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4 years ago

How do I prove that (1+sinx)/cosx=cos(-x)/(1+sine(-x))

Mathematics

1 answer:

stira [4]4 years ago

5 0

$\dfrac{1+\sin x}{\cos x}=\dfrac{\cos (-x)}{1+\sin(-x)}\\
\dfrac{1+\sin x}{\cos x}=\dfrac{\cos x}{1-\sin x}\\
\cos^2 x=(1+\sin x)(1-\sin x)\\
\cos^2 x=1-\sin^2 x\\
\sin^2 x+\cos^2 x=1
$

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Which input in this table is incorrect?

Elenna [48]

Answer:

A) 10.00

Step-by-step explanation:

10.00 x 0.8 is equal to 8 giving a total of 18.00. But it is incorrectly done in the table.

6 0

2 years ago

An individual who has automobile insurance from a certain company is randomly selected. Let y be the number of moving violations

Hoochie [10]

Answer:

a) $E(Y)= \sum_{i=1}^n Y_i P(Y_i)$

And replacing we got:

$E(Y) = 0*0.45 +1*0.2 +2*0.3 +3*0.05= 0.95$

b) $E(80Y^2) =80[ 0^2*0.45 +1^2*0.2 +2^2*0.3 +3^2*0.05]= 148$

Step-by-step explanation:

Previous concepts

In statistics and probability analysis, the expected value "is calculated by multiplying each of the possible outcomes by the likelihood each outcome will occur and then summing all of those values".

The variance of a random variable Var(X) is the expected value of the squared deviation from the mean of X, E(X).

And the standard deviation of a random variable X is just the square root of the variance.

Solution to the problem

Part a

We have the following distribution function:

Y 0 1 2 3

P(Y) 0.45 0.2 0.3 0.05

And we can calculate the expected value with the following formula:

$E(Y)= \sum_{i=1}^n Y_i P(Y_i)$

And replacing we got:

$E(Y) = 0*0.45 +1*0.2 +2*0.3 +3*0.05= 0.95$

Part b

For this case the new expected value would be given by:

$E(80Y^2)= \sum_{i=1}^n 80Y^2_i P(Y_i)$

And replacing we got

$E(80Y^2) =80[ 0^2*0.45 +1^2*0.2 +2^2*0.3 +3^2*0.05]= 148$

5 0

3 years ago

A manufacturing process produces semiconductor chips with a known failure rate of . If a random sample of chips is selected, app

AleksandrR [38]

Answer:

The probability that at least 14 of the chips will be defective is 0.6664.

Step-by-step explanation:

The complete question is:

A manufacturing process produces semiconductor chips with a known failure rate of 5.4%. If a random sample of 300 chips is selected, approximate the probability that at least 14 will be defective. Use the normal approximation to the binomial with a correction for continuity .

Solution:

Let X = number of defective chips.

The probability that a chip is defective is, p = 0.054.

A random sample of n = 300 chips is selected.

A chip is defective or not is independent of the other chips.

The random variable X follows a Binomial distribution with parameters n = 300 and p = 0.054.

But the sample selected is too large.

So a Normal approximation to binomial can be applied to approximate the distribution of X if the following conditions are satisfied:

np ≥ 10
n(1 - p) ≥ 10

Check the conditions as follows:

$np=300\times 0.054=16.2>10\\n(1-p)=300\times (1-0.054)=283.8>10$

Thus, a Normal approximation to binomial can be applied.

So, $X\sim N(\mu =16.2,\ \sigma^{2}=15.3252)$ .

Compute the probability that at least 14 of the 300 chips will be defective as follows:

Use continuity correction:

P (X ≥ 14) = P (X > 14 + 0.50)

= P (X > 14.50)

$=P(\frac{X-\mu}{\sigma}>\frac{14.50-16.20}{\sqrt{15.3252}})$

$=P(Z>-0.43)\\=P(Z$

*Use a z-table for the probability.

Thus, the probability that at least 14 of the chips will be defective is 0.6664.

5 0

3 years ago

A human gene carries a certain disease from the mother to the child with a probability rate of 34%. That is, there is a 34% chan

eimsori [14]

Answer:

The probability that at least one of the children get the disease from their mother is P=0.7125.

Step-by-step explanation:

This can be modeled as a binomial random variable.

The parameter p=0.34 is the probability of a child being infected, and is constant and independent of the other events.

The parameter n=3 is the number of children (sample size).

Then, we have to calculate the probabilty that at least one of the children get the disease from their mother. That is:

$P(x\geq1)$

The probability of exactly k children being infected can be calculated as:

$P(x=k) = \dbinom{n}{k} p^{k}(1-p)^{n-k}$

Then, the easiest way to calculate this probability is using the complement: the value of the probability is 1 (or 100%) less the probability that no children gets infected.

$P(x\geq1)=1-P(x=0)\\\\\\P(x=0) = \dbinom{3}{0} p^{0}(1-p)^{3}=1*1*0.2875=0.2875\\\\\\P(x\geq1)=1-0.2875=0.7125$

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3 years ago

(x + 1)^2 -8 (x-1 +16)

tia_tia [17]

Answer: x2−6x−119

Step-by-step explanation:

provided in the screenshot below

5 0

3 years ago