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andrew-mc [135]

3 years ago

15

(1.1/1.2: Interpolating polynomials) Say we want to find a polynomialf(x) ofdegree 3,f(x) =a0+a1x+a2x2+a3x3,satisfying some inte

rpolation conditions. In each case below, write a system of linearequations whose solutions are (a0,a1,a2,a3). You don’t need to solve the system.(a) We wantf(x) to pass through the points (−1,−1),(1,2),(2,1) and (3,5).(b) We wantf(x) to pass through (1,0) with derivative +2 and (2,3) withderivative−1

1 answer:

Hunter-Best [27]3 years ago

7 0

(a) If

<em>f(x)</em> = <em>a</em>₀ + <em>a</em>₁ <em>x </em>+<em> a</em>₂ <em>x</em> ² + <em>a</em>₃ <em>x</em> ³

then from the given conditions we get the system of equations,

<em>f</em> (-1) = <em>a</em>₀ - <em>a</em>₁<em> </em>+<em> a</em>₂ - <em>a</em>₃ = -1

<em>f</em> (1) = <em>a</em>₀ + <em>a</em>₁<em> </em>+<em> a</em>₂ + <em>a</em>₃ = 2

<em>f</em> (2) = <em>a</em>₀ + 2<em>a</em>₁<em> </em>+ 4<em>a</em>₂ + 8<em>a</em>₃ = 1

<em>f</em> (3) = <em>a</em>₀ + 3<em>a</em>₁<em> </em>+<em> </em>9<em>a</em>₂ + 27<em>x</em> ³ = 5

(b) Similarly, if

<em>f(x)</em> = <em>a</em>₀ + <em>a</em>₁ <em>x </em>+<em> a</em>₂ <em>x</em> ² + <em>a</em>₃ <em>x</em> ³

then

<em>f'(x)</em> = <em>a</em>₁<em> </em>+<em> </em>2<em>a</em>₂ <em>x</em> + 3<em>a</em>₃ <em>x</em> ²

so that the given conditions yield the system,

<em>f</em> (1) = <em>a</em>₀ + <em>a</em>₁<em> </em>+<em> a</em>₂ + <em>a</em>₃ = 0

<em>f'</em> (1) = <em>a</em>₁<em> </em>+<em> </em>2<em>a</em>₂ + 3<em>a</em>₃ = 2

<em>f</em> (2) = <em>a</em>₀ + 2<em>a</em>₁<em> </em>+<em> </em>4<em>a</em>₂ + 27<em>a</em>₃ = 3

<em>f'</em> (2) = <em>a</em>₁<em> </em>+<em> </em>4<em>a</em>₂ + 12<em>a</em>₃ = -1

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What is the form of the two squares identity?

zvonat [6]

Answer:

D

Step-by-step explanation:

Consider all options:

A. False, because

$(a^2+b^2)(c^2+d^2)=a^2c^2+a^2d^2+b^2c^2+b^2d^2\\ \\(ab-cd)^2+(ac+bd)^2=a^2b^2-2abcd+c^2d^2+a^2c^2+2abcd+b^2d^2=\\ =a^2b^2+c^2d^2+a^2c^2+b^2d^2\\ \\a^2c^2+a^2d^2+b^2c^2+b^2d^2\neq a^2b^2+c^2d^2+a^2c^2+b^2d^2$

B. False, because

$(a^2-b^2)(c^2+d^2)=a^2c^2+a^2d^2-b^2c^2-b^2d^2\\ \\(ac-bd)^2-(ad+bc)^2=a^2c^2-2abcd+b^2d^2+a^2d^2+2abcd+b^2c^2=\\=a^2c^2+b^2d^2+a^2d^2+b^2c^2\\ \\a^2c^2+a^2d^2-b^2c^2-b^2d^2\neq a^2c^2+b^2d^2+a^2d^2+b^2c^2$

C. False, because

$(a^2+b^2)(c^2-d^2)=a^2c^2-a^2d^2+b^2c^2-b^2d^2\\ \\(ac+bd)^2-(ad+bc)^2=a^2c^2+2abcd+b^2d^2-a^2d^2-2abcd-b^2c^2=\\=a^2c^2+b^2d^2-a^2d^2-b^2c^2\\ \\a^2c^2-a^2d^2+b^2c^2-b^2d^2\neq a^2c^2+b^2d^2-a^2d^2-b^2c^2$

D. True, because

$(a^2+b^2)(c^2+d^2)=a^2c^2+a^2d^2+b^2c^2+b^2d^2\\ \\(ac-bd)^2-(ad+bc)^2=a^2c^2-2abcd+b^2d^2+a^2d^2+2abcd+b^2c^2=\\=a^2c^2+b^2d^2+a^2d^2+b^2c^2\\ \\a^2c^2+a^2d^2+b^2c^2+b^2d^2= a^2c^2+b^2d^2+a^2d^2+b^2c^2$

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3 years ago

a particular truck in a transportation fleet was driven 194,070 miles last year versus 329,919 miles this year . by what percent

Ann [662]

We take as the 100% the 194070 miles from last year. Then, we apply a rule of three:

$\begin{gathered} 194070\rightarrow100\text{ \%} \\ 329919\rightarrow x\text{ \%} \\ \Rightarrow x=\frac{329919\cdot100}{194070}=\frac{32991900}{194070}=170 \\ x=170\text{ \%} \end{gathered}$

Since 329,919 is the 170% of last year's miles, then the increase was of 70%

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1 year ago

Cheryl collected data for her mathematics project. She noted that the data set was approximately normal.

nadya68 [22]

Answer:

$X_(r) >> X_(n)$

The mean for this case would increase since is defined as:

$\bar X= \frac{\sum_{i=1}^n X_i}{n}$

The interquartile range would not change since the definition for the IQR is $IQR =Q_3 -Q_1$ and the quartiles are the same.

The standard deviation would not remain the same since by definition is:

$s = \sqrt{\frac{\sum_{i=1}^n (X_i -\bar X)^2}{n-1}}$

And since we change the largest value the deviation would increase considerably.

And for the last option is not always true since if we select a value so much higher then the distribution would be skewed to the right.

So the best option for this case is:

Mean would increase.

Step-by-step explanation:

For this case we assume that we have a random sample given $X_(1), X_(2) ,..., X_(n)$ and for each observation $X_i \sim N(\mu, \sigma)$ since the problem states that the data is approximately normal.

Let's assume that the largest value on this sample is $X_(n)$ and for this case we are going to replace this value by another one extremely higher so we satisfy this condition:

$X_(r) >> X_(n)$

The mean for this case would increase since is defined as:

$\bar X= \frac{\sum_{i=1}^n X_i}{n}$

The interquartile range would not change since the definition for the IQR is $IQR =Q_3 -Q_1$ and the quartiles are the same.

The standard deviation would not remain the same since by definition is:

$s = \sqrt{\frac{\sum_{i=1}^n (X_i -\bar X)^2}{n-1}}$

And since we change the largest value the deviation would increase considerably.

And for the last option is not always true since if we select a value so much higher then the distribution would be skewed to the right.

So the best option for this case is:

Mean would increase.

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3 years ago