C. Between E and F and between G and H the elevator could be travelling from a higher floor to a lower floor.
<u>We have to justify all the options with respect to the </u><u>graph </u><u>:-</u>
<u>Option A</u> : Between A and B, the elevator would be travelling from a lower floor to a higher floor. Also between C and D, the elevator would be travelling from a lower floor to a slightly higher floor.
So, this option is <u>not correct.</u>
<u>Option B</u> : Between B and C, the elevator wouldn't go higher. And also the same would happen in between D and E.
So, this option is <u>not correct.</u>
<u>Option C</u> : Between E and F, the elevator would be travelling from higher floor to lower and also between G and H.
So, this option is <u>correct</u>.
<u>Option 4</u> : Between F and G, there would not have much changes. So it will be omitted.
So, this option is <u>not correct.</u>
<u />
Learn more about graph here :
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Your answer to this question will be y^2-4y+4
Answer:
4/5
Step-by-step explanation:
the area of rectangle is length × breadth
length = 2 2/5 breadth = 1/3
2 2/5 × 1/3 = 12/15
reducing to the lowest term =4/5
Step-by-step explanation:
There are the combinations that result in a total less than 7 and at least one die showing a 3:
[3, 3] [3,2] [2,1] [1,3] [2,3]
The probability of each of these is 1/6 * 1/6 = 1/36
There is a little ambiguity here about whether or not we should count [3,3] as the problem says "and one die shows a 3." Does this mean that only one die shows a 3 or at least one die shows a 3? Assuming the latter, the total probability is the sum of the individual probabilities:
1/36 + 1/36 + 1/36 + 1/36 + 1/36 = 5/36
Therefore, the required probability is: 5/36
