A car travelling from Ibadan to Lagos at 90 km/hr

takes 1 hour 20 min. How fast must one travel to

cover the distance in one hour?

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:

0.001

Step-by-step explanation:

Here, the aim is to support the null hypothesis, Ha. Where Ha: p > 0.5. Which means we are to reject null hypothesis H0. Where H0: p = 0.5.

The higher the pvalue, the higher the evidence of success. We know If the pvalue is less than level of significance, the null hypothesis H0 is rejected.

Hence the smallest possible value 0.001 is preferred as the pvalue because it corresponds to the sample evidence that most strongly supports the alternative hypothesis that the method is effective

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:

√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:

6

Step-by-step explanation:

nerd physics

Step-by-step explanation:

pnotgrt8rthan4 = 3 ÷ 7 × 100

= 42.8571428571 / 43%

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:

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:

Step-by-step explanation:

k = 3 ; 2k + 2 = 2*3 + 2 = 6 + 2 = 8

k = 4;  2k + 2 = 2*4 + 2 = 8 +2 = 10

k =5; 2k + 2 = 2*5 +2 = 10+2 = 12

k=6;  2k +2 = 2*6 + 2 = 12+2 = 14

k = 7 ; 2k + 2 = 2*7 +2 = 14 +2 = 16

k = 8 ; 2k + 2 = 2*8 + 2 = 16 +2 = 18

∑ (2k + 2) = 8 + 10 + 12 + 14 + 16 + 18 = 78

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:

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