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Troyanec [42]
2 years ago
9

PLEASEEE HELP MEEEENO LINKSSS

Mathematics
1 answer:
Artemon [7]2 years ago
3 0

Answer:

\huge\boxed{(2,\ 3);\ (-1,\ 6)}

Step-by-step explanation:

f(x)=x^2-2x+3\to y=x^2-2x+3\\f(x)=-x+5\to y=-x+5\\\\\left\{\begin{array}{ccc}y=x^2-2x+3&(1)\\y=-x+5&(2)\end{array}\right

substitute (1) to (2):

x^2-2x+3=-x+5     add x to both sides

x^2-2x+x+3=-x+x+5

x^2-x+3=5        subtract 5 from both sides

x^2-x+3-5=5-5

x^2-x-2=0\\\\x^2+x-2x-2=0\\\\x(x+1)-2(x+1)=0\\\\(x+1)(x-2)=0\iff x+1=0\ \vee\ x-2=0\\\\\boxed{x=-1\ \vee\ x=2}

substitute the values of x to (2):

x=-1\\\\y=-(-1)+5=1+5=6\\\\\huge\boxed{(-1,\ 6)}\\\\x=2\\\\y=-2+5=3\\\\\huge\boxed{(2,\ 3)}

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Sam had a finish time with a z-score of -2.93.

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Sam:

The mean finish time for a yearly amateur auto race was 186.94 minutes with a standard deviation of 0.372 minute. The winning car, driven by Sam, finished in 185.85 minutes.

So \mu = 186.94, \mu = 0.372, X = 185.85

Z = \frac{X - \mu}{\sigma}

Z = \frac{185.85 - 186.94}{0.372}

Z = -2.93

Sam finishing time was 2.93 standard deviations below the mean.

Sam had a finish time with a z-score of -2.93.

Karen:

The previous year's race had a mean finishing time of 110.7 with a standard deviation of 0.115 minute. The winning car that year, driven by Karen, finished in 110.48 minutes.

So \mu = 110.7, \sigma = 0.115, X = 110.48

Z = \frac{X - \mu}{\sigma}

Z = \frac{110.48 - 110.7}{0.115}

Z = -1.91

Karen finishing time was 1.91 standard deviations below the mean.

Karen had a finish time with a z-score -1.91.

Who had the more convincing victory?

Sam finishing time was 2.93 standard deviations below the mean.

Karen finishing time was 1.91 standard deviations below the mean.

Sam finished more standard deviations below the mean than Karen, that is, he was faster relative to his competition than Karen.

So Sam had the more convincing victory.

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