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pochemuha
2 years ago
13

Zachary weighs 37 kilograms 95 grams. Gabe weighs 4,650 grams less than Zachary. Harry weighs 2,905 grams less than Gabe. How mu

ch does Harry weigh?
Mathematics
1 answer:
Archy [21]2 years ago
5 0
First make the numbers the same.
37 kilograms is 37000 grams, for there are 1000 grams in a kilograms.
Zachary weighs 37095 grams
37095-4650= Gabes weight = 32445 grams
32445-2905= Harry's weight = 29540 grams or 29,54 kilograms
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Given: Circle X with a Radius r and circle Y with radius s Prove: Circle x is similar to circle y
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Two figures are similar if one is the scaled version of the other.

This is always the case for circles, because their geometry is fixed, and you can't modify it in anyway, otherwise it wouldn't be a circle anymore.

To be more precise, you only need two steps to prove that every two circles are similar:

  1. Translate one of the two circles so that they have the same center
  2. Scale the inner circle (for example) unit it has the same radius of the outer one. You can obviously shrink the outer one as well

Now the two circles have the same center and the same radius, and thus they are the same. We just proved that any two circles can be reduced to be the same circle using only translations and scaling, which generate similar shapes.

Recapping, we have:

  1. Start with circle X and radius r
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So, we passed from X to X' to X'', and they are all similar to each other, and in the end we have X''=Y, which ends the proof.

8 0
2 years ago
Read 2 more answers
A bin of 5 transistors is known to contain 2 that are defective. The transistors are to be tested, one at a time, until the defe
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Answer:

P(N_1 = a , N_2 = b)= \frac{1}{5-a C 1} * \frac{5-a C 1}{5C2} = \frac{1}{5C2}=\frac{1}{10}

Step-by-step explanation:

For the random variable N_1 we define the possible values for this variable on this case [1,2,3,4,5] . We know that we have 2 defective transistors so then we have 5C2 (where C means combinatory) ways to select or permute the transistors in order to detect the first defective:

5C2 = \frac{5!}{2! (5-2)!}= \frac{5*4*3!}{2! 3!}= \frac{5*4}{2*1}=10

We want the first detective transistor on the ath place, so then the first a-1 places are non defective transistors, so then we can define the probability for the random variable N_1 like this:

P(N_1 = a) = \frac{5-a C 1}{5C2}

For the distribution of N_2 we need to take in count that we are finding a conditional distribution. N_2 given N_1 =a, for this case we see that N_2 \in [1,2,...,5-a], so then exist 5-a C 1 ways to reorder the remaining transistors. And if we want b additional steps to obtain a second defective transistor we have the following probability defined:

P(N_2 =b | N_1 = a) = \frac{1}{5-a C 1}

And if we want to find the joint probability we just need to do this:

P(N_1 = a , N_2 = b) = P(N_2 = b | N_1 = a) P(N_1 =a)

And if we multiply the probabilities founded we got:

P(N_1 = a , N_2 = b)= \frac{1}{5-a C 1} * \frac{5-a C 1}{5C2} = \frac{1}{5C2}=\frac{1}{10}

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Answer:

A) Proportion.

Step-by-step explanation:

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