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4 years ago

11

The local Tupperware dealers earned the following commissions last month. What was the mean commission earned? Round your answe

r to the nearest cent. $2894.21 $1777.15 $2144.77 $4096.37 $4046.29 $1786.37 $3296.69 $4086.27 $2784.22 $4027.79
A. $3437.79
B. $3094.01
C. $3088.01
D. $3867.52

1 answer:

kirza4 [7]4 years ago

5 0

Answer:

B. $ 3094.01

Step-by-step explanation:

Given observations are,

$2894.21, $1777.15, $2144.77, $4096.37, $4046.29, $1786.37, $3296.69, $4086.27, $2784.22

Number of observations = 10,

Sum of the observations = 2894.21+1777.15+2144.77+4096.37+4046.29+1786.37+3296.69+4086.27+2784.22+4027.79 = 30940.13,

Hence,

$Mean = \frac{\text{Sum of observations}}{\text{Number of observation}}$

$=\frac{30940.13}{10}$

$=3094.013$

$\approx 3094.01$

Option 'B' is correct.

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All three sides of a triangle are initially 4 m in length. One of the triangle's sides is oriented horizontally. The triangle is

Arlecino [84]

Answer:

the new height of the triangle = 2.449

Step-by-step explanation:

Given that:

the sides of the triangle are 4m in length i.e they are equal. It shows that the triangle is known to be an equilateral triangle.

Let say the triangle is a triangle IJK

Let the length of the side to be i = 4

Definitely

IJ = JK = IK = i = 4

If a midline is drawn and cuts the equilateral triangle in two equal halves of a right-angle triangle. Then, suppose the midline is L

Then ;

$JL = \dfrac{1}{2} JK$

$JL = \dfrac{1}{2} i$

Let consider triangle IJL

(IL)² = (IJ)² - (JL)²

$(IL)^2 = i^2 - \dfrac{i^2}{4}$

$(IL)^2 = \dfrac{3i^2}{4}$

$(IL) =\sqrt{ \dfrac{3i^2}{4}}$

$IL =\sqrt{3} \dfrac{i}{2}$

Area of triangle IJK can be expressed as:

$Area = \dfrac{1}{2}\times Base \times Height$

$Area = \dfrac{1}{2}\times i \times \sqrt{3}\dfrac{i}{2}$

$Area = \sqrt{3}\dfrac{i^2}{4}$

where, i = 4

Then:

$Area = \sqrt{3}\dfrac{4^2}{4}$

$Area = \sqrt{3} \dfrac{16}{4}$

$Area =4 \sqrt{3}$

when the area is exactly half of the original triangle's area, the new height is :

$A = \dfrac{1}{2} \times Area$

$\sqrt{3} \dfrac{i^2}{4} = \dfrac{1}{2} \times 4\sqrt{3}$

$\dfrac{i^2}{4} = \dfrac{1}{2} \times 4$

$\dfrac{i^2}{4} =2$

$i^2 = 8$

$i= \sqrt{8}$

$i= \sqrt{4 \times 2}$

$i= 2 \sqrt{2}$

Finally, the new height of the new triangle is:

$IL =\sqrt{3} \dfrac{i}{2}$

$IL =\sqrt{3} \dfrac{2 \sqrt{2}}{2}$

$IL =\sqrt{3 \times 2}$

$IL =\sqrt{6}$

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<em><u>Solution:</u></em>

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