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Schach [20]
3 years ago
7

Write the expression in terms of sine and cosine, and then simplify so that no quotients appear in the final expression. sec bet

a - cos beta.
A.1-cos^2beta
B.sin^2beta
C.cos^2beta
D.1
E.sin beta tan beta
Mathematics
1 answer:
Ede4ka [16]3 years ago
3 0

Answer:

E

Step-by-step explanation:

The expression given is:  sec β - cos β ( i will write "beta" instead of the symbol β)

<em>We know that sec(x)=\frac{1}{cos(x)}</em>

<em>Substituting this into the the expression and doing some algebra manipulation, we have:</em>

sec(beta)-cos(beta)\\=\frac{1}{cos(beta)}-cos(beta)\\=\frac{1-cos^{2}(beta)}{cos(beta)}

<em>Using the identity  cos^{2}(x)+sin^{2}(x)=1\\ (also sin^{2}(x)=1-cos^{2}(x)), we can now write:</em>

\frac{1-cos^{2}(beta)}{cos(beta)}\\=\frac{sin^2(beta)}{cos(beta)}\\=\frac{sin(beta)*sin(beta)}{cos(beta)}\\=\frac{sin(beta)}{cos(beta)}*sin(beta)

<em>We know tan(x)=\frac{sin(x)}{cos(x)}, substituting this, we have:</em>

\frac{sin(beta)}{cos(beta)}*sin(beta)\\=tan(beta)*sin(beta)


Answer choice E is the right answer.

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Explanation:

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\bf X=X\qquad thus\qquad \implies \cfrac{24000}{a}=\cfrac{36000}{b}\implies \cfrac{24000}{a}=\cfrac{36000}{\boxed{3+a}} \\\\\\ (3+a)24000=36000a\implies \cfrac{3+a}{a}=\cfrac{36000}{24000}\implies \cfrac{3-a}{a}=\cfrac{3}{2} \\\\\\ 6-2a=3a\implies 6=5a\implies \cfrac{6}{5}=a\implies 1\frac{1}{5}=a\implies \blacktriangleright 1.2 = x\blacktriangleleft

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3 0
3 years ago
Nitrates are groundwater contaminants derived from fertilizer, septic tank seepage and other sewage. Nitrate poisoning is partic
o-na [289]

Answer:

a) n=\frac{0.5(1-0.5)}{(\frac{0.02}{1.96})^2}=2401  

And rounded up we have that n=2401

b) n=\frac{0.07(1-0.07)}{(\frac{0.02}{1.96})^2}=625.22  

And rounded up we have that n=626

And we see that if we have previous info about p the sample size varies a lot.

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".  

The margin of error is the range of values below and above the sample statistic in a confidence interval.  

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".  

Solution to the problem

In order to find the critical value we need to take in count that we are finding the interval for a proportion, so on this case we need to use the z distribution. Since our interval is at 95% of confidence, our significance level would be given by \alpha=1-0.95=0.05 and \alpha/2 =0.025. And the critical value would be given by:

z_{\alpha/2}=-1.96, z_{1-\alpha/2}=1.96

Part a

The margin of error for the proportion interval is given by this formula:  

ME=z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}}    (a)  

And on this case we have that ME =\pm 0.02 and we are interested in order to find the value of n, if we solve n from equation (a) we got:  

n=\frac{\hat p (1-\hat p)}{(\frac{ME}{z})^2}   (b)  

Since we don't have prior information about the population proportion we can use ase estimator \hat p =0.5. And replacing into equation (b) the values from part a we got:

n=\frac{0.5(1-0.5)}{(\frac{0.02}{1.96})^2}=2401  

And rounded up we have that n=2401

Part b

For thi case the estimator for p is \hat p =0.07

n=\frac{0.07(1-0.07)}{(\frac{0.02}{1.96})^2}=625.22  

And rounded up we have that n=626

And we see that if we have previous info about p the sample size varies a lot.

4 0
3 years ago
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