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deff fn [24]
3 years ago
8

Mr. Malloy wants to make sure Amo and Javier get the best possible grade after five tests and can either give a grade based on t

he median or the mean.
Test
Number
Amo’s Scores
Javier’s Scores
1
97
68
2
92
97
3
89
65
4
55
92
5
90
65

What measure of center should Mr. Malloy use for Amo and for Javier for the best possible grade?

The teacher should use the mean for both Amo and Javier.
The teacher should use the median for both Amo and Javier.
The teacher should use the median for Amo and the mean for Javier.
The teacher should use the mean for Amo and the median for Javier.

Mathematics
2 answers:
joja [24]3 years ago
8 0
The data has been properly arranged in tabular form and is shown below in the image.

First we need to find the mean and median of scores of both students.

1) For Amo:
Mean = \frac{97+92+89+55+90}{5}=84.6
Median = Middle Value when data is arranged in ascending order = 90

2) For Javier:
Mean =\frac{68+97+65+92+95}{5}=83.4
Median = Middle Value when data is arranged in ascending order = 92

For both the students, value of Median is larger then the mean. So in order to give the best possible grade Mr. Malloy should use the median score for both students.

AleksAgata [21]3 years ago
5 0

Answer:

The answer is C

Step-by-step explanation:

Edg

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Step-by-step explanation:

<em>Step 1: Determine the dimensions of the rectangle</em>

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3 0
2 years ago
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Evaluate the triple integral ∭EzdV where E is the solid bounded by the cylinder y2+z2=81 and the planes x=0,y=9x and z=0 in the
dem82 [27]

Answer:

I = 91.125

Step-by-step explanation:

Given that:

I = \int \int_E \int zdV where E is bounded by the cylinder y^2 + z^2 = 81 and the planes x = 0 , y = 9x and z = 0 in the first octant.

The initial activity to carry out is to determine the limits of the region

since curve z = 0 and y^2 + z^2 = 81

∴ z^2 = 81 - y^2

z = \sqrt{81 - y^2}

Thus, z lies between 0 to \sqrt{81 - y^2}

GIven curve x = 0 and y = 9x

x =\dfrac{y}{9}

As such,x lies between 0 to \dfrac{y}{9}

Given curve x = 0 , x =\dfrac{y}{9} and z = 0, y^2 + z^2 = 81

y = 0 and

y^2 = 81 \\ \\ y = \sqrt{81}  \\ \\  y = 9

∴ y lies between 0 and 9

Then I = \int^9_{y=0} \int^{\dfrac{y}{9}}_{x=0} \int^{\sqrt{81-y^2}}_{z=0} \ zdzdxdy

I = \int^9_{y=0} \int^{\dfrac{y}{9}}_{x=0} \begin {bmatrix} \dfrac{z^2}{2} \end {bmatrix}    ^ {\sqrt {{81-y^2}}}_{0} \ dxdy

I = \int^9_{y=0} \int^{\dfrac{y}{9}}_{x=0} \begin {bmatrix}  \dfrac{(\sqrt{81 -y^2})^2 }{2}-0  \end {bmatrix}     \ dxdy

I = \int^9_{y=0} \int^{\dfrac{y}{9}}_{x=0} \begin {bmatrix}  \dfrac{{81 -y^2} }{2} \end {bmatrix}     \ dxdy

I = \int^9_{y=0}  \begin {bmatrix}  \dfrac{{81x -xy^2} }{2} \end {bmatrix} ^{\dfrac{y}{9}}_{0}    \ dy

I = \int^9_{y=0}  \begin {bmatrix}  \dfrac{{81(\dfrac{y}{9}) -(\dfrac{y}{9})y^2} }{2}-0 \end {bmatrix}     \ dy

I = \int^9_{y=0}  \begin {bmatrix}  \dfrac{{81 \  y -y^3} }{18} \end {bmatrix}     \ dy

I = \dfrac{1}{18} \int^9_{y=0}  \begin {bmatrix}  {81 \  y -y^3}  \end {bmatrix}     \ dy

I = \dfrac{1}{18}  \begin {bmatrix}  {81 \ \dfrac{y^2}{2} - \dfrac{y^4}{4}}  \end {bmatrix}^9_0

I = \dfrac{1}{18}  \begin {bmatrix}  {40.5 \ (9^2) - \dfrac{9^4}{4}}  \end {bmatrix}

I = \dfrac{1}{18}  \begin {bmatrix}  3280.5 - 1640.25  \end {bmatrix}

I = \dfrac{1}{18}  \begin {bmatrix}  1640.25  \end {bmatrix}

I = 91.125

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3 years ago
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