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Oksi-84 [34.3K]
3 years ago
8

-/2 points

Mathematics
1 answer:
Rom4ik [11]3 years ago
6 0

Answer:

Part 1) The domain of the quadratic function is the interval  (-∞,∞)

Part 2) The range is the interval  (-∞,1]

Step-by-step explanation:

we have

f(x)=-x^2+14x-48

This is a quadratic equation (vertical parabola) open downward (the leading coefficient is negative)

step 1

Find the domain

The domain of a function is the set of all possible values of x

The domain of the quadratic function is the interval

(-∞,∞)

All real numbers

step 2

Find the range

The range of a function is the complete set of all possible resulting values of y, after we have substituted the domain.

we have a vertical parabola open downward

The vertex is a maximum

Let

(h,k) the vertex of the parabola

so

The range is the interval

(-∞,k]

Find the vertex

f(x)=-x^2+14x-48

Factor -1 the leading coefficient

f(x)=-(x^2-14x)-48

Complete the square

f(x)=-(x^2-14x+49)-48+49

f(x)=-(x^2-14x+49)+1

Rewrite as perfect squares

f(x)=-(x-7)^2+1

The vertex is the point (7,1)

therefore

The range is the interval

(-∞,1]

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Answer:

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Step-by-step explanation:

Let x = the amount Deshaun invested in an account that paid 9% interest

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{\large{\textsf{\textbf{\underline{\underline{Given :}}}}}}

‣ A coin is tossed three times.

{\large{\textsf{\textbf{\underline{\underline{To \: Find :}}}}}}

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1) Exactly 3 tails

2) At most 2 heads

3) At least 2 tails

4) Exactly 2 heads

5) Exactly 3 heads

{\large{\textsf{\textbf{\underline{\underline{Using \: Formula :}}}}}}

\star \: \tt  P(E)= {\underline{\boxed{\sf{\red{  \dfrac{ Favourable \:  outcomes }{Total \:  outcomes}  }}}}}

{\large{\textsf{\textbf{\underline{\underline{Solution :}}}}}}

★ When three coins are tossed,

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Here

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<u>3) At least 2 tails </u>

[It means there can be two or more tails]

Here

• Favourable outcomes = {TTH, TTT, HTT, THT} = 4

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\longrightarrow   \sf Probability_{(at \: least \: 2 \:  tails)}  =  \dfrac{4}{8}

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<u>4) Exactly 2 heads </u>

Here

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<u>5) Exactly 3 heads</u>

Here

• Favourable outcomes = {HHH} = 1

• Total outcomes = 8

\therefore  \sf Probability_{(exactly \: 3 \:  heads)}  =  \purple{ \dfrac{1}{8}}

\rule{280pt}{2pt}

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