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3 years ago

What is the median of the data set 10,7,9,9,4, 6?

Mathematics

1 answer:

lilavasa [31]3 years ago

4 0

Answer:

Step-by-step explanation:

The median is the middle of the numbers in a given data set.

To find the median, line up all of the numbers from greatest to least.

4, 6, 7, 9, 9, 10

Now, find the middle of the numbers. Since there is an even number of numbers in the data set, find the two middle numbers.

4, 6, 7, 9, 9, 10

Next, find the average of these two numbers. (To do this, add the numbers and divide that sum by 2.)

7 + 9 = 16

16/2 = 8

Therefore, the median is 8.

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What are some of the applications for exponential functions?

Pavel [41]

Answer:

Option B and D are correct

Step-by-step explanation:

Exponential functions are really useful in the real world situation. They can be used to solve following

a) Population Models

b) Determination of area and perimeters

c) Determine time related things such as half life, time of happening of an event etc.

d) Useful for solving financial problems such as computing investments etc.

Hence, option B and D

8 0

3 years ago

How would you write an equation for a non proportional table????

Bad White [126]

Answer:

i need the table

Step-by-step explanation:

But usually it has something to do with an exponent

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3 years ago

Simplify each expression. Assume that all variables are positive.

kozerog [31]

Q1. The answer is $\frac{8x^{3}y^{6} }{27}$

$( \frac{16 x^{5} y^{10}}{81x y^{2} } )^{ \frac{3}{4} }= ( \frac{16}{81}* \frac{ x^{5} }{x}* \frac{ y^{10} }{y^{2}} )^{ \frac{3}{4} } \\ \\ 
 \frac{ x^{a} }{ x^{b} }= x^{a-b} \\ \\ 
( \frac{16}{81}* \frac{ x^{5} }{x}*\frac{ y^{10} }{y^{2}} )^{ \frac{3}{4} }}=( \frac{16}{81 }* x^{5-1}* y^{10-2})^{ \frac{3}{4} }=( \frac{16}{81 }* x^{4}* y^{8})^{ \frac{3}{4} }= \\ \\ = (\frac{16}{18} )^{ \frac{3}{4} }*(x^{4})^{ \frac{3}{4} }*(y^{8})^{ \frac{3}{4} }=$
$\frac{(16)^{ \frac{3}{4} }}{(18)^{ \frac{3}{4} }}*(x^{4})^{ \frac{3}{4} }*(y^{8})^{ \frac{3}{4} }=\frac{( 2^{4} )^{ \frac{3}{4} }}{( 3^{4} )^{ \frac{3}{4} }}*(x^{4})^{ \frac{3}{4} }*(y^{8})^{ \frac{3}{4} } \\ \\ 
 (x^{a} )^{b} = x^{a*b} \\ \\ 
\frac{( 2^{4} )^{ \frac{3}{4} }}{( 3^{4} )^{ \frac{3}{4} }}*(x^{4})^{ \frac{3}{4} }*(y^{8})^{ \frac{3}{4} } = \frac{ 2^{4* \frac{3}{4} } }{ 3^{4* \frac{3}{4} } } * x^{4* \frac{3}{4} } * y^{8*\frac{3}{4}} = \frac{ 2^{3} }{ 3^{3} } * x^{3} *y^{6} =$
$= \frac{8x^{3}y^{6} }{27}$

Q2. The answer is 1/16

$(-64) ^ \frac{-2}{3} =(-1* 2^{6} ) ^ \frac{-2}{3}=(-1)^ \frac{-2}{3} *(2^{6} ) ^ \frac{-2}{3} \\\\x^{-a} = \frac{1}{ x^{a} } \\\\(-1)^ \frac{-2}{3} *(2^{6} ) ^ \frac{-2}{3} = \frac{1}{(-1)^ \frac{2}{3}} *\frac{1}{(2^{6})^ \frac{2}{3}} \\ \\ (x^{a} )^{b}=x^{a*b} \\\\x^{ \frac{a}{b} = \sqrt[b]{ x^{a} } } \\ \\ 
$
$\frac{1}{(-1)^ \frac{2}{3}} *\frac{1}{2^{6*\frac{2}{3}}} = \frac{1}{ \sqrt[3]{(-1)^{2} } } * \frac{1}{ 2^{4} } = \frac{1}{ \sqrt[3]{1} } * \frac{1}{16} = \frac{1}{1} * \frac{1}{16}= \frac{1}{16}$

Q3. The answer is $a^{ \frac{7}{6} }$

$a^{ \frac{2}{3} } * a^{ \frac{1}{2} } \\ \\ 
 x^{a}* x^{b} =x^{a+b} \\ \\ 
a^{ \frac{2}{3} } * a^{ \frac{1}{2} }= a^{ \frac{2}{3} + \frac{1}{2} } =a^{ \frac{2*2}{3*2} + \frac{1*3}{2*3} }=a^{ \frac{4}{6} + \frac{3}{6} }=a^{ \frac{4+3}{6} }=a^{ \frac{7}{6} }$

7 0

3 years ago

What do i do if i have a problem like this

yarga [219]

X^5 = 1024 = 4^5
Therefore, x = 4.

Given, (-1/7)^-x = -1/343
(1/7)^-x = 1/343
(1/7)^-x = (1/7)^3
-x = 3
x = -3

5 0

3 years ago

(-4, 7) and (1, 2) point slope form and slope intercept form

Monica [59]

I’m not sure but I think graphing it would help I really don’t know what to say

3 0

3 years ago