Out of 1,200 students, 700 are boys. Therefore, the fraction would be
.
To convert this into a percentage, divide 700 by 1,200.
= 0.583333 (or just 0.583 if you round)
Multiply 0.583
by 100 to get a percentage.
0.583 · 100 =
58.3So, the percent of boys at the school is
58.3%.
A good place to start is to set 
to y. That would mean we are looking for 
to be an integer. Clearly, 
, because if y were greater the part under the radical would be a negative, making the radical an imaginary number, not an integer. Also note that since 
is a radical, it only outputs values from ![[0,\infty]](https://tex.z-dn.net/?f=%20%5B0%2C%5Cinfty%5D%20)
, which means y is on the closed interval: ![[0,120]](https://tex.z-dn.net/?f=%20%5B0%2C120%5D%20)
.
With that, we don't really have to consider y anymore, since we know the interval that is on.
Now, we don't even have to find the x values. Note that only 11 perfect squares lie on the interval , which means there are at most 11 numbers that x can be which make the radical an integer. All of the perfect squares are easily constructed. We can say that if k is an arbitrary integer between 0 and 11 then:
Which is strictly positive so we know for sure that all 11 numbers on the closed interval will yield a valid x that makes the radical an integer.
Answer:
1000 gallons of tank
Step-by-step explanation:
I believe the answer is D.
3^2 is 9 and 4^2 is 16, and because 9/16 was the third term in the sequence, you raise both the numerator and the denominator by the number of the term minus 1.
So to find the seventh term, you raise the numerator and denominator by 7 - 1 = 6.
So the answer is 729/4096
The statement that is most likely true is that the median is in the 6-10 interval and the mean is in the 6-10
There were a total of 30 different pieces of data collected, so the 15th and 16th pieces of data that would create the median. Both the 15th and 16th numbers would be in the 6 - 10.
If you find the average of the middle data point in each interval the mean would be approximately 7.2. This is in the 2nd interval (6-10).
