Answer:
There are 9 teachers in the auditorium.
Step-by-step explanation:
81*1/9=9
-5 > -6
I think this is what it would be because -5 is greater than (>) -6
Hope this is right and it helps :)
The pencil costs $0.25.
One way to deduce this is to divide 1.50 by 2 (equaling 0.75) and then subtracting .5 to get 0.25 because if it's $1 more, then the ruler would cost 0.75 + 0.5 and the pencil would cost 0.75 - 0.5, making a total of $1.50.
Another way is to make it an equation. Let's say the ruler is r dollars and the pencil is p dollars.
r + p = 1.50
r = p + 1.00
If we know these two equations, then you can substitute p + 1.00 in the 1st equation (instead of r). This gets us:
p + 1.00 + p = 1.50
You can simplify this into:
2p + 1.00 = 1.50
Then you subtract 1.00 from both sides:
2p = 0.50
And you divide 2 from both sides:
p = 0.25
Getting you the answer of:
The pencil costs $0.25
Hope this helps! :)
Given that:
mean,μ=35.6 min
std deviation,σ=10.3 min
we are required to find the value of x such that 22.96% of the 60 days have a travel time that is at least x.
using z-table, the z-score that will give us 0.2296 is:-1.99
therefore:
z-score is given by:
(x-μ)/σ
hence:
-1.99=(x-35.6)/10.3
-20.497=x-35.6
x=35.6-20.497
x=15.103
Answer:
A Pipe that is 120 cm long resonates to produce sound of wavelengths 480 cm, 160 cm and 96 cm but does not resonate at any wavelengths longer than these. This pipe is:
A. closed at both ends
B. open at one end and closed at one end
C. open at both ends.
D. we cannot tell because we do not know the frequency of the sound.
The right choice is:
B. open at one end and closed at one end
.
Step-by-step explanation:
Given:
Length of the pipe,
= 120 cm
Its wavelength
= 480 cm
= 160 cm and
= 96 cm
We have to find whether the pipe is open,closed or open-closed or none.
Note:
- The fundamental wavelength of a pipe which is open at both ends is 2L.
- The fundamental wavelength of a pipe which is closed at one end and open at another end is 4L.
So,
The fundamental wavelength:
⇒ 
It seems that the pipe is open at one end and closed at one end.
Now lets check with the subsequent wavelengths.
For one side open and one side closed pipe:
An odd-integer number of quarter wavelength have to fit into the tube of length L.
⇒
⇒ 
⇒
⇒ 
⇒
⇒ 
⇒
⇒
So the pipe is open at one end and closed at one end
.