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zvonat [6]
3 years ago
15

A service station has both self-service and full-service islands. On each island, there is a single regular unleaded pump with t

wo hoses. Let X denote the number of hoses being used on the self-service island at a particular time, and let Y denote the number of hoses on the full-service island in use at that time. The joint pmf of X and Y appears in the accompanying tabulation
Mathematics
1 answer:
Annette [7]3 years ago
8 0

Answer:

ANSWER LINK

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LUCKY_DIMON [66]
Y=mx+b is slope intercept form
m=-8
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y=-8x+2 

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What sum will amount to rupees 4000 in 3 years at 6% p.a. compound interest​
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Answer:

\begin{gathered}{\Large{\textsf{\textbf{\underline{\underline{\color{purple}{Given:}}}}}}}\end{gathered}

  • ⇢ Principle = Rs.4000
  • ⇢ Rate = 6%
  • ⇢ Time = 3 year

\begin{gathered}\end{gathered}

\begin{gathered}{\Large{\textsf{\textbf{\underline{\underline{\color{purple}{To Find:}}}}}}}\end{gathered}

  • ⇢ Amount

\begin{gathered}\end{gathered}

\begin{gathered}{\Large{\textsf{\textbf{\underline{\underline{\color{purple}{Using Formula:}}}}}}}\end{gathered}

{\dag{\underline{\boxed{\sf{Amount  ={P{\bigg(1 + \dfrac{R}{100}{\bigg)}^{T}}}}}}}}

\dag{\underline{\boxed{\sf{Compound \: Interest = Amount- Principle }}}}

\begin{gathered}\end{gathered}

\begin{gathered}{\Large{\textsf{\textbf{\underline{\underline{\color{purple}{Solution:}}}}}}}\end{gathered}

{\bigstar \:{\underline{\pmb{\frak{\red{Firstly,Finding  \: the  \: Amount }}}}}}

\quad {:\implies{\sf{Amount  = \bf{P{\bigg(1  +  \dfrac{R}{100}{\bigg)}^{T}}}}}}

  • Substituting the values

\quad {:\implies{\sf{Amount  = \bf{4000{\bigg(1  +  \dfrac{6}{100}{\bigg)}^{3}}}}}}

\quad {:\implies{\sf{Amount  = \bf{4000{\bigg(1 \times 100  +  \dfrac{6}{100}{\bigg)}^{3}}}}}}

\quad {:\implies{\sf{Amount  = \bf{4000{\bigg( \dfrac{100 + 6}{100}{\bigg)}^{3}}}}}}

\quad {:\implies{\sf{Amount  = \bf{4000{\bigg( \dfrac{106}{100}{\bigg)}^{3}}}}}}

\quad {:\implies{\sf{Amount  = \bf{4000{\bigg({\cancel{\dfrac{106}{100}}{\bigg)}}^{3}}}}}}

\quad {:\implies{\sf{Amount  = \bf{4000{\bigg( \dfrac{53}{50}{\bigg)}^{3}}}}}}

\quad {:\implies{\sf{Amount  = \bf{4000{\bigg( \dfrac{53}{50} \times \dfrac{53}{50} \times \dfrac{53}{50}{\bigg)}}}}}}

\quad {:\implies{\sf{Amount  = \bf{4000{\bigg( \dfrac{148877}{125000}{\bigg)}}}}}}

\quad {:\implies{\sf{Amount  = \bf{4000 \times  \dfrac{148877}{125000}}}}}

\quad {:\implies{\sf{Amount  = \bf{4{\cancel{000}} \times  \dfrac{148877}{125{\cancel{000}}}}}}}

\quad {:\implies{\sf{Amount  = \bf{\dfrac{148877 \times 4}{125}}}}}

\quad {:\implies{\sf{Amount  = \bf{\dfrac{595508}{125}}}}}

\quad {:\implies{\sf{Amount  = \bf{\cancel{\dfrac{595508}{125}}}}}}

\quad {:\implies{\sf{Amount  = \bf{4764.064}}}}

\begin{gathered} \dag{\boxed{\textsf{\textbf{\underline{\color{green}{Amount = {Rs.4764.064}}}}}}}\end{gathered}

  • Hence, The Amount is Rs.4764.064

\begin{gathered}\end{gathered}

{\bigstar \:{\underline{\pmb{\frak{\red{ Now,Finding  \: The \:  Compound \:  Interest }}}}}}

\quad{: \implies{\sf{Compound \: Interest =  \bf{Amount- Principle }}}}

  • Substituting the values

\quad{: \implies{\sf{Compound \: Interest = \bf{4764.064- 4000 }}}}

\quad{: \implies{\sf{Compound \: Interest =\bf{764.064}}}}

\begin{gathered} \dag{\boxed{\textsf{\textbf{\underline{\color{green}{Compound Interest  = Rs.764.064}}}}}}\end{gathered}

  • Henceforth,The Compound Interest is Rs.764064

\begin{gathered}\end{gathered}

\begin{gathered}{\Large{\textsf{\textbf{\underline{\underline{\color{purple}{Learn More:}}}}}}}\end{gathered}

\begin{gathered}\begin{gathered}\begin{gathered} \dag \: \underline{\bf{More \: Useful \: Formula}}\\ {\boxed{\begin{array}{cc}\dashrightarrow {\sf{Amount = Principle + Interest}} \\ \\ \dashrightarrow \sf{ P=Amount - Interest }\\ \\ \dashrightarrow \sf{ S.I = \dfrac{P \times R \times T}{100}} \\ \\ \dashrightarrow \sf{P = \dfrac{Interest \times 100 }{Time \times Rate}} \\ \\ \dashrightarrow \sf{P = \dfrac{Amount\times 100 }{100 + (Time \times Rate)}} \\ \end{array}}}\end{gathered}\end{gathered}\end{gathered}

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Find the domain and range with a vertex of (1,-2)
klemol [59]

Answer:

see explanation

Step-by-step explanation:

the domain ( values of x ) that a quadratic can have is all real numbers

domain : - ∞ < x < ∞

the range ( values of y ) are from the vertex upwards , that is

range : y ≥ - 2

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Since a polynomial is where we have like terms such as (1 x 10²) and (4 x 10²), we can add these up using the distributive property to get (5 x 10²) but still keep the 10². For example, it's similar to if we had 2x²+3x²=5x². The x² is still there, but we add up the 2 and 3. Similarly, we can add these up for 10^1 and 10^0
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