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2 years ago

Which sport is the most popular in the word?

Mathematics

2 answers:

Colt1911 [192]2 years ago

3 0

Answer:

Football/Soccer

The simple sport of soccer only needing a ball and two goals, is a fun easy game to do. And nationally, soccer has the most positive impact on people

Keith_Richards [23]2 years ago

3 0

Answer:

association football also known as soccer

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mezya [45]

Hello!

As you can see, 22 is the opposite angle. We are looking for the adjacent angle. Let's just call it x. Using SOH CAH TOA, we see that the tangent is the opposite divided by the adjacent side. Let's use that.

tan54≈1.38

This gives us the equation below.

1.38=22/x

Let's solve this equation for x.

1.38x=22

x=22/1.38

x=15.94

Therefore, the length of the shadow is about 15.94 meters.

I hope this helps!

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Answer:

a) $P(X=0)=(4C0)(0.75)^0 (1-0.75)^{4-0}=0.0039$

$P(X=1)=(4C1)(0.75)^1 (1-0.75)^{4-1}=0.0469$

$P(X=2)=(4C2)(0.75)^2 (1-0.75)^{4-2}=0.211$

$P(X=3)=(4C3)(0.75)^2 (1-0.75)^{4-3}=0.422$

$P(X=4)=(4C4)(0.75)^2 (1-0.75)^{4-4}=0.316$

b) $E(X) = np = 4*0.75=3$

c) $Sd(X) =\sqrt{np(1-p)}=\sqrt{4*0.75*(1-0.75)}=0.866$

d) $P(X \geq 3) \geq 0.98$

And the dsitribution that satisfy this is $X\sim Binom(n=9,p=0.75$

Step-by-step explanation:

Previous concepts

The binomial distribution is a "DISCRETE probability distribution that summarizes the probability that a value will take one of two independent values under a given set of parameters. The assumptions for the binomial distribution are that there is only one outcome for each trial, each trial has the same probability of success, and each trial is mutually exclusive, or independent of each other".

Let X the random variable of interest, on this case we now that:

$X \sim Binom(n=4, p=1-0.25=0.75)$

The probability mass function for the Binomial distribution is given as:

$P(X)=(nCx)(p)^x (1-p)^{n-x}$

Where (nCx) means combinatory and it's given by this formula:

$nCx=\frac{n!}{(n-x)! x!}$

Part a

$P(X=0)=(4C0)(0.75)^0 (1-0.75)^{4-0}=0.0039$

$P(X=1)=(4C1)(0.75)^1 (1-0.75)^{4-1}=0.0469$

$P(X=2)=(4C2)(0.75)^2 (1-0.75)^{4-2}=0.211$

$P(X=3)=(4C3)(0.75)^2 (1-0.75)^{4-3}=0.422$

$P(X=4)=(4C4)(0.75)^2 (1-0.75)^{4-4}=0.316$

Part b

The expected value is givn by:

$E(X) = np = 4*0.75=3$

Part c

For the standard deviation we have this:

$Sd(X) =\sqrt{np(1-p)}=\sqrt{4*0.75*(1-0.75)}=0.866$

Part d

For this case the sample size needs to be higher or equal to 9. Since we need a value such that:

$P(X \geq 3) \geq 0.98$

And the dsitribution that satisfy this is $X\sim Binom(n=9,p=0.75$

We can verify this using the following code:

"=1-BINOM.DIST(3,9,0.75,TRUE)" and we got 0.99 and the condition is satisfied.

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3 years ago