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Tems11 [23]
3 years ago
11

The weights of five oranges are 7.47 ounces, 7.23 ounces, 6.46 ounces, 7.48 ounces, and 6.81 ounces. Using the clustering estima

tion technique, what is the approximate total weight of the oranges?
34 ounces
35 ounces
35.45 ounces
35.5 ounces
Mathematics
2 answers:
topjm [15]3 years ago
8 0
The answer is b: 35 ounces. not a, c or d. its b
alexandr402 [8]3 years ago
5 0
Using the clustering estimation technique the approximate total weight oranges is 35 ounces.
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Airida [17]

Answer:

89.1° or -1.4°  

Step-by-step explanation:

1. Location:

You are on the Mont-Saint-Jean escarpment, near the Belgian town of Waterloo.

The French troops are about 50 m below you and 1.2 km distant.

2. Finding the firing angle

Data:

R = 1200 m

u = 600 m/s

h = -50 m (the height of the target)

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Vertical distance

(2) -50 = 600t sinθ - 4.9t²

Divide each side of (1) by 600cosθ.

(3) \, t =\dfrac{2}{\cos \theta}

Substitute (3) into (2)

-50 = 600t \sin \theta - 4.9t^{2} =  600 \left( \dfrac{2}{\cos \theta} \right ) \sin \theta - 4.9 \left( \dfrac{2}{\cos \theta} \right )^{2}\\\\(4) \, -50 = 1200 \tan \theta - \dfrac{19.6}{\cos^{2} \theta}

Recall that

(5) sec²θ = 1/cos²θ = tan²θ + 1

Substitute (5) into (4)

-50 = 1200 \tan \theta - 19.6 \left(\tan^{2} \theta}+ 1\right )

Set up a quadratic equation

\begin{array}{rcl}-50 & = & 1200 \tan \theta - 19.6\tan^{2} \theta -19.6 \\0 & = & 1200 \tan \theta - 19.6\tan^{2} \theta + 30.4\\0 & =&19.6\tan^{2} \theta - 1200 \tan \theta - 30.4\\0 & =&\tan^{2} \theta - 61.224 \tan \theta - 1.551\\\end{array}

Solve for θ

Use the quadratic formula.

tanθ = 61.249 or -0.025

θ = arctan(61.249) = 89.1° or

θ = arctan(-0.025) = -1.4°

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Answer:

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Step-by-step explanation:

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