What is the probability that a randomly chosen four-digit integer (no leading zeros) has distinct digits and ends with the digit 7? For example, 1237 qualifies for the event space but 1227 does not. All of the possible answers are rounded to four decimal digits.

a. 0.3484
b. 0.0448
c. 0.0498
d. 0.0720

c. 0.0498

Step-by-step explanation:

First of all, let us find the number of four digit numbers possible.

There are 4 digits and let us have a look at the possibility of each digit.

Number of possible options for Unit's digit = 10

Number of possible options for ten's digit = 10

Number of possible options for hundred's digit = 10

Number of possible options for thousand's digit = 9 (because 0 can not be there to make it a 4 digit number)

Total number of possible outcomes =

As per the given condition, unit's digit is 7.

So, number of possible options for unit's digit = 1

Number of possible options for thousand's digit = 8 (7 and 0 can not be there to make it a 4 digit number)

Number of possible options for hundred's digit = 8 (7 and one digit used in thousand's place can not be there)

Number of possible options for ten's digit = 7 (7, two digits used at thousand's and hundred's places)

Number of possible outcomes as per given conditions =

Formula for probability of an event E can be observed as:

first one is a: renaming a fraction. second one is c: de-c

step-by-step explanation:

my explanation is that these were the answers on edg : )

let's add {f(x)=x+1}f(x)=x+1 and {g(x)=2x}g(x)=2x together to make a new function.

f(x)+g(x) =(x+1)+(2x)=x+1+2x=3x+1

let's call this new function hh. so we have:

{h(x)}={f(x)}+{g(x)}{=3x+1}h(x)=f(x)+g(x)=3x+1

we can also evaluate combined functions for particular inputs. let's evaluate function hh above for x=2x=2. below are two ways of doing this.

method 1: substitute x=2x=2 into the combined function hh.

h(x)

h(2)

​    

=3x+1

=3(2)+1

=7

​ since h(x)=f(x)+g(x)h(x)=f(x)+g(x), we can also find h(2)h(2) by finding f(2) +g(2)f(2)+g(2).

first, let's find f(2)f(2):

f(x)

f(2)

​    

=x+1

=2+1

=3

​  

now, let's find g(2)g(2):

g(x)

g(2)

​    

=2x

=2⋅2

=4

​  

so f(2)+g(2)=3+4=\greend7f(2)+g(2)=3+4=7.

notice that substituting x =2x=2 directly into function hh and finding f(2) + g(2)f(2)+g(2) gave us the same answer!