represents all sets of points on the z-axis in the R3 plane that lies bewteen z = 0 and z= 7.
It is represents a line segment joining two point z= 0 and z=7 along z- axis in the R3 plane, while x=0 and y=0 on the x-y plane.
Check the tree diagram, of all the possible scenarios and the probabilities.
All the possible scenarios are:
1st.non defective, 2nd.non defective
1st.non defective, 2nd. defective
1st. defective, 2nd. non defective
1st. defective, 2nd. defective
consider the case 1st.non defective, 2nd.non defective
the probability that the first one is non defective is 8/10, since 8 out of 10 are non defective.
In this scenario, the second one is non defective as well. The probability for this to happen is 7/9, since now we have 7 non defective pots, and 9 in total.
This means that the probability of the first to be non defective, and the second non-defective is (8/10) * (7/9)=0.622
similarly we calculate the other cases, as shown in the picture.
P(at least one is defective)=
P(1st.non defective, 2nd. defective)+P(1st. defective, 2nd. non defective)+P(1st. defective, 2nd. defective)=
0.178+0.178+0.022=0.378
Answer: 0.378
All the possible scenarios are:
1st.non defective, 2nd.non defective
1st.non defective, 2nd. defective
1st. defective, 2nd. non defective
1st. defective, 2nd. defective
consider the case 1st.non defective, 2nd.non defective
the probability that the first one is non defective is 8/10, since 8 out of 10 are non defective.
In this scenario, the second one is non defective as well. The probability for this to happen is 7/9, since now we have 7 non defective pots, and 9 in total.
This means that the probability of the first to be non defective, and the second non-defective is (8/10) * (7/9)=0.622
similarly we calculate the other cases, as shown in the picture.
P(at least one is defective)=
P(1st.non defective, 2nd. defective)+P(1st. defective, 2nd. non defective)+P(1st. defective, 2nd. defective)=
0.178+0.178+0.022=0.378
Answer: 0.378