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zheka24 [161]
3 years ago
14

the average of johnnys two tests is a 92. what must he get to on a third test to raise his average to a 94

Mathematics
1 answer:
ASHA 777 [7]3 years ago
8 0
N=3 tests and the average is 94
3*94= 282
92+92 (add the average scores) = 184
282-184=98
He needs a 98 to raise his average to a 94

Hopefully this helps...
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Describe the translation of the function from the parent function f(x) = x².
kkurt [141]
\bf \qquad \qquad \qquad \qquad \textit{function transformations}
\\ \quad \\\\

\begin{array}{rllll} 
% left side templates
f(x)=&{{  A}}({{  B}}x+{{  C}})+{{  D}}
\\ \quad \\
y=&{{  A}}({{  B}}x+{{  C}})+{{  D}}
\\ \quad \\
f(x)=&{{  A}}\sqrt{{{  B}}x+{{  C}}}+{{  D}}
\\ \quad \\
f(x)=&{{  A}}(\mathbb{R})^{{{  B}}x+{{  C}}}+{{  D}}
\\ \quad \\
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\end{array}

\bf \begin{array}{llll}
% right side info
\bullet \textit{ stretches or shrinks horizontally by  } {{  A}}\cdot {{  B}}\\\\
\bullet \textit{ flips it upside-down if }{{  A}}\textit{ is negative}
\\\\

\end{array}\\

\bf \begin{array}{llll}
\bullet \textit{ horizontal shift by }\frac{{{  C}}}{{{  B}}}\\
\qquad  if\ \frac{{{  C}}}{{{  B}}}\textit{ is negative, to the right}\\\\
\qquad  if\ \frac{{{  C}}}{{{  B}}}\textit{ is positive, to the left}\\\\
\bullet \textit{ vertical shift by }{{  D}}\\
\qquad if\ {{  D}}\textit{ is negative, downwards}\\\\
\qquad if\ {{  D}}\textit{ is positive, upwards}\\\\
\bullet \textit{ period of }\frac{2\pi }{{{  B}}}
\end{array}


now, notice the template above... now let's see your function \bf y=x^2+4\implies 
\begin{array}{llllll}
y=&1(&1x&+&0)^2+&4\\
&A&B&&C&D
\end{array}

so.. what do you think was the shift then?
8 0
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LenaWriter [7]
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8 0
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Evgesh-ka [11]

Answer:

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Step-by-step explanation:

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Here, A = ($77000)(1 + 0.07/2)^(2*3), or

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          A  =  $77000(1.229), or

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7 0
3 years ago
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Andrei [34K]
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JulsSmile [24]

Answer:

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Step-by-step explanation:

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