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Virty [35]
1 year ago
12

Street sweeping vehicles can clean 3 miles of streets per hour. A city owns two street sweepers, and each sweeper can be used fo

r three hours before it comes in for an hour to refuel. During an 18-hour shift, how many miles of street can be cleaned?
Mathematics
1 answer:
Vanyuwa [196]1 year ago
4 0
We know that
In a period of 4 hours (3 hours of work and 1 hour to refuel)
each street sweepers clean------> 3 miles*3 =9 miles

divide 18 hours by 4
18/4=4.5
4.5 is equal to 4 periods of 4 hours plus 2 hours

Multiply 4 by 9 miles
4*9=36 miles

in the period of 2 hours (<span>in this period there is no refuel)
</span>3*2=6 miles

each street sweepers clean in 18 hours-----> (36+6)-----> 42 miles
two street sweepers clean in 18 hours=2*42------> 84 miles

the answer is
84 miles
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A sample of 900 college freshmen were randomly selected for a national survey. Among the survey participants, 372 students were
Rufina [12.5K]

Answer:

a) ME=2.58 \sqrt{\frac{0.413(1-0.413)}{900}}=0.0423

b) 0.413 - 2.58 \sqrt{\frac{0.413(1-0.413)}{900}}=0.371

Step-by-step explanation:

1) Data given and notation  

n=900 represent the random sample taken    

X=372 represent the students were pursuing liberal arts degrees

\hat p=\frac{372}{900}=0.413 estimated proportion of students were pursuing liberal arts degrees

\alpha=0.01 represent the significance level

z would represent the statistic (variable of interest)    

p_v represent the p value (variable of interest)    

p= population proportion of students were pursuing liberal arts degrees

2) Solution to the problem

The confidence interval would be given by this formula

\hat p \pm z_{\alpha/2} \sqrt{\frac{\hat p(1-\hat p)}{n}}

For the 99% confidence interval the value of \alpha=1-0.99=0.01 and \alpha/2=0.005, with that value we can find the quantile required for the interval in the normal standard distribution.

z_{\alpha/2}=2.58

The margin of error is given by:

ME=z_{\alpha/2} \sqrt{\frac{\hat p(1-\hat p)}{n}}

If we replace we have:

ME=2.58 \sqrt{\frac{0.413(1-0.413)}{900}}=0.0423

And replacing into the confidence interval formula we got:

0.413 - 2.58 \sqrt{\frac{0.413(1-0.413)}{900}}=0.371

0.413 + 2.58 \sqrt{\frac{0.413(1-0.413)}{900}}=0.455

And the 99% confidence interval would be given (0.371;0.455).

We are confident (99%) that about 37.1% to 45.5% of students were pursuing liberal arts degrees.

7 0
2 years ago
To aid in sea navigation, Little Gull Island Lighthouse shines a light from a height of 91 feet above sea level with an unknown
algol13

Answer:

\approx 6^\circ

Step-by-step explanation:

Given that:

Little Gull Island Lighthouse shines a light from a height of 91 feet above the sea level.

The angle of depression is unknown.

Distance of the point at sea surface from the base of lighthouse is 865 ft.

This situation can be modeled or can be represented as the figure attached in  the answer area.

The situation can be represented by a right angled \triangle ABC in which we are given the base and the height of the triangle.

And we have to find the value of \angle BAD \ or \ \angle C (Because they are the internal vertically opposite angles).

Using tangent ratio:

tan\theta = \dfrac{Perpendicular}{Base}

tanC = \dfrac{AC}{BC}\\\Rightarrow tanC = \dfrac{91}{865}\\\Rightarrow tanC = 0.105\\\Rightarrow \angle C \approx 6^\circ

Therefore, the angle of depression is: \approx 6^\circ

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A plane intersects the center of a sphere with a volume of about 113.1 m3. What is the area of the cross section? Round to the n
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Answer:

Step-by-step explanation:

if ~radius=r \\volume =\frac{4}{3} \pi r^3\\113.1=\frac{4}{3} \pi r^3\\113.1 \times \frac{3}{4 \pi } =r^3\\\frac{339.3 }{4 \times 3.14} =r^3\\r^3=\frac{339.3}{12.56} \approx 27\\r=3\\area =\pi r^2=\pi *3^2=9\pi =9*3.14=28.26 \approx 28.3 ~m^2

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A taco costs $5
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Which geometric solids would model the tent?
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a cylinder and a cone. the cone would go on the top, and the cylinder on the bottom.

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