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Misha Larkins [42]
3 years ago
7

Given that Cosecant (t) = negative StartFraction 13 Over 5 EndFraction for Pi less-than t less-than StartFraction 3 pi Over 2 En

dFraction use an appropriate identity to find the value of cot(t).
Mathematics
1 answer:
Sergio [31]3 years ago
4 0

Answer:

\bold{cot(t) =\dfrac{12}{5}}

Step-by-step explanation:

Given that:

Cosec (t) = -\frac{13}5

for \pi < t < \frac{3 \pi}2

That means, angle t is in the 3rd quadrant.

To find:

Value of cot(t)

Solution:

First of all, let us recall what trigonometric ratios are positive and what trigonometric ratios are negative in 3rd quadrant.

In 3rd quadrant, tangent and cotangent are positive.

All other trigonometric ratios are negative.

Let us have a look at the following identity:

cosec^2\theta -cot^2\theta =1

here, \theta =t

So, cosec^2t-cot^2t=1

\Rightarrow (-\dfrac{13}{5})^2-cot^2t=1\\\Rightarrow (\dfrac{169}{25})-cot^2t=1\\\Rightarrow \dfrac{169}{25}-1=cot^2t\\\Rightarrow \dfrac{169-25}{25}=cot^2t\\\Rightarrow \dfrac{144}{25}=cot^2t\\\Rightarrow cot(t)=\pm\sqrt{\dfrac{144}{25}}\\\Rightarrow cot(t)=\pm\dfrac{12}{5}

But, angle t is in 3rd quadrant, so value of

\bold{cot(t) =\dfrac{12}{5}}

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A particular concentration of a chemical found in polluted water has been found to be lethal to 20% of the fish that are exposed
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a) P(X=14)=(20C14)(0.8)^{14} (1-0.8)^{20-14}=0.109

b) P(X\geq 10) = 1-P(X \leq 9) = 0.9994

c) P(X \leq 16)= 1-P(X>16) =1-P(X \geq 17)= 1- [P(X=17) +...+P(X=20)]=0.589

d) E(X)= np = 20 *0.8 = 16

Var(X) = np(1-p) = 20*0.8*(1-0.8) = 3.2

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=20, p=1-0.2=0.8)

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

For this case we want this probability:

P(X=14)

And using the mass function we have this:

P(X=14)=(20C14)(0.8)^{14} (1-0.8)^{20-14}=0.109

Part b

For this case we want this probability:

P(X \geq 10)

And we can find this using the complement rule:

P(X\geq 10) = 1-P(X

P(X=0)=(20C0)(0.8)^{0} (1-0.8)^{20-0}=1.05x10^{-14}

P(X=1)=(20C1)(0.8)^{1} (1-0.8)^{20-1}=8.39x10^{-13}

P(X=2)=(20C2)(0.8)^{2} (1-0.8)^{20-2}=3.19x10^{-11}

P(X=3)=(20C3)(0.8)^{3} (1-0.8)^{20-3}=7.65x10^{-10}

P(X=4)=(20C4)(0.8)^{4} (1-0.8)^{20-4}=1.30x10^{-8}

P(X=5)=(20C5)(0.8)^{5} (1-0.8)^{20-5}=1.66x10^{-7}

P(X=6)=(20C6)(0.8)^{6} (1-0.8)^{20-6}=1.66x10^{-6}

P(X=7)=(20C7)(0.8)^{7} (1-0.8)^{20-7}=1.33x10^{-5}

P(X=8)=(20C8)(0.8)^{8} (1-0.8)^{20-8}=8.65x10^{-5}

P(X=9)=(20C9)(0.8)^{9} (1-0.8)^{20-9}=0.00046

And if we replace we got:

P(X\geq 10) = 1-P(X \leq 9) = 0.9994

Part c

For this case we want this probability:

P(X \leq 16)

And we can use the complement rule like this:

P(X \leq 16)= 1-P(X>16) =1-P(X \geq 17)= 1- [P(X=17) +...+P(X=20)]

P(X=17)=(20C17)(0.8)^{17} (1-0.8)^{20-17}=0.205

P(X=18)=(20C18)(0.8)^{18} (1-0.8)^{20-18}=0.137

P(X=19)=(20C19)(0.8)^{19} (1-0.8)^{20-19}=0.0576

P(X=20)=(20C20)(0.8)^{20} (1-0.8)^{20-20}=0.0115

And if we replace we got:

P(X \leq 16)= 1-P(X>16) =1-P(X \geq 17)= 1- [P(X=17) +...+P(X=20)]=0.589

Part d

The expected value is given by:

E(X)= np = 20 *0.8 = 16

Var(X) = np(1-p) = 20*0.8*(1-0.8) = 3.2

3 0
3 years ago
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