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agasfer [191]
2 years ago
15

Figure B is a scaled copy of Figure A What is the scale factor from Figure A to Figureb

Mathematics
2 answers:
Rus_ich [418]2 years ago
7 0

Answer:

Below.

Step-by-step explanation:

About x3 bigger then figure b.

Arturiano [62]2 years ago
7 0
<h3>Explanation:</h3>
  • We already know that Figure A is the pre-image and Figure B is the image. Since the image is smaller than the pre-image, the scale factor will be less than 1.
  • We also know that the pre-image is three times greater than the image. Since the scale factor is less than 1, 3 will go in the denominator.

The scale factor from Figure A to Figure B is 1/3.

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Five men and 5 women are ranked according to their scores on an examination. Assume that no two scores are alike and all 10! pos
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Answer:

P(X=1)=\frac{1}{2}\\P(X=2)=\frac{5}{18}\\P(X=3)=\frac{5}{36}\\P(X=4)=\frac{5}{84}\\P(X=5)=\frac{5}{252}\\P(X=6)=\frac{1}{252} \\P(X>6)=0

Step-by-step explanation:

Let's define the following event :

X : ''Highest ranking achieved by a woman''

There are five men and five women so the lowest ranking possible is 6, this means that all five men achieved a better ranking than all five women.

In terms of the variable :

X can assume the following values :

X=1\\X=2\\X=3\\X=4\\X=5\\X=6

⇒ P(X=7)=0\\P(X=8)=0\\P(X=9)=0\\P(X=10)=0

⇒ P(X>6)=0

Now let's calculate the probability function for X

P(X=1)=\frac{(5)(9!)}{(10!)}=0.5=\frac{1}{2}

We write on the denominator the total ways to arrange ten different persons in a line. This number is 10!

On the numerator we write the total cases in which the first person is a woman ⇒ (5).(9!)

Because they are five different women and when you pick one of then they are 9! ways to arrange the 9 remaining persons

Following this logic, we continue calculating the probability function :

P(X=2)=\frac{(5)(5)(8!)}{(10!)}=\frac{5}{18}

Again : Five different ways to choose between the five men in the first position.Then, 5 different ways to choose between the five women.And finally, 8! ways to arrange the remaining persons

P(X=3)=\frac{(5)(4)(5)(7!)}{(10!)}=\frac{5}{36}

Here we have : five different ways to choose between five men.Four different ways to choose between the four remaining men.Five different ways to choose between the five women for the third position and finally 7! ways to arrange the remaining persons.

P(X=4)=\frac{(5)(4)(3)(5)(6!)}{(10!)}=\frac{5}{84}

P(X=5)=\frac{(5)(4)(3)(2)(5)(5!)}{(10!)}=\frac{5}{252}

P(X=6)=\frac{(5)(4)(3)(2)(1)(5)(4!)}{(10!)}=\frac{1}{252}

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