Quadratic function ( in standard form ):
y = a x² + b x + c
x = 0:
635 = 0 + 0 + c
c = 635
x = 1:
644 = a + b + 635
a + b = 9
x = 2:
719 = 4 a + 2 b + 635
4 a + 2 b = 84
and now we have a system:
a + b = 9 / * ( - 2 )
4 a + 2 b = 84
-------------------------
- 2 a + 2 b = - 18
+
4 a + 2 b = 84
-------------------------
2 a = 66
a = 33; b = - 24;
Function is:
f ( x ) = 33 x² - 24 x + 635
f ( 8 ) = 33 * 64 - 24 * 8 + 635 = 2,112 - 192 + 635 = 2,555
Answer: The number of waterfowl on week 8 would be 2,555.
Answer:
C) The third table. K= 5

Step-by-step explanation:
1) Below there are the missing data:
A proportional relationship through a constant k. It is obtained when we divide:

2)In this case, when we divide the temperature (T) by the time (h).

3)So, examining the table below we are searching for a ratio k common to all measures (temperatures over hour).

Difference between 25 and 35 = 35 -25
= 10
Then
percentage by which 25 is less than 35 =(10/35) * 100
= (2/7) * 100
= 200/7
= 28.57 percent
So 25 is 28.57% less than 35.This is the only clear way of solving these kind of problems. I hope you
have understood the method used to solve this problem. Hopefully you
can do such type of problems without needing any help.
We assume the lunch prices we observe are drawn from a normal distribution with true mean
and standard deviation 0.68 in dollars.
We average
samples to get
.
The standard deviation of the average (an experiment where we collect 45 samples and average them) is the square root of n times smaller than than the standard deviation of the individual samples. We'll write

Our goal is to come up with a confidence interval (a,b) that we can be 90% sure contains
.
Our interval takes the form of
as
is our best guess at the middle of the interval. We have to find the z that gives us 90% of the area of the bell in the "middle".
Since we're given the standard deviation of the true distribution we don't need a t distribution or anything like that. n=45 is big enough (more than 30 or so) that we can substitute the normal distribution for the t distribution anyway.
Usually the questioner is nice enough to ask for a 95% confidence interval, which by the 68-95-99.7 rule is plus or minus two sigma. Here it's a bit less; we have to look it up.
With the right table or computer we find z that corresponds to a probability p=.90 the integral of the unit normal from -z to z. Unfortunately these tables come in various flavors and we have to convert the probability to suit. Sometimes that's a one sided probability from zero to z. That would be an area aka probability of 0.45 from 0 to z (the "body") or a probability of 0.05 from z to infinity (the "tail"). Often the table is the integral of the bell from -infinity to positive z, so we'd have to find p=0.95 in that table. We know that the answer would be z=2 if our original p had been 95% so we expect a number a bit less than 2, a smaller number of standard deviations to include a bit less of the probability.
We find z=1.65 in the typical table has p=.95 from -infinity to z. So our 90% confidence interval is

in other words a margin of error of
dollars
That's around plus or minus 17 cents.
Answer:
11+7x
Step-by-step explanation:
Simplifying
9x + 10 + -2x + 1
Reorder the terms:
10 + 1 + 9x + -2x
Combine like terms: 10 + 1 = 11
11 + 9x + -2x
Combine like terms: 9x + -2x = 7x
11 + 7x
Final result:
11 + 7x