Answer:
B 26.3% increase
Step-by-step explanation:
1 - 3.50/4.75 = .263 = 26.3%
Answer:
The 99% confidence interval for the mean sodium content in Oriental Spice Sauce is between 454.67 milligrams and 1722.61 milligrams
Step-by-step explanation:
We have the standard deviation for the sample, so we use the t-distribution to solve this question.
The first step to solve this problem is finding how many degrees of freedom, we have. This is the sample size subtracted by 1. So
df = 40 - 1 = 39
99% confidence interval
Now, we have to find a value of T, which is found looking at the t table, with 39 degrees of freedom(y-axis) and a confidence level of
. So we have T = 2.7079
The margin of error is:
M = T*s = 2.7079*234.12 = 633.97
In which s is the standard deviation of the sample.
The lower end of the interval is the sample mean subtracted by M. So it is 1088.64 - 633.97 = 454.67 milligrams
The upper end of the interval is the sample mean added to M. So it is 1088.64 + 633.97 = 1722.61 milligrams
The 99% confidence interval for the mean sodium content in Oriental Spice Sauce is between 454.67 milligrams and 1722.61 milligrams
If there are 70 chairs but only 20 rows then that means he will have to set up 3.5 chairs per row, or Brook could set up 14 rows with 5 chairs in each row.
Srry if that didn't help but that's what i got from that.
Answer:
See explanation.
General Formulas and Concepts:
<u>Pre-Algebra</u>
Order of Operations: BPEMDAS
- Brackets
- Parenthesis
- Exponents
- Multiplication
- Division
- Addition
- Subtraction
<u>Algebra I</u>
Functions
- Exponential Property [Rewrite]:

- Exponential Property [Root Rewrite]:
![\displaystyle \sqrt[n]{x} = x^{\frac{1}{n}}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Csqrt%5Bn%5D%7Bx%7D%20%3D%20x%5E%7B%5Cfrac%7B1%7D%7Bn%7D%7D)
<u>Calculus</u>
Differentiation
- Derivatives
- Derivative Notation
Derivative Property [Multiplied Constant]: ![\displaystyle \frac{d}{dx} [cf(x)] = c \cdot f'(x)](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cfrac%7Bd%7D%7Bdx%7D%20%5Bcf%28x%29%5D%20%3D%20c%20%5Ccdot%20f%27%28x%29)
Derivative Property [Addition/Subtraction]: ![\displaystyle \frac{d}{dx}[f(x) + g(x)] = \frac{d}{dx}[f(x)] + \frac{d}{dx}[g(x)]](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cfrac%7Bd%7D%7Bdx%7D%5Bf%28x%29%20%2B%20g%28x%29%5D%20%3D%20%5Cfrac%7Bd%7D%7Bdx%7D%5Bf%28x%29%5D%20%2B%20%5Cfrac%7Bd%7D%7Bdx%7D%5Bg%28x%29%5D)
Basic Power Rule:
- f(x) = cxⁿ
- f’(x) = c·nxⁿ⁻¹
Derivative Rule [Chain Rule]: ![\displaystyle \frac{d}{dx}[f(g(x))] =f'(g(x)) \cdot g'(x)](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cfrac%7Bd%7D%7Bdx%7D%5Bf%28g%28x%29%29%5D%20%3Df%27%28g%28x%29%29%20%5Ccdot%20g%27%28x%29)
Step-by-step explanation:
We are given the following and are trying to find the second derivative at <em>x</em> = 2:


We can differentiate the 1st derivative to obtain the 2nd derivative. Let's start by rewriting the 1st derivative:

