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Answer:

The 95 percent confidence interval for the true mean metal thickness is between 0.2903 mm and 0.2907 mm

Step-by-step explanation:

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

[tex]\alpha = \frac{1-0.95}{2} = 0.025[/tex]

Now, we have to find z in the Ztable as such z has a pvalue of [tex]1-\alpha[/tex].

So it is z with a pvalue of [tex]1-0.025 = 0.975[/tex], so [tex]z = 1.96[/tex]

Now, find the margin of error M as such

[tex]M = z*\frac{\sigma}{\sqrt{n}}[/tex]

In which [tex]\sigma[/tex] is the standard deviation of the population and n is the size of the sample.

[tex]M = 1.96\frac{0.000586}{\sqrt{59}} = 0.0002[/tex]

The lower end of the interval is the sample mean subtracted by M. So it is 0.2905 - 0.0002 = 0.2903 mm

The upper end of the interval is the sample mean added to M. So it is 0.2905 + 0.0002 = 0.2907 mm

The 95 percent confidence interval for the true mean metal thickness is between 0.2903 mm and 0.2907 mm

Answer:

√75 = 5√3 and √12 = 2√3 so √75 + √12 = 5√3 + 2√3 = 7√3.

Answer:

[tex] 7\sqrt{3} [/tex]

Step-by-step explanation:

[tex] \sqrt{12} \: can \: be \: simplified \: as \: 2 \sqrt{3} \: and \: \sqrt{75} \: canbe \: simplified \: as \: 5 \sqrt{3} \\ after \: simplifying \: we \: can \: add \: them \: up \\ 2 \sqrt{3} + 5 \sqrt{3} = 7 \sqrt{3} [/tex]

Answer:

Step-by-step explanation:

a) If we assume the annual withdrawals are at the beginning of the year, we can use the formula for an annuity due to compute the necessary savings.

The principal P that must be invested at rate r for n annual withdrawals of amount A is ...

  P = A(1+r)(1 -(1 +r)^-n)/r

  P = $60,000(1.08)(1 -1.08^-20)/0.08 = $636,215.95

__

b) 20 withdrawals of $60,000 each total ...

  20×$60,000 = $1,200,000

__

c) The excess over the amount deposited is interest:

  $1,200,000 -636,215.95 = $563,784.05

Answer:

6

Step-by-step explanation:

nerd physics

Answer:

n = 5

Step-by-step explanation:

Since this is a right triangle, we can use trig functions

tan theta = opp/ adj

tan 30 = n/ 5 sqrt(3)

5 sqrt(3) tan 30 = n

5 sqrt(3) * 1/ sqrt(3) = n

5 = n

Answer:

The curvature is modelled by [tex]\kappa = \frac{e^{-t}}{6\sqrt{2}}[/tex].

Step-by-step explanation:

The equation of the curvature is:

[tex]\kappa = \frac{|\dot {x}\cdot \ddot {y}-\dot{y}\cdot \ddot{x}|}{[\dot{x}^{2}+\dot{y}^{2}]^{\frac{3}{2} }}[/tex]

The parametric componentes of the curve are:

[tex]x = 6\cdot e^{t} \cdot \cos t[/tex] and [tex]y = 6\cdot e^{t}\cdot \sin t[/tex]

The first and second derivative associated to each component are determined by differentiation rules:

First derivative

[tex]\dot{x} = 6\cdot e^{t}\cdot \cos t - 6\cdot e^{t}\cdot \sin t[/tex] and [tex]\dot {y} = 6\cdot e^{t}\cdot \sin t + 6\cdot e^{t} \cdot \cos t[/tex]

[tex]\dot x = 6\cdot e^{t} \cdot (\cos t - \sin t)[/tex] and [tex]\dot {y} = 6\cdot e^{t}\cdot (\sin t + \cos t)[/tex]

Second derivative

[tex]\ddot{x} = 6\cdot e^{t}\cdot (\cos t-\sin t)+6\cdot e^{t} \cdot (-\sin t -\cos t)[/tex]

[tex]\ddot x = -12\cdot e^{t}\cdot \sin t[/tex]

[tex]\ddot {y} = 6\cdot e^{t}\cdot (\sin t + \cos t) + 6\cdot e^{t}\cdot (\cos t - \sin t)[/tex]

[tex]\ddot{y} = 12\cdot e^{t}\cdot \cos t[/tex]

Now, each term is replaced in the the curvature equation:

[tex]\kappa = \frac{|6\cdot e^{t}\cdot (\cos t - \sin t)\cdot 12\cdot e^{t}\cdot \cos t-6\cdot e^{t}\cdot (\sin t + \cos t)\cdot (-12\cdot e^{t}\cdot \sin t)|}{\left\{\left[6\cdot e^{t}\cdot (\cos t - \sin t)\right]^{2}+\right[6\cdot e^{t}\cdot (\sin t + \cos t)\left]^{2}\right\}^{\frac{3}{2}}} }[/tex]

And the resulting expression is simplified by algebraic and trigonometric means:

[tex]\kappa = \frac{72\cdot e^{2\cdot t}\cdot \cos^{2}t-72\cdot e^{2\cdot t}\cdot \sin t\cdot \cos t + 72\cdot e^{2\cdot t}\cdot \sin^{2}t+72\cdot e^{2\cdot t}\cdot \sin t \cdot \cos t}{[36\cdot e^{2\cdot t}\cdot (\cos^{2}t -2\cdot \cos t \cdot \sin t +\sin^{2}t)+36\cdot e^{2\cdot t}\cdot (\sin^{2}t+2\cdot \cos t \cdot \sin t +\cos^{2} t)]^{\frac{3}{2} }}[/tex]

[tex]\kappa = \frac{72\cdot e^{2\cdot t}}{[72\cdot e^{2\cdot t}]^{\frac{3}{2} } }[/tex]

[tex]\kappa = [72\cdot e^{2\cdot t}]^{-\frac{1}{2} }[/tex]

[tex]\kappa = 72^{-\frac{1}{2} }\cdot e^{-t}[/tex]

[tex]\kappa = \frac{e^{-t}}{6\sqrt{2}}[/tex]

The curvature is modelled by [tex]\kappa = \frac{e^{-t}}{6\sqrt{2}}[/tex].

Answer:

23.63

Step-by-step explanation:

multiply the cost by the pounds

Answer:

$23.63

Step-by-step explanation:

3.4 X 6.95 = 23.63