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3 years ago

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About 42% of people who get their eyes examined are found to be nearsighted (unable to see objects at a distance). What is the p

robability of an ophthalmologist examining five people until he finds the first patient who is nearsighted? (0.42)5 (0.58)4(0.42)1 (0.42)4(0.58)1 (0.58)5 1 − (0.42)4

1 answer:

Law Incorporation [45]3 years ago

3 0

Answer:

(0.58)⁴ (0.42)¹

Step-by-step explanation:

42% of people who get their eyes examined are nearsighted.

So 58% of people who get their eyes examined are not nearsighted.

The probability that the ophthalmologist examines 4 people who are not near sighted, followed by 1 person who is nearsighted, is:

P = (0.58)⁴ (0.42)¹

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$V_1 =$

$V_1 =$

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$V_2 =$

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$V_2 =$

$n = V_1 * V_2$

$V_1 * V_2 =\left[\begin{array}{ccc}i&j&k\\0&2&8\\8&7&4\end{array}\right]$

The product is always of the form + - +

So:

$V_1 * V_2 = i\left[\begin{array}{cc}2&8\\7&4\end{array}\right]$ $-j\left[\begin{array}{cc}0&8\\8&4\end{array}\right]$ $+k\left[\begin{array}{cc}0&2\\8&7\end{array}\right]$

Calculate the product

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<em>Hence, the equation of the plane is</em> $3x - 4y + z = 1$ <em></em>

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