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Vladimir [108]
2 years ago
10

What percentage grade should a road have if the angle of elevation of the road is 6 degrees? (The percentage grade is defined as

the change in the altitude of the road over a 100-foot horizontal distance. For example a 5% grade means that the road rises 5 feet for every 100 feet of horizontal distance.)
Mathematics
1 answer:
Andreas93 [3]2 years ago
3 0

Answer:

  10.5%

Step-by-step explanation:

The tangent of the angle is the ratio of rise to run.

  tan(angle) = grade

  tan(6°) ≈ 0.105 = 10.5%

The grade will be 10.5% for an angle of elevation of 6°.

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A random sample of 35 undergraduate students who completed two years of college were asked to take a basic mathematics test. The
lisov135 [29]

Answer:

(75.1 -72.1) -1.66 \sqrt{\frac{12.8^2}{35} +\frac{14.6^2}{50}}= -1.96

(75.1 -72.1) +1.66 \sqrt{\frac{12.8^2}{35} +\frac{14.6^2}{50}}= 7.96

And the confidence interval would be given by:

-1.96 \leq \mu_1 -\mu_2 \leq 7.96

And since the confidence interval contains the value 0 we have enough evidence to conclude that we don't have significant differences between the two means at 10% of significance.

Step-by-step explanation:

For this case we have the following info given :

\bar X_1= 75.1 represent the sample mean for the scores of the undergraduate students

s_1 = 12.8 represent the standard deviation for the undergraduate students

n_1 =35 the sample size for the undergraduate

\bar X_2= 72.1 represent the sample mean for the scores of the high school students

s_2 = 14.6 represent the standard deviation for the high school students

n_2 =50 the sample size for the high school

The confidence interval for the true difference of means is given by:

(\bar X_1 -\bar X_2) \pm t_{\alpha/2} \sqrt{\frac{s^2_1}{n_1} +\frac{s^2_2}{n_2}}

The degrees of freedom are given by:

df=n_1 +n_2 -2= 35+50-2=83

The confidence level is 90% and the significance level is \alpha=0.1 and \alpha/2 =0.05 then the critical value would be:

t_{\alpha/2}= 1.99

And replacing the info we got:

(75.1 -72.1) -1.66 \sqrt{\frac{12.8^2}{35} +\frac{14.6^2}{50}}= -1.96

(75.1 -72.1) +1.66 \sqrt{\frac{12.8^2}{35} +\frac{14.6^2}{50}}= 7.96

And the confidence interval would be given by:

-1.96 \leq \mu_1 -\mu_2 \leq 7.96

And since the confidence interval contains the value 0 we have enough evidence to conclude that we don't have significant differences between the two means at 10% of significance.

8 0
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A gym charges $40 per month after an initial service fee. One member paid a total of $255 after 6 months. Find the equation that
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Y = 40x +15 because it charges 15$ as it's initial service fee and it charges an additional $40 a month so you multiply 40 times the number of months and add 15 to get the total after x amount of months
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3/4 divides by 5

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given the points A(-3,-5) and B (5,0), find the coordinates of the point P on a directed line segment AB that partitions AB in t
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\bf ~~~~~~~~~~~~\textit{internal division of a line segment} \\\\\\ A(-3,-5)\qquad B(5,0)\qquad \qquad \stackrel{\textit{ratio from A to B}}{2:3} \\\\\\ \cfrac{A\underline{P}}{\underline{P} B} = \cfrac{2}{3}\implies \cfrac{A}{B} = \cfrac{2}{3}\implies 3A=2B\implies 3(-3,-5)=2(5,0)\\\\[-0.35em] \rule{31em}{0.25pt}\\\\ P=\left(\frac{\textit{sum of "x" values}}{\textit{sum of ratios}}\quad ,\quad \frac{\textit{sum of "y" values}}{\textit{sum of ratios}}\right)\\\\[-0.35em] \rule{31em}{0.25pt}


\bf P=\left(\cfrac{(3\cdot -3)+(2\cdot 5)}{2+3}\quad ,\quad \cfrac{(3\cdot -5)+(2\cdot 0)}{2+3}\right) \\\\\\ P=\left( \cfrac{-9+10}{5}~~,~~\cfrac{-15+0}{5} \right)\implies P=\left(\frac{1}{5}~,~-3  \right)

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B is da answer homie
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