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3 years ago

In triangle ABC below, which trigonometric ratio equals 3/2?

Mathematics

2 answers:

TEA [102]3 years ago

6 0

Answer:

The answer is the option C

$tan(C)=\frac{3}{2}$

Step-by-step explanation:

see the attached figure to better understand the problem

In the right triangle of the figure

$tan(C)=\frac{AB}{AC}$

we have that

$AB=3\ units$ -----> opposite side angle C

$AC=2\ units$ ------> adjacent side angle C

substitute

$tan(C)=\frac{3}{2}$

Vlada [557]3 years ago

5 0

The figure is right triangle with base =segment AB = 3 and height = segment AC = 2.

The angle B has tangent, tan (B) = 2 / 3

The angle C, has tangent, tan (C) = 3 / 2

Then, the answer is option C, tan C

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Answer:

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

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

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

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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