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4 years ago

I need help one this one

Mathematics

1 answer:

8_murik_8 [283]4 years ago

3 0

56 in squared I believe!

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Verify that:

Lelu [443]

Answer:

See Below.

Step-by-step explanation:

Problem 1)

We want to verify that:

$\displaystyle \left(\cos(x)\right)\left(\cot(x)\right)=\csc(x)-\sin(x)$

Note that cot(x) = cos(x) / sin(x). Hence:

$\displaystyle \left(\cos(x)\right)\left(\frac{\cos(x)}{\sin(x)}\right)=\csc(x)-\sin(x)$

Multiply:

$\displaystyle \frac{\cos^2(x)}{\sin(x)}=\csc(x)-\sin(x)$

Recall that Pythagorean Identity: sin²(x) + cos²(x) = 1 or cos²(x) = 1 - sin²(x). Substitute:

$\displaystyle \frac{1-\sin^2(x)}{\sin(x)}=\csc(x)-\sin(x)$

Split:

$\displaystyle \frac{1}{\sin(x)}-\frac{\sin^2(x)}{\sin(x)}=\csc(x)-\sin(x)$

Simplify:

$\csc(x)-\sin(x)=\csc(x)-\sin(x)$

Problem 2)

We want to verify that:

$\displaystyle (\csc(x)-\cot(x))^2=\frac{1-\cos(x)}{1+\cos(x)}$

Square:

$\displaystyle \csc^2(x)-2\csc(x)\cot(x)+\cot^2(x)=\frac{1-\cos(x)}{1+\cos(x)}$

Convert csc(x) to 1 / sin(x) and cot(x) to cos(x) / sin(x). Thus:

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

Factor out the sin²(x) from the denominator:

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

Factor (perfect square trinomial):

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

Using the Pythagorean Identity, we know that sin²(x) = 1 - cos²(x). Hence:

$\displaystyle \frac{(\cos(x)-1)^2}{1-\cos^2(x)}=\frac{1-\cos(x)}{1+\cos(x)}$

Factor (difference of two squares):

$\displaystyle \frac{(\cos(x)-1)^2}{(1-\cos(x))(1+\cos(x))}=\frac{1-\cos(x)}{1+\cos(x)}$

Factor out a negative from the first factor in the denominator:

$\displaystyle \frac{(\cos(x)-1)^2}{-(\cos(x)-1)(1+\cos(x))}=\frac{1-\cos(x)}{1+\cos(x)}$

Cancel:

$\displaystyle \frac{\cos(x)-1}{-(1+\cos(x))}=\frac{1-\cos(x)}{1+\cos(x)}$

Distribute the negative into the numerator. Therefore:

$\displaystyle \frac{1-\cos(x)}{1+\cos(x)}=\displaystyle \frac{1-\cos(x)}{1+\cos(x)}$

3 0

3 years ago

Please help me solve this

Leokris [45]

D is the answer I think

5 0

3 years ago

Read 2 more answers

A sample of n = 8 scores has ss = 50. if these same scores were a population, then the ss value for the population would be ____

katen-ka-za [31]

A sample of n=8 has an ss=50 if these were a population the ss value of the population would be (50/8) of the sample = 6.25 of sample. Usually, ss value of the population i s calculated by dividing ss by the sample values. This is the way to calculate ss value of the population.

3 0

4 years ago

Read 2 more answers

a) LOOCV is not random because you fit all n possible models that leave out a single observation. k-fold CV is random because we

Daniel [21]

Answer:

1. Randomly divide the available set of observations into two

parts, a training set and a validation set or hold-out set.

2. Fit the model on the training set.

3. Use the resulting fitted model to predict the responses for the

observations in the validation set.

4. The resulting validation set error rate is typically assessed using

the MSE in the case of a quantitative response. This provides

an estimate of the test error rate.

5 0

4 years ago

In 2010, the Census Bureau estimated the proportion of all Americans who own their homes to be 0.669. An urban economist wants t

Alexxandr [17]

Answer:

Sample size making use of the Census Bureau: 1,499 American adults.

Sample size without making use of the Census Bureau: 1,692 American adults

ii)

Step-by-step explanation:

The sample size n in Simple Random Sampling is given by

$\bf n=\frac{z^2p(1-p)}{e^2}$

where

z = 1.645 is the critical value for a 90% confidence level (*) 

p= 0.669 is the population proportion given by the Census

e = 0.02 is the margin of error

$\bf n=\frac{(1.645)^2*0.669*0.331}{0.02^2}=1,498.05\approx 1,499$

rounded up to the nearest integer.

(*)This is a point z such that the area under the Normal curve N(0,1) 1nside the interval [-z, z] equals 90% = 0.9

It can be obtained with tables or in Excel or OpenOffice Calc with

NORMSINV(0.95)

If she ignores the Census estimate, the she has to take the largest sample possible that meets the requirements.

Let's show it is obtained when p = 0.5

As we said, the sample size n is

$\bf n=\frac{z^2p(1-p)}{e^2}$

where

e = 0.02 is the error proportion

z = 1.645

hence

$\bf n=\frac{(1.645)^2p(1-p)}{(0.02)^2}=6765.0625p(1-p)=6765.0625p-6765.0625p^2$

taking the derivative with respect to p, we get

n'(p)=6765.0625-2*6765.0625p

and

n'(p) = 0 when p=0.5

By taking the second derivative we see n''(p)<0, so p=0.5 is a maximum of n

This means that if we set p=0.5, we get the maximum sample size for the confidence level required for the proportion error 0.02

Replacing p with 0.5 in the formula for the sample size we get

$\bf n=6765.0625*0.5-6765.0625(0.5)^2=1691.27\approx 1,692$

rounded to the nearest integer.

ii)

When we do not have a proportion but a variable whose approximate standard deviation s is known, then the sample size n in Simple Random Sampling is given by

$\bf n=\frac{z^2s^2}{e^2}$

where

z = 2.241 is the critical value for a 95% confidence level (*) 

s = 7.5 is the estimated population standard deviation

e = 2 hours is the margin of error

$\bf n=\frac{z^2s^2}{e^2}=\frac{(2.241)^2(7.5)^2}{(2)^2}=70.62\approx 71$

(*)This is a point z such that the area under the Normal curve N(0,1) inside the interval [-z, z] equals 95% = 0.95

It can be obtained in Excel or OpenOffice Calc with

NORMSINV(0.9875)

5 0

4 years ago