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3 years ago

Points E,D, and H are the midpoints of the sides of triangle TUV. UV = 80, TV = 100 and HD = 80

1 answer:

8 0

Without the diagram, there are 6 possible arrangements for points E, D, H. Let's see if we can reduce that to something smaller. HD can not be parallel to either TV or UV since it would have to be half of either of those sides in length. So HD has to be parallel to side TU. This means that point E has to be the midpoint of side TU. However, it's indeterminate about which point is H, or which point is D. Therefore there's 2 possible answers to this question. Without the diagram, it's impossible to determine which of the two possibilities is correct, so you'll have to choose the correct one yourself. Option 1. D is midpoint of TV, H is midpoint of UV. Triangle TUV is similar to triangle DHV. Since the ratio of the corresponding sides is 2:1, the length of HE is half of the length of TV. So 100/2 = 50. And because of that 2:1 ratio, the length of TU is twice that of HD, so 2 * 80 = 160. In summary: HE = 50 TU = 160 Option 2. H is midpoint of TV, D is midpoint of UV. We still have a similar triangle, but H&D are swapped. The length of TU will be the same as in option 1 above. And the length of HE will be half that of UV instead of TV. So 80/2 = 40. So in summary: HE = 40 TU = 160

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3 years ago

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Answer:

$P(A) = \frac{1}{9}$

$P(B) = \frac{1}{3}$

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Step-by-step explanation:

We know that:

Only employees are hired during the first 3 days of the week with equal probability.

2 employees are selected at random.

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A. The probability that an employee has been hired on a Monday is:

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B. We now look for the probability that two selected employees have been hired on the same day of the week.

The probability that both are hired on a Monday, for example, we know is $P(A) = \frac{1}{9}$ . We also know that the probability of being hired on a Monday is equal to the probability of being hired on a Tuesday or on a Wednesday. But if both were hired on the same day, then it could be a Monday, a Tuesday or a Wednesday.

$P(B) = \frac{1}{9} + \frac{1}{9} + \frac{1}{9}\\\\P(B) = \frac{1}{3}$ .

C. If the probability that two people have been hired on a specific day of the week is $3(\frac{1}{3}) ^ 2$ , then the probability that 7 people have been hired on the same day is:

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D. The probability is $\frac{1}{729}$ . This number is quite close to zero. Therefore it is an unlikely bastate event.

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3 years ago

Number three . Show work plz

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13/15 - 1/3 Find a common denominator
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