Answer:
1920 in³
Step-by-step explanation:
The volume of the water can be gotten by calculation the volume of the aquarium it fills/occupies.
Since the water fills to exactly three inches below the top of the aquarium, hence the height of the water in the aquarium is given as:
height of water = 13 in - 3 in = 10 in
Therefore the volume of the water in the tank is given by the formula:
Volume of water = height of water * breadth of tank * length of tank
Volume of water = 10 in * 8 in * 24 in = 1920 in³
A=20
5a-4=3a+36 (subtract 3a from both sides and add 4 to both sides)
2a=40. Divide 40by 2
A=20
9514 1404 393
Answer:
(a, b, c) = (-0.425595, 11.7321, 2.16667)
f(x) = -0.425595x² +11.7321x +2.16667
f(1) ≈ 13.5
Step-by-step explanation:
A suitable tool makes short work of this. Most spreadsheets and graphing calculators will do quadratic regression. All you have to do is enter the data and make use of the appropriate built-in functions.
Desmos will do least-squares fitting of almost any function you want to use as a model. It tells you ...
a = -0.425595
b = 11.7321
c = 2.16667
so
f(x) = -0.425595x² +11.7321x +2.16667
and f(1) ≈ 13.5
_____
<em>Additional comment</em>
Note that a quadratic function doesn't model the data very well if you're trying to extrapolate to times outside the original domain.
Answer:
Step-by-step explanation:
Given that sample size is 130 >30. Also by central limit theorem, we know that mean (here proportion) of all means of different samples would tend to become normal with mean = average of all means(here proportions)
Hence we can assume normality assumptions here.
Proportion sample given = 92/130 = 0.7077
The mean proportion of different samples for large sample size will follow normal with mean = sample proportion and std error = square root of p(1-p)/n
Hence mean proportion p= 0.7077
q = 1-p =0.2923
Std error = 0.0399
For 95% confidence interval we find that z critical for 95% two tailed is 1,.96
Hence margin of error = + or - 1.96(std error)
= 0.0782
Confidence interval = (p-margin of error, p+margin of error)
= (0.7077-0.0782,0.7077+0.0782)
=(0.6295, 0.7859)
We are 95% confident that average of sample proportions of different samples would lie within these values in the interval for large sample sizes.
