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GarryVolchara [31]

4 years ago

6

Using Pythagorean inequalities, determine which of the following sets of values can represent the lengths of the sides of a righ

t triangle I know it's now B
A. 4,5,6
B. 5,13,19
C 12, 14, 25
D 21,72,75

1 answer:

Cerrena [4.2K]4 years ago

7 0

Use Pythagorean theorem.
a^2 + b^2 = c^2
D.
21^2 + 72^2 = 75^2
441 + 5184 = 5625

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You are running a race. The probability that you win is 3/5. What is the probability that you lose at most 2 out of your 6 races

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Answer:

The probability that you lose at most 2 out of your 6 races is 0.54432.

Step-by-step explanation:

We are given that you are running a race. The probability that you win is 3/5.

There are total of 6 races.

The above situation can be represented through binomial distribution;

$P(X = r) = \binom{n}{r} \times p^{r} \times (1-p)^{n-r} ; x = 0,1,2,3,.....$

where, n = number of trials (samples) taken = 6 races

r = number of success = at most 2 lost

p = probability of success which in our question is probability that

you lose a race = 1 - (3/5) = 2/5 or 0.4

Let X = <u><em>Number of races lost </em></u>

So, X ~ Binom(n = 6, p = 0.40)

Now, the probability that you lose at most 2 out of your 6 races is given by = P(X $\leq$ 2)

P(X $\leq$ 2) = P(X = 0) + P(X = 1) + P(X = 2)

= $\binom{6}{0} \times 0.40^{0} \times (1-0.40)^{6-0} + \binom{6}{1} \times 0.40^{1} \times (1-0.40)^{6-1} + \binom{6}{2} \times 0.40^{2} \times (1-0.40)^{6-2}$

= $1 \times1 \times 0.60^{6} + 6 \times 0.40^{1} \times 0.60^{5} +15\times 0.40^{2} \times 0.60^{4}$

= <u>0.54432</u>

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