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3 years ago

7

Teams of 4 are competing in a 1/4 mile relay race. Each runner must run the same exact distance. What is the distance each teamm

ate runs?

1 answer:

PSYCHO15rus [73]3 years ago

4 0

As we know 1/4 mile = 402.336 m

So, 402.336 ÷ 4 = 100.584 m

100.584 m = 0.0625 mile or 1/16 mile

Hope it helps you

Please mark me as Brainsliest

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Suppose P(E) = 81%; P(E and F) = 9%; and P(E or F) = 65%,

PIT_PIT [208]

Using Venn probabilities, it is found that P(F) = -0.07, which is not a probability, as it is a negative value.

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We have two events, E and F.
Their probabilities are: $P(E) = 0.81, P(E \cap F) = 0.09, P(E \cup F) = 0.65$
The probabilities are related by the following equation:

$P(E \cap F) = P(E) + P(F) - P(E \cup F)$

Replacing into the equation, we can find P(F).

$0.09 = 0.81 + P(F) - 0.65$

$0.16 + P(F) = 0.09$

$P(F) = -0.07$

P(F) = -0.07, which is not a probability, as it is a negative value.

A similar problem is given at brainly.com/question/21421475

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3 years ago

Exercise 7.3.5 The following is a Markov (migration) matrix for three locations        1 5 1 5 2 5 2 5 2 5 1 5 2 5 2 5 2

fredd [130]

Answer:

Both get the same results that is,

$\left[\begin{array}{ccc}140\\160\\200\end{array}\right]$

Step-by-step explanation:

Given :

$\bf M=\left[\begin{array}{ccc}\frac{1}{5}&\frac{1}{5}&\frac{2}{5}\\\frac{2}{5}&\frac{2}{5}&\frac{1}{5}\\\frac{2}{5}&\frac{2}{5}&\frac{2}{5}\end{array}\right]$

and initial population,

$\bf P=\left[\begin{array}{ccc}130\\300\\70\end{array}\right]$

a) - After two times, we will find in each position.

$P_2=[P].[M]^2=[P].[M].[M]$

$M^2=\left[\begin{array}{ccc}\frac{1}{5}&\frac{1}{5}&\frac{2}{5}\\\frac{2}{5}&\frac{2}{5}&\frac{1}{5}\\\frac{2}{5}&\frac{2}{5}&\frac{2}{5}\end{array}\right]\times \left[\begin{array}{ccc}\frac{1}{5}&\frac{1}{5}&\frac{2}{5}\\\frac{2}{5}&\frac{2}{5}&\frac{1}{5}\\\frac{2}{5}&\frac{2}{5}&\frac{2}{5}\end{array}\right]$

$=\frac{1}{25} \left[\begin{array}{ccc}7&7&7\\8&8&8\\10&10&10\end{array}\right]$

$\therefore\;\;\;\;\;\;\;\;\;\;\;P_2=\left[\begin{array}{ccc}7&7&7\\8&8&8\\10&10&10\end{array}\right] \times\left[\begin{array}{ccc}130\\300\\70\end{array}\right] = \left[\begin{array}{ccc}140\\160\\200\end{array}\right]$

b) - With in migration process, 500 people are numbered. There will be after a long time,

$After\;inifinite\;period=[M]^n.[P]$

$Then,\;we\;get\;the\;same\;result\;if\;we\;measure [M]^n=\frac{1}{25} \left[\begin{array}{ccc}7&7&7\\8&8&8\\10&10&10\end{array}\right]$

$=\left[\begin{array}{ccc}140\\160\\200\end{array}\right]$

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