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3 years ago

12. The probability a basketball player making a free-throw

Mathematics

1 answer:

Rasek [7]3 years ago

5 0

Answer:

The odds against making a free throw are: 0.1905 or 0.16:0.84

Step-by-step explanation:

The odds can be categorized into two categories

Odds-against
Odds-in favour

The odds against is the ratio of probability of failure to probability of success.

Given

Probability of making a free throw = P = 0.84

Probability of not making a free throw = Q = 1-P = 1-0.84 = 0.16

The formula for odds-against is:

$Odds-against = \frac{0.16}{0.84}\\= 0.19047..\\=0.1905$

Hence,

The odds against making a free throw are: 0.1905 or 0.16:0.84

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PtichkaEL [24]

Answer:

(a) $y = \frac{4}{x^2}$

(b) $x = \frac{2}{5}$

Step-by-step explanation:

Given

Variation: Inverse proportional.

This is represented as:

$y\ \alpha\ \frac{1}{x^2}$

See attachment for table

Solving (a):

First convert variation to equation

$y = k\frac{1}{x^2}$

From the table:

$(x,y) = (1,4)$

So, we have:

$4 = k * \frac{1}{1^2}$

$4 = k * \frac{1}{1}$

$4 = k * 1$

$4 = k$

$k = 4$

Substitute 4 for k in $y = k\frac{1}{x^2}$

$y = 4 * \frac{1}{x^2}$

$y = \frac{4}{x^2}$

Solving (b): x when y = 25.

Substitute 25 for y in $y = \frac{4}{x^2}$

$25 = \frac{4}{x^2}$

Cross Multiply

$25 * x^2 = 4$

Divide through by 25

$x^2 = \frac{4}{25}$

Take positive square roots of both sides

$x = \sqrt{\frac{4}{25}$

$x = \frac{2}{5}$

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