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3 years ago

Write the equation of a cosine function with the following properties

Mathematics

1 answer:

alukav5142 [94]3 years ago

6 0

★ Analytical Trigonometry ★

Let it's general format be -
$f(x) = a \: cos \: b(x - c) + d \\$
Having frequency as -
$f = \frac{b}{2\pi}$
According to provided data -

amplitude : 3
Hence , a = 3

$frequency = \frac{1}{2\pi}$
Hence b = 1

Phase Shift to the left of 2 : c = -2 [ left to the y-axis ]

Midline of y = 4

Hence d = 4

Now finally substituting every value obtained -

The equation of Cos function is -

$y = 3cos1(x - 2) + 4$

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Assume that X is a binomial random variable with n = 27 and p = 0.90. Calculate the following probabilities. (Do not round inter

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Answer:

a) $P(X=26) = 27C26 (0.9)^{26} (1-0.9)^{27-26}= 0.1744$

b) $P(X=25) = 27C25 (0.9)^{25} (1-0.9)^{27-25}= 0.2520$

c) $P(X=25) = 27C25 (0.9)^{25} (1-0.9)^{27-25}= 0.2520$

$P(X=26) = 27C26 (0.9)^{26} (1-0.9)^{27-26}= 0.1744$

$P(X=27) = 27C27 (0.9)^{27} (1-0.9)^{27-27}= 0.0582$

And addind the values we got:

$P(X\geq 25)= P(X=25)+P(X=26)+P(X=27)=0.2520+0.1744+0.0582=0.4846$

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".

Solution to the problem

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

$X \sim Binom(n=27, p=0.9)$

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

We want this probability:

$P(X=26)$

And we can use the probability mass function and we got:

$P(X=26) = 27C26 (0.9)^{26} (1-0.9)^{27-26}= 0.1744$

Part b

We want this probability:

$P(X=25)$

And we can use the probability mass function and we got:

$P(X=25) = 27C25 (0.9)^{25} (1-0.9)^{27-25}= 0.2520$

Part c

We want this probability:

$P(X\geq 25)= P(X=25)+P(X=26)+P(X=27)$

And we can use the probability mass function and we got:

$P(X=25) = 27C25 (0.9)^{25} (1-0.9)^{27-25}= 0.2520$

$P(X=26) = 27C26 (0.9)^{26} (1-0.9)^{27-26}= 0.1744$

$P(X=27) = 27C27 (0.9)^{27} (1-0.9)^{27-27}= 0.0582$

And addind the values we got:

$P(X\geq 25)= P(X=25)+P(X=26)+P(X=27)=0.2520+0.1744+0.0582=0.4846$

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