Answer:
There is 95% confidence that the population proportion of people from Vermont who exercised for at least 30 minutes a day 3 days a week is between 55.9% and 74.7%.
Step-by-step explanation:
We have to answer the population proportion for Vermont.
We can only do it by a confidence interval, as we only have information from a sample.
This sample, of size n=100, has a proportion p=0.653.
The degrees of freedom are:
We will calculate a 95% confidence interval, which for df=99 has a critical value of t of t=1.984.
The margin of error can be calculated as:
Then, the upper and lower bounds of the confidence interval are:
Then, we can say that there is 95% confidence that the population proportion of people from Vermont who exercised for at least 30 minutes a day 3 days a week is between 55.9% and 74.7%.
There will be 2 cubes left over, because 14X7 is 98 and you can't go any higher than that.
Let p(x) be a polynomial, and suppose that a is any real
number. Prove that
lim x→a p(x) = p(a) .
Solution. Notice that
2(−1)4 − 3(−1)3 − 4(−1)2 − (−1) − 1 = 1 .
So x − (−1) must divide 2x^4 − 3x^3 − 4x^2 − x − 2. Do polynomial
long division to get 2x^4 − 3x^3 − 4x^2 – x – 2 / (x − (−1)) = 2x^3 − 5x^2 + x –
2.
Let ε > 0. Set δ = min{ ε/40 , 1}. Let x be a real number
such that 0 < |x−(−1)| < δ. Then |x + 1| < ε/40 . Also, |x + 1| <
1, so −2 < x < 0. In particular |x| < 2. So
|2x^3 − 5x^2 + x − 2| ≤ |2x^3 | + | − 5x^2 | + |x| + | − 2|
= 2|x|^3 + 5|x|^2 + |x| + 2
< 2(2)^3 + 5(2)^2 + (2) + 2
= 40
Thus, |2x^4 − 3x^3 − 4x^2 − x − 2| = |x + 1| · |2x^3 − 5x^2
+ x − 2| < ε/40 · 40 = ε.
Answer:
40.1%
Step-by-step explanation:
I am assuming that 192 is in 100%.
100% = 192
I then represent the value that we are looking for with 
.
x% = 77
By dividing both equations (100% = 192 and x% = 77) and remembering that both left hand sides of BOTH equations have the percentage unit (%).
Now, of course, we take the reciprocal, or inverse, of both sides:
x = 40.1%
Thus making the answer: 40.1% of 192 is 77.
