Answer:
And the best option would be:
c. 1450 +/- 12
Step-by-step explanation:
Information provided
represent the sample mean for the SAT scores
population mean (variable of interest)
represent the sample variance given
n=25 represent the sample size
Solution
The confidence interval for the true mean is given by :
(1)
The sample deviation would be 
The degrees of freedom are given by:
The Confidence is 0.954 or 95.4%, the value of 
and 
, assuming that we can use the normal distribution in order to find the quantile the critical value would be 
The confidence interval would be
And the best option would be:
c. 1450 +/- 12
They gave you an eqution with variables and then they gave you numbers that all you have to do is put in
k(h-j)+h
5(5-4)+5
5(1)+5
5+5
10
Answer:
One convergence criteria that is useful here is that, if aₙ is the n-th term of this sequence, then we must have:
Iaₙ₊₁I < IaₙI
This means that the absolute value of the terms must decrease as n increases.
Then we must have:
We can write this as:
If we assume that n is a really big number, then:
n + 1 ≈ 1
And we can write:
Then we have the inequality
And remember that this must be in absolute value, then we will have that:
-1 < (x - 2)/3 < 1
-3 < x - 2 < 3
-3 + 2 < x < 3 + 2
-1 < x < 5
The first option looks like this, but it uses the symbols ≤≥, so it is not the same as this, then the correct option will be the second.
Answer:
Using either method, we obtain: 
Step-by-step explanation:
a) By evaluating the integral:
![\frac{d}{dt} \int\limits^t_0 {\sqrt[8]{u^3} } \, du](https://tex.z-dn.net/?f=%5Cfrac%7Bd%7D%7Bdt%7D%20%5Cint%5Climits%5Et_0%20%7B%5Csqrt%5B8%5D%7Bu%5E3%7D%20%7D%20%5C%2C%20du)
The integral itself can be evaluated by writing the root and exponent of the variable u as: ![\sqrt[8]{u^3} =u^{\frac{3}{8}](https://tex.z-dn.net/?f=%5Csqrt%5B8%5D%7Bu%5E3%7D%20%3Du%5E%7B%5Cfrac%7B3%7D%7B8%7D)
Then, an antiderivative of this is: 
which evaluated between the limits of integration gives:
and now the derivative of this expression with respect to "t" is:
b) by differentiating the integral directly: We use Part 1 of the Fundamental Theorem of Calculus which states:
"If f is continuous on [a,b] then
is continuous on [a,b], differentiable on (a,b) and 
Since this this function 
is continuous starting at zero, and differentiable on values larger than zero, then we can apply the theorem. That means:
Answer:
<h2>The required answer is 218 degree.</h2>
Step-by-step explanation:
∠AOB = 90 degree.
∠BOC = 52 degree.
Arc CDE = 180 degree, since CE is the diameter.
Hence, Arc EAC = 180 degree.
Besides, Arc EAC = Arc EA + Arc AB + Arc BC = Arc EA + 90 + 52 = Arc EA + 142.
Thus, Arc EA = 180 - 142 = 38 degree.
Arc ADC = 180 + 38 = 218 degree.
