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3 years ago

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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

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

‣ A coin is tossed three times.

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

‣ The probability of getting,

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2) At most 2 heads

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

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

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

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