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n200080 [17]
3 years ago
6

What is the answer for number one and possibly number two please

Mathematics
1 answer:
zloy xaker [14]3 years ago
6 0

1. Answer: 24

<u>Step-by-step explanation:</u>

There are 4 cities to choose from.  After visiting the 1st city, there are only 3 cities to choose from, etc (see below).

1st city  and   2nd city   and     3rd city   and   4th city

   4        x          3            x             2           x         1        =   24

2. Answer: 30,240

<u>Step-by-step explanation:</u>

There are 10 digits to choose from.  After choosing the first digit, there are only 9 digits to choose from (because you can't have duplicate digits in the combination) (see below).

1st digit  and  2nd digit  and  3rd digit  and  4th digit  and  5th digit

   10        x           9           x         8           x          7           x          6       = 30,240

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Number 18. Please?!!!!
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Step-by-step explanation:

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The numbers​ 1, 2,​ 3, 4, and 5 are written on slips of​ paper, and 2 slips are drawn at random one at a time without replacemen
mamaluj [8]
<h2>Answer:</h2>

(a)

The probability is :  1/2

(b)

The probability is :  1/2

<h2>Step-by-step explanation:</h2>

The numbers​ 1, 2,​ 3, 4, and 5 are written on slips of​ paper, and 2 slips are drawn at random one at a time without replacement.

The total combinations that are possible are:

(1,2)   (1,3)    (1,4)    (1,5)

(2,1)   (2,3)   (2,4)   (2,5)

(3,1)   (3,2)   (3,4)   (3,5)

(4,1)   (4,2)   (4,3)   (4,5)

(5,1)   (5,2)   (5,3)   (5,4)

i.e. the total outcomes are : 20

(a)

Let A denote the event that the first number is 4.

and B denote the event that the sum is: 9.

Let P denote the probability of an event.

We are asked to find:

               P(A|B)

We know that it could be calculated by using the formula:

P(A|B)=\dfrac{P(A\bigcap B)}{P(B)}

Hence, based on the data we have:

P(A\bigcap B)=\dfrac{1}{20}

( Since, out of a total of 20 outcomes there is just one outcome which comes in A∩B and it is:  (4,5) )

and

P(B)=\dfrac{2}{20}

( since, there are just two outcomes such that the sum is: 9

(4,5) and (5,4) )

Hence, we have:

P(A|B)=\dfrac{\dfrac{1}{20}}{\dfrac{2}{20}}\\\\i.e.\\\\P(A|B)=\dfrac{1}{2}

(b)

Let A denote the event that the first number is 3.

and B denote the event that the sum is: 8.

Let P denote the probability of an event.

We are asked to find:

               P(A|B)

Hence, based on the data we have:

P(A\bigcap B)=\dfrac{1}{20}

( since, the only outcome out of 20 outcomes is:  (3,5) )

and

P(B)=\dfrac{2}{20}

( since, there are just two outcomes such that the sum is: 8

(3,5) and (5,3) )

Hence, we have:

P(A|B)=\dfrac{\dfrac{1}{20}}{\dfrac{2}{20}}\\\\i.e.\\\\P(A|B)=\dfrac{1}{2}

7 0
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Answer:

y=ln(x/(1-x))

Step-by-step explanation:

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y(1+e^x)=e^x

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Put anything with x on with side and everything without x on opposing side:

y=e^x-ye^x

Factor right hand side

y=(1-y)e^x

Divide both sides by (1-y)

y/(1-y)=e^x

Use natural log.

ln(y/(1-y))=x

The inverse is

y=ln(x/(1-x))

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