luke is making 4 first aid kits. he wants to put 3 large and 4 small bandages in each kit. how many bandages does he need for th
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Answer:
There are 16 yellow houses in the neighborhood.
Step-by-step explanation:
We have that:
2/5 = 40% of the houses are painted yellow
1 - 2/5 = 3/5 = 60% of the houses are not painted yellow
If there are 24 houses that are not painted yellow, how many yellow houses are in the neighborhood?
The first step is finding the total number of houses in the neighborhood.
We have that 24 are not painted yellow, and this is 60%(0.6 decimal). How much is 100%?
We solve a rule of three
24 houses - 0.6
x houses - 1
There are 40 houses. Of those, 40% are painted yellow.
0.4*40 = 16.
There are 16 yellow houses in the neighborhood.
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