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Triss [41]
3 years ago
6

Verify the identity: 2 sec^2 x − 2 sec^2 x sin^2 x − sin^2 x − cos^2 x = 1

Mathematics
1 answer:
Amiraneli [1.4K]3 years ago
8 0

\mathrm{Solving}\: \mathrm{prove\:2\sec ^2\left(x\right)-2\sec ^2\left(x\right)\sin ^2\left(x\right)-\sin ^2\left(x\right)-\cos ^2\left(x\right)=1}\\steps \\2\sec ^2\left(x\right)-2\sec ^2\left(x\right)\sin ^2\left(x\right)-\sin ^2\left(x\right)-\cos ^2\left(x\right)=1\\\mathrm{Manipulating\:left\:side}\\2\sec ^2\left(x\right)-2\sec ^2\left(x\right)\sin ^2\left(x\right)-\sin ^2\left(x\right)-\cos ^2\left(x\right)\\=-1+2\cos ^2\left(x\right)\sec ^2\left(x\right)\\=1Answer:

Step-by-step explanation:

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Read 2 more answers
Determine whether a probability distribution is given. If a probability distribution is given, find its mean and standard deviat
drek231 [11]

Answer:

E(X) = \sum_{i=1}^n X_i P(X_i) = 0*0.031 +1*0.156+ 2*0.313+3*0.313+ 4*0.156+ 5*0.031 = 2.5

We can find the second moment given by:

E(X^2) = \sum_{i=1}^n X^2_i P(X_i) = 0^2*0.031 +1^2*0.156+ 2^2*0.313+3^2*0.313+ 4^2*0.156+ 5^2*0.031 =7.496

And we can calculate the variance with this formula:

Var(X) =E(X^2) -[E(X)]^2 = 7.496 -(2.5)^2 = 1.246

And the deviation is:

Sd(X) = \sqrt{1.246}= 1.116

Step-by-step explanation:

For this case we have the following probability distribution given:

X          0            1        2         3        4         5

P(X)   0.031   0.156  0.313  0.313  0.156  0.031

The expected value of a random variable X is the n-th moment about zero of a probability density function f(x) if X is continuous, or the weighted average for a discrete probability distribution, if X is discrete.

The variance of a random variable X represent the spread of the possible values of the variable. The variance of X is written as Var(X).  

We can verify that:

\sum_{i=1}^n P(X_i) = 1

And P(X_i) \geq 0, \forall x_i

So then we have a probability distribution

We can calculate the expected value with the following formula:

E(X) = \sum_{i=1}^n X_i P(X_i) = 0*0.031 +1*0.156+ 2*0.313+3*0.313+ 4*0.156+ 5*0.031 = 2.5

We can find the second moment given by:

E(X^2) = \sum_{i=1}^n X^2_i P(X_i) = 0^2*0.031 +1^2*0.156+ 2^2*0.313+3^2*0.313+ 4^2*0.156+ 5^2*0.031 =7.496

And we can calculate the variance with this formula:

Var(X) =E(X^2) -[E(X)]^2 = 7.496 -(2.5)^2 = 1.246

And the deviation is:

Sd(X) = \sqrt{1.246}= 1.116

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