When we differentiate this, we must follow the Chain Rule: ![\displaystyle \frac{d^2y}{dx^2} = \frac{d}{dx} \Big[ 6(x^2 + 3y^2)^\big{\frac{1}{2}} \Big] \cdot \frac{d}{dx} \Big[ (x^2 + 3y^2) \Big]](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cfrac%7Bd%5E2y%7D%7Bdx%5E2%7D%20%3D%20%5Cfrac%7Bd%7D%7Bdx%7D%20%5CBig%5B%206%28x%5E2%20%2B%203y%5E2%29%5E%5Cbig%7B%5Cfrac%7B1%7D%7B2%7D%7D%20%5CBig%5D%20%5Ccdot%20%5Cfrac%7Bd%7D%7Bdx%7D%20%5CBig%5B%20%28x%5E2%20%2B%203y%5E2%29%20%5CBig%5D)
Use the Basic Power Rule:

We know that y' is the notation for the 1st derivative. Substitute in the 1st derivative equation:
![\displaystyle \frac{d^2y}{dx^2} = 3(x^2 + 3y^2)^\big{\frac{-1}{2}} \big[ 2x + 6y(6\sqrt{x^2 + 3y^2}) \big]](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cfrac%7Bd%5E2y%7D%7Bdx%5E2%7D%20%3D%203%28x%5E2%20%2B%203y%5E2%29%5E%5Cbig%7B%5Cfrac%7B-1%7D%7B2%7D%7D%20%5Cbig%5B%202x%20%2B%206y%286%5Csqrt%7Bx%5E2%20%2B%203y%5E2%7D%29%20%5Cbig%5D)
Simplifying it, we have:
![\displaystyle \frac{d^2y}{dx^2} = 3(x^2 + 3y^2)^\big{\frac{-1}{2}} \big[ 2x + 36y\sqrt{x^2 + 3y^2} \big]](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cfrac%7Bd%5E2y%7D%7Bdx%5E2%7D%20%3D%203%28x%5E2%20%2B%203y%5E2%29%5E%5Cbig%7B%5Cfrac%7B-1%7D%7B2%7D%7D%20%5Cbig%5B%202x%20%2B%2036y%5Csqrt%7Bx%5E2%20%2B%203y%5E2%7D%20%5Cbig%5D)
We can rewrite the 2nd derivative using exponential rules:
![\displaystyle \frac{d^2y}{dx^2} = \frac{3\big[ 2x + 36y\sqrt{x^2 + 3y^2} \big]}{\sqrt{x^2 + 3y^2}}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cfrac%7Bd%5E2y%7D%7Bdx%5E2%7D%20%3D%20%5Cfrac%7B3%5Cbig%5B%202x%20%2B%2036y%5Csqrt%7Bx%5E2%20%2B%203y%5E2%7D%20%5Cbig%5D%7D%7B%5Csqrt%7Bx%5E2%20%2B%203y%5E2%7D%7D)
To evaluate the 2nd derivative at <em>x</em> = 2, simply substitute in <em>x</em> = 2 and the value f(2) = 2 into it:
![\displaystyle \frac{d^2y}{dx^2} \bigg| \limits_{x = 2} = \frac{3\big[ 2(2) + 36(2)\sqrt{2^2 + 3(2)^2} \big]}{\sqrt{2^2 + 3(2)^2}}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cfrac%7Bd%5E2y%7D%7Bdx%5E2%7D%20%5Cbigg%7C%20%5Climits_%7Bx%20%3D%202%7D%20%3D%20%5Cfrac%7B3%5Cbig%5B%202%282%29%20%2B%2036%282%29%5Csqrt%7B2%5E2%20%2B%203%282%29%5E2%7D%20%5Cbig%5D%7D%7B%5Csqrt%7B2%5E2%20%2B%203%282%29%5E2%7D%7D)
When we evaluate this using order of operations, we should obtain our answer:

Topic: AP Calculus AB/BC (Calculus I/I + II)
Unit: Differentiation
Answer: $24.50
Post office:£1 = $1.27
£350 = $1.27 × 350 = $444.50
Bank:£1 = $1.34
£350 = $1.34 × 350 = $469
Therefore, the difference will be:
= $469 - $444.50
= $24.50