A histogram titled Number of texts has time on the x-axis and texts on the y-axis. From 6 a m to 7:59 a m there were 15 texts, 8 to 9:59 a m: 5, 10 a m to 11:59 p m: 0, 12 p m to 1:59 p m: 0, 2 p m to 3:59 p m: 0, 4 to 5:59 p m: 29, 6 to 7:59 p m: 19, 8 to 8:59 p m, 14, 10 p m to 11:59 a m: 5 The histogram shows the number of text messages sent by two high school juniors on one Monday. Which statement most reasonably explains the hours when 0 texts were sent? The students were asleep and did not have phones turned on. The students only had 100 text messages available. The students were in school and were not allowed to text. Texting is done in clusters.

The students were in school and were not allowed to text

The time-span from 10 AM to 3:59 PM is when 0 messages were sent in total. This is likely the duration of either the entire school day or much of it. So it's likely the school does not allow texting or does not allow phones to be in use.

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!

b) 1/3 * 1/5

step-by-step explanation:

each smallest part is 1/15 since the written one is 1/5 and it's divided by 3 equal parts. (1/5 )/3 = 1/15 = 1/3 * 1/5