I NEED HELP PLEASE, THANKS! :)

Mathematics

1 answer:

Elena L [17]3 years ago

7 0

Answer:

First option is the right choice.

Step-by-step explanation:

$\frac{1}{\cos \left(x\right)+1}+\frac{1}{\cos \left(x\right)-1}\\\\\frac{\cos \left(x\right)-1}{\left(\cos \left(x\right)+1\right)\left(\cos \left(x\right)-1\right)}+\frac{\cos \left(x\right)+1}{\left(\cos \left(x\right)+1\right)\left(\cos \left(x\right)-1\right)}\\\\=\frac{\cos \left(x\right)-1+\cos \left(x\right)+1}{\left(\cos \left(x\right)+1\right)\left(\cos \left(x\right)-1\right)}\\\\=\frac{2cos\left(x\right)}{cos^2\left(x\right)-1}\\\\=\frac{2cos\left(x\right)}{sin^2\left(x\right)}$

$=2\cdot \frac{cos\left(x\right)}{sin\left(x\right)}\cdot \frac{1}{sin\left(x\right)}\\\\=-2cot(x)csc(x)$

Best Regards!

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An unbalanced die is manufactured so that there is a 20% chance of rolling a “six." The die is rolled 6

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Answer:

Probability of rolling at least 4 sixes is 0.01696.

Step-by-step explanation:

We are given that an unbalanced die is manufactured so that there is a 20% chance of rolling a “six." The die is rolled 6 times.

The above situation can be represented through binomial distribution;

$P(X = r) = \binom{n}{r} \times p^{r} \times (1-p)^{n-r};x=0,1,2,3,.......$

where, n = number trials (samples) taken = 6 trials

r = number of success = at least 4

p = probability of success which in our question is probability of

rolling a “six", i.e; p = 0.20

Let X = Number of sixes on a die

So, X ~ Binom(n = 6, p = 0.20)

Now, Probability of rolling at least 4 sixes is given by = P(X $\geq$ 4)

P(X $\geq$ 4) = P(X = 4) + P(X = 5) + P(X = 6)

= $\binom{6}{4} \times 0.20^{4} \times (1-0.20)^{6-4}+\binom{6}{5} \times 0.20^{5} \times (1-0.20)^{6-5}+\binom{6}{6} \times 0.20^{6} \times (1-0.20)^{6-6}$