1.)Based on the average.how many people will usethe ramp each week? Setup ratio.2.) What will the surface areaof the ramp be? Writeequation & solve. Roundto the nearest hundredth.3.) What percentage of thepeople do not need theramp?4.) How high will the ramp be9 feet from the front walk?Draw a diagram & solve.5.) How high will the ramp be6 feet fromthe front walk?Draw a diagram & solve.Please help with these questions.thank you.

1Based on the averagehow many people will usethe ramp each week Setup ratio2 What will the surface areaof the ramp be Writeequation amp solve Roundto the neares class=

Respuesta :

1)

Let:

N = Total people every week = 207

r = People which needs a ramp = 1:9 = 1/9

So:

[tex]N\cdot r=207\cdot\frac{1}{9}=23[/tex]

Answer: 23 people

2)

[tex]\begin{gathered} SA=b\cdot h+pw \\ _{\text{ }}where\colon \\ b=18ft \\ h=4ft \\ p=b+h+l \\ l=\sqrt[]{h^2+b^2} \\ l=\sqrt[]{340} \\ w=4.75ft \end{gathered}[/tex]

Therefore:

[tex]\begin{gathered} SA=72+192.09 \\ SA=264.09ft^2 \end{gathered}[/tex]

3)

The percentage of the people that don't need the ramp will be given by:

[tex]1-r=1-\frac{1}{9}=\frac{8}{9}\approx0.89[/tex]

Answer: Approximately 89%

4)

[tex]\begin{gathered} h=\sqrt[]{(\sqrt[]{340})^2-9^2} \\ h=\sqrt[]{259} \\ h\approx16.09ft \end{gathered}[/tex]

5)

[tex]\begin{gathered} h=\sqrt[]{(\sqrt[]{340})^2-6^2} \\ h=\sqrt[]{340} \\ h\approx17.44ft \end{gathered}[/tex]

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