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SVEN [57.7K]
3 years ago
9

A student wants to report on the number of meals his friends buy each week. The collected data are below: 4 19 3 3 2 3 2 4 Which

measure of center is most appropriate for this situation and what is its value? (2 points) Group of answer choices Mean; 3 Mean; 5 Median; 3 Median; 5
Mathematics
2 answers:
lbvjy [14]3 years ago
4 0

median;

i tried mean 3, and mean 5 is wrong for sure, meaning it is median

explanation

arranged least to greatest

2, 2, 3, <em>3, 3,</em> 4, 4, 19 (8 numbers)

3+3/2=3

median=3

I am Lyosha [343]3 years ago
4 0

Answer:

Median 3

Step-by-step explanation:

I just took the test and got it right :)

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PLEASE HELP !! ILL GIVE BRAINLIEST !!
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2 and 6, 4 and 8, 1 and 5, 3 and 7

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Lucy has $19$ dollars and $23$ cents. She wants to buy as many popsicles as she can with her money. The popsicles are priced at
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Answer:

103                                                                                                                                                                               im pretty sure.

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Can someone please help me with my maths question​
DIA [1.3K]

Answer:

a. \  \dfrac{625 \cdot m}{27 \cdot n^{11}}

b. \  \dfrac{x^{3 \cdot m - 2}}{y^{ 3 + n}}

Step-by-step explanation:

The question relates with rules of indices

(a) The give expression is presented as follows;

\dfrac{m^3 \times \left (n^{-2} \right )^4 \times (5 \cdot m)^4}{\left (3 \cdot m^2 \cdot n \right )^3}

By expanding the expression, we get;

\dfrac{m^3 \times n^{-8} \times 5^4 \times m^4}{\left 3^3 \times m^6 \times n^3}

Collecting like terms gives;

\dfrac{m^{(3 + 4 - 6)}  \times 5^4}{ 3^3 \times n^{3 + 8}} = \dfrac{625 \cdot m}{27 \cdot n^{11}}

\dfrac{m^3 \times \left (n^{-2} \right )^4 \times (5 \cdot m)^4}{\left (3 \cdot m^2 \cdot n \right )^3}= \dfrac{625 \cdot m}{27 \cdot n^{11}}

(b) The given expression is presented as follows;

x^{3 \cdot m + 2} \times \left (y^{n - 1} \right )^3 \div (x \cdot y^n)^4

Therefore, we get;

x^{3 \cdot m + 2} \times \left (y^{n - 1} \right )^3 \times  x^{-4} \times y^{-4 \cdot n}

Collecting like terms gives;

x^{3 \cdot m + 2 - 4} \times \left (y^{3 \cdot n - 3 -4 \cdot n}} \right ) = x^{3 \cdot m - 2} \times \left (y^{ - 3 -n}} \right ) = x^{3 \cdot m - 2} \div \left (y^{ 3 + n}} \right )

x^{3 \cdot m - 2} \div \left (y^{ 3 + n}} \right ) = \dfrac{x^{3 \cdot m - 2}}{y^{ 3 + n}}

x^{3 \cdot m + 2} \times \left (y^{n - 1} \right )^3 \times  x^{-4} \times y^{-4 \cdot n} =\dfrac{x^{3 \cdot m - 2}}{y^{ 3 + n}}

4 0
3 years ago
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