MATHEMATICS COLLEGE

Q 3.28: To create a confidence interval from a bootstrap distribution using percentiles, we keep the middle values and chop off a certain percent from each tail. Indicate what percent of values must be chopped off from each tail for a 97% confidence level. A : We keep the middle 97% of values by chopping off 1.5% from each tail. B : We keep the middle 1.5% of values by chopping off 97% from each tail. C : We keep the middle 3% of values by chopping off 97% from each tail. D : We keep the middle 3% of values by chopping off 1.5% from each tail. E : We keep the middle 97% of values by chopping off 3% from each tail.

Answers

Answer 1

Answer:

Answer:

A : We keep the middle 97% of values by chopping off 1.5% from each tail.

Step-by-step explanation:

We have that to find our $\alpha$ level, that is the subtraction of 1 by the confidence interval divided by 2. So:

$\alpha = (1-0.97)/(2) = 0.015$

This means that for a 97% confidence interval, 1.5% of each tail is removed, while the middle 97% of values are kept.

So the corect answer is:

A : We keep the middle 97% of values by chopping off 1.5% from each tail.

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Use the product rule to calculate the derivatives of
( ax² + bx + c ) ( cx + d )

Answers

$\n \sf\longmapsto (d)/(dx)(ax^2+bx+c)(cx+d)$

$\boxed{\sf (d)/(dx)f(x).g(x)=f(x)(d)/(dx)g(x)+g(x)(d)/(dx)f(x)}$

c and d are constants

$\n \sf\longmapsto (ax^2+bx+c)(d)/(dx)(cx+d)+(cx+d)(d)/(dx)(ax^2+bx+c)$

$\n \sf\longmapsto (ax^2+bx+c)(c)+(cx+d)(2ax+b)$

$\n \sf\longmapsto acx^2+bcx+c^2+2acx^2+bcx+2adx+bd$

$\n \sf\longmapsto 3acx^2+2bcx+2adx+bd+c^2$

Answer:

• Product rule is as below:

${ \boxed{ \tt{ \: (dy)/(dx) = { \huge{v}} (du)/(dx) + { \huge{u}} (dv)/(dx) }}} \n$

u is (ax² + bx + c)
v is (cx + d)
du/dx is 2ax + bx
dv/dx is c

$\hookrightarrow \: { \rm{ (dy)/(dx) = (cx + d)(2ax + b) + (ax {}^(2) + bx + c)(c) }} \n \n { \rm{ (dy)/(dx) = (2ac {x}^(2) + bcx + 2adx + db) + (ac {x}^(2) + bcx + {c}^(2) )}} \n \n { \boxed{ \rm{ (dy)/(dx) = 3ac {x}^(2) + \{2bcx + 2adx \}x + (db + {c}^(2)) }}}$

Pleeease open the image and hellllp me

Answers

1. Rewrite the expression in terms of logarithms:

$y=x^x=e^(\ln x^x)=e^(x\ln x)$

Then differentiate with the chain rule (I'll use prime notation to save space; that is, the derivative of y is denoted y' )

$y'=e^(x\ln x)(x\ln x)'=x^x(x\ln x)'$

$y'=x^x(x'\ln x+x(\ln x)')$

$y'=x^x\left(\ln x+\frac xx\right)$

$y'=x^x(\ln x+1)$

2. Chain rule:

$y=\ln(\csc(3x))$

$y'=\frac1{\csc(3x)}(\csc(3x))'$

$y'=\sin(3x)\left(-\cot^2(3x)(3x)'\right)$

$y'=-3\sin(3x)\cot^2(3x)$

Since $\cot x=(\cos x)/(\sin x)$ , we can cancel one factor of sine:

$y'=-3(\cos^2(3x))/(\sin(3x))=-3\cos(3x)\cot(3x)$

3. Chain rule:

$y=e^{e^(\sin x)}$

$y'=e^{e^(\sin x)}\left(e^(\sin x)\right)'$

$y'=e^{e^(\sin x)}e^(\sin x)(\sin x)'$

$y'=e^{e^(\sin x)+\sin x}\cos x$

4. If you're like me and don't remember the rule for differentiating logarithms of bases not equal to e, you can use the change-of-base formula first:

$\log_2x=(\ln x)/(\ln2)$

Then

$(\log_2x)'=\left((\ln x)/(\ln 2)\right)'=\frac1{\ln 2}$

So we have

$y=\cos^2(\log_2x)$

$y'=2\cos(\log_2x)\left(\cos(\log_2x)\right)'$

$y'=2\cos(\log_2x)(-\sin(\log_2x))(\log_2x)'$

$y'=-\frac2{\ln2}\cos(\log_2x)\sin(\log_2x)$

and we can use the double angle identity and logarithm properties to condense this result:

$y'=-\frac1{\ln2}\sin(2\log_2x)=-\frac1{\ln2}\sin(\log_2x^2)$

5. Differentiate both sides:

$\left(x^2-y^2+\sin x\,e^y+\ln y\,x\right)'=0'$

$2x-2yy'+\cos x\,e^y+\sin x\,e^yy'+\frac{xy'}y+\ln y=0$

$-\left(2y-\sin x\,e^y-\frac xy\right)y'=-\left(2x+\cos x\,e^y+\ln y\right)$

$y'=(2x+\cos x\,e^y\ln y)/(2y-\sin x\,e^y-\frac xy)$

$y'=(2xy+\cos x\,ye^y\ln y)/(2y^2-\sin x\,ye^y-x)$

6. Same as with (5):

$\left(\sin(x^2+\tan y)+e^(x^3\sec y)+2x-y+2\right)'=0'$

$\cos(x^2+\tan y)(x^2+\tan y)'+e^(x^3\sec y)(x^3\sec y)'+2-y'=0$

$\cos(x^2+\tan y)(2x+\sec^2y y')+e^(x^3\sec y)(3x^2\sec y+x^3\sec y\tan y\,y')+2-y'=0$

$\left(\cos(x^2+\tan y)\sec^2y+x^3\sec y\tan y\,e^(x^3\sec y)-1\right)y'=-\left(2x\cos(x^2+\tan y)+3x^2\sec y\,e^(x^3\sec y)+2\right)$

$y'=-(2x\cos(x^2+\tan y)+3x^2\sec y\,e^(x^3\sec y)+2)/(\cos(x^2+\tan y)\sec^2y+x^3\sec y\tan y\,e^(x^3\sec y)-1)$

7. Looks like

$y=x^2-e^(2x)$

Compute the second derivative:

$y'=2x-2e^(2x)$

$y''=2-4e^(2x)$

Set this equal to 0 and solve for x :

$2-4e^(2x)=0$

$4e^(2x)=2$

$e^(2x)=\frac12$

$2x=\ln\frac12=-\ln2$

$x=-\frac{\ln2}2$

16. The original price of a tie is $12.50. The new price of the tie is now$7.50. By what percentage was the tie marked down? *

Answers

The tie was marked down by 40%

PLSS I NEED HELP ASAP BECAUSE ITS DUE SOONNick has to build a brick wall. Each row of the wall requires 62 bricks. There are 10 rows in the wall. How many bricks will Nick require to build the wall?
A.
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B.
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C.
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D.
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Answers

D. 10 x 62

If one row needs 62 bricks and there's 10 rows, you multiply them to get how many bricks nick needs

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Answers

Answer:

Step-by-step explanation:

Answer the following...

Answers

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