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Amanda [17]
3 years ago
8

What is .9 times 10 to the -6th power? i really suck at math haha

Mathematics
1 answer:
dexar [7]3 years ago
3 0
0.0000009



.9 times 10 is 9, so then you have to move it back -6 times behind the decimal.
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When solving a system of equations, how do you determine which method to use?
bazaltina [42]
Substitution is best to use one one or both equations is already solved for one of the variables. Elimitation can be use if all variables have a coefficient other than 1.
5 0
2 years ago
The following data set represents the age of
Julli [10]

\huge\text{Hey there!}


\huge\textsf{Breaking the meaning down in simpler terms}
\huge\textbf{Median:}\\\large\text{is understood to be \bf the number in the center (also known as}\\\large\textbf{the middle number)}


\huge\textbf{Mean:}\\\large\text{is understood to be the \bf total.}

\huge\textsf{Formulas for each term}

\huge\text{Median:}\\\large\textsf{To find the median you have to put each of your numbers in your data}\\\large\textsf{set from LEAST (smallest) to GREATEST (biggest). }


\huge\text{Mean:}\\\mathsf{\dfrac{sum\ of\ all\ terms}{number\ of\ terms}= mean}


\huge\textbf{SOLVING FOR THE QUESTION}

\huge\textsf{Median:}\\\\\large\textbf{Original data set: }\large\text{27, 52, 64, 41, 33, 38, 42, 60, 72, 68}\\\\\large\textbf{Conversion: }\large\text{27, 33, 38, 41, 42, 52, 64, 68, 72}\\\\\large\textsf{Make sure it is even between both sides of the data plot.}\\\\\large\text{It seems to even on BOTH sides of  \boxed{\textsf{14}} so it could possibly be your}\\\large\text{median.}

\huge\textsf{Mean:}\\\\\large\textbf{Original data set: }\large\text{27, 52, 64, 41, 33, 38, 42, 60, 72, 68}\\\\\large\textsf{Your equation: }\mathsf{\dfrac{27 + 52 + 64 + 41 + 33 + 38 + 42 + 60 +72 + 68}{10}}\\\\\mathsf{\dfrac{79 + 64 + 41 + 33 + 38 + 42 + 60 + 72 + 68}{10}}\\\\\mathsf{\dfrac{143 + 41 + 33 + 38 + 42 + 60 + 72 + 68}{10}}\\\\\mathsf{\dfrac{184 + 33 + 38 + 42 + 60 + 72 + 68}{10}}\\\\\mathsf{\dfrac{217 + 38 + 42 + 60 + 72 + 68}{10}}\\\\\mathsf{\dfrac{255 + 42 + 60 + 72 + 68}{10}}

\mathsf{\dfrac{297 + 60 + 72 + 68}{10}}\\\\\mathsf{\dfrac{357 + 72 + 68}{10}}\\\\\mathsf{\dfrac{429 + 68}{10}}\\\\\mathsf{\dfrac{497}{10}}\\\\\mathsf{\approx 49.70}\\\\\large\text{From the looks of it \boxed{\rm{\dfrac{497}{10}}}\ or \boxed{\text{49.70}} could possibly be your mean.}


\huge\text{Good luck on your assignment \& enjoy your day!}

<h3>
~\frak{Amphitrite1040:)}</h3>
8 0
2 years ago
Alan and Margot each drive from City A to City B, a distance of 147 miles. They take the same route and drive at constant speeds
Westkost [7]

Answer:

Margot gets from A to B faster

Step-by-step explanation:

Alan takes 2 hr and 35 min = 2. 58... hr

Margot takes 147/64 hr = 2.296.. hr

7 0
3 years ago
In the figure, m&lt;2 = 83, and m&lt;21 = 68. Find the measure of each angle
zhuklara [117]

Answer:

m<1 = 97

m<2 = 83

m<3 = 112

m<4 = 68

m<5 = 112

m<6 = 68

m<7 = 97

m<8 = 83

m<9 = 97

m<10 = 83

m<11 = 112

m<12 = 68

m<13 = 112

m<14 = 68

m<15 = 97

m<16 =  83

Step-by-step explanation:

8 0
3 years ago
Enter the coefficients of the fifth Taylor polynomial T5(x) for the function f(x) = x5−3x4+2x2+5x−2 based at b=1. T5(x)= + (x−1)
DENIUS [597]

Compute the necessary values/derivatives of f(x) at x=1:

f(1)=3

f'(1)=2

f''(1)=-12

f'''(1)=-12

f^{(4)}(1)=48

f^{(5)}(1)=120

Taylor's theorem then says we can "approximate" (in quotes because the Taylor polynomial for a polynomial is another, exact polynomial) f(x) at x=1 by

T_5(x)=\dfrac3{0!}+\dfrac2{1!}(x-1)-\dfrac{12}{2!}(x-1)^2-\dfrac{12}{3!}(x-1)^3+\dfrac{48}{4!}(x-1)^4+\dfrac{120}{5!}(x-1)^5

T_5(x)=3+2(x-1)-6(x-1)^2-2(x-1)^3+2(x-1)^4+(x-1)^5

###

Another way of doing this would be to solve for the coefficients a,b,c,d,e,g in

f(x)=a+b(x-1)+c(x-1)^2+d(x-1)^3+e(x-1)^4+g(x-1)^5

by expanding the right hand side and matching up terms with the same power of x.

5 0
3 years ago
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