The population of Adamsville grew from 6000 to 13000 in 7 years. Assuming uninhibited exponential growth, what is the expected population in an additional 3 years?

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:

2.24 s

Step-by-step explanation:

Given:

Δy = 80 ft

v₀ = 0 ft/s

a = 32 ft/s²

Find: t

Δy = v₀ t + ½ at²

80 ft = (0 ft/s) t + ½ (32 ft/s²) t²

t = 2.24 s

Answer:

6

Step-by-step explanation:

nerd physics

Answer:

Step-by-step explanation:

When two variables vary inversely, it means that an increase in one would lead to a decrease in the other and vice versa. Since the volume of a gas, V in a container varies inversely with the pressure on the gas, P, if we introduce a constant of proportionality, k, the expression would be

V = k/P

If V = 370 in³ and P = 15psi, then

370 = k/15

k = 370 × 15 = 5550

The equation that relates the volume, V, to the pressure, P would be

V = 5550/P

if the pressure was increased to 25psi, the volume would be

V = 5550/25 = 222 in³

Answer:

v=5550/p

222

Step-by-step explanation:

Answer:

150 km

Step-by-step explanation:

1 liter ............ 40 km

15/4 liter .........x km

x = 15/4×40/1 = 600/4 = 150 km

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

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

Complete Question:

You want to install a 1 yd wide walk around a circular swimming pool. The diameter of the pool is 23 yd. What is the area of the​ walk? Use 3.14 for pi π.

Answer:

75.36 square yard

Step-by-step explanation:

From the question,

The diameter of this circular pool inside is 23 yd.

This means that the radius = Diameter/2 = 23yd/2 = 11.5 yd.

The formula for the area of a circle =

A = πr²

A = π(11.5)²

A =3.14 × 11.5²

A = 415.265 yd²

This is the Area of the inner circle.

We were told in the question also that he wants to install a walk of 1 yard

Hence, the radius of outer circle =

radius of inner circle +length of the walk

11.5yard + 1 yard

= 12.5 yard

A = πr²

A = 3.14 × (12.5)²

A = 490.625yd²

Area of the walk = Area of the Outer circle - Area of the inner circle

= (490.625 - 415.265)yd = 75.36 yd²

Therefore, the area of the walk is 75.36 square yards.

Answer:

a. The 95% confidence interval for the mean is (33.52, 35.48).

b. The 95% confidence interval for the mean is (34.02, 34.98).  

c. n=49 ⇒ Width = 1.95

n=196 ⇒ Width = 0.96

Note: it should be a factor of 2 between the widths, but the different degrees of freedom affects the critical value for each interval, as the sample size is different. It the population standard deviation had been used, the factor would have been exactly 2.

d. 5. Quadrupling the sample size while holding the confidence coefficient fixed decreases the width of the confidence interval by a factor of 2.

Step-by-step explanation:

a. We have to calculate a 95% confidence interval for the mean.

The population standard deviation is not known, so we have to estimate it from the sample standard deviation and use a t-students distribution to calculate the critical value.

The sample mean is M=34.5.

The sample size is N=49.

When σ is not known, s divided by the square root of N is used as an estimate of σM:

[tex]s_M=\dfrac{s}{\sqrt{N}}=\dfrac{3.4}{\sqrt{49}}=\dfrac{3.4}{7}=0.486[/tex]

The degrees of freedom for this sample size are:

[tex]df=n-1=49-1=48[/tex]

The t-value for a 95% confidence interval and 48 degrees of freedom is t=2.011.

The margin of error (MOE) can be calculated as:

[tex]MOE=t\cdot s_M=2.011 \cdot 0.486=0.98[/tex]

Then, the lower and upper bounds of the confidence interval are:

[tex]LL=M-t \cdot s_M = 34.5-0.98=33.52\\\\UL=M+t \cdot s_M = 34.5+0.98=35.48[/tex]

The 95% confidence interval for the mean is (33.52, 35.48).

b. We have to calculate a 95% confidence interval for the mean.

When σ is not known, s divided by the square root of N is used as an estimate of σM:

[tex]s_M=\dfrac{s}{\sqrt{N}}=\dfrac{3.4}{\sqrt{196}}=\dfrac{3.4}{14}=0.243[/tex]

The degrees of freedom for this sample size are:

[tex]df=n-1=196-1=195[/tex]

The t-value for a 95% confidence interval and 195 degrees of freedom is t=1.972.

The margin of error (MOE) can be calculated as:

[tex]MOE=t\cdot s_M=1.972 \cdot 0.243=0.48[/tex]

Then, the lower and upper bounds of the confidence interval are:

[tex]LL=M-t \cdot s_M = 34.5-0.48=34.02\\\\UL=M+t \cdot s_M = 34.5+0.48=34.98[/tex]

The 95% confidence interval for the mean is (34.02, 34.98).

c. The width of the intervals is:

[tex]n=49\rightarrow UL-LL=33.52-35.48=1.95\\\\n=196\rightarrow UL-LL=34.02-34.98=0.96[/tex]

d. The width of the intervals is decreased by a factor of √4=2 when the sample size is quadrupled, while the others factors are fixed.

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].