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3 years ago

Can I take a photo of my question lol

Mathematics

1 answer:

Vikentia [17]3 years ago

3 0

Yes, you can take a photo of your question. You push the camera button at the bottom of the page.

You might be interested in

Why am i ugly please tell me

DIA [1.3K]

Answer:

your not ugly you have beauty or handsomeness in your own way you just don't show or give effert trust me I used to say that until someone told me what I told you

3 0

2 years ago

Read 2 more answers

What are the coordinates of A', the image of A(-3, 4), after a rotation of 180 about the 10. origin? (4, -3) B (-4, -3) C (3, 4)

nataly862011 [7]

Answer:

( 4 , -3)

Step-by-step explanation:

4 0

2 years ago

The point (1/3,1/4) lies on the terminal said of an angle. Find the exact value of the six trig functions and explain which func

katrin2010 [14]

Answer:

sine and cosec are inverse of each other.

cosine and sec are inverse of each other.

tan and cot are inverse of each other.

Step-by-step explanation:

Given point on terminal side of an angle ( $\frac{1}{3},\frac{1}4$ ).

Kindly refer to the attached image for the diagram of the given point.

Let it be point A( $\frac{1}{3},\frac{1}4$ )

Let O be the origin i.e. (0,0)

Point B will be ( $\frac{1}{3},0$ )

Now, let us consider the right angled triangle $\triangle OBA$ :

Sides:

$Base, OB = \frac{1}{3}\\Perpendicular, AB = \frac{1}{4}$

Using Pythagorean theorem:

$\text{Hypotenuse}^{2} = \text{Base}^{2} + \text{Perpendicular}^{2}\\\Rightarrow OA^{2} = OB^{2} + AB^{2}\\\Rightarrow OA^{2} = \frac{1}{3}^{2} + \frac{1}{4}^{2}\\\Rightarrow OA = \sqrt{\frac{1}{3}^{2} + \frac{1}{4}^{2}}\\\Rightarrow OA = \sqrt{\frac{4^2+3^2}{3^{2}.4^2 }}\\\Rightarrow OA = \frac{5}{12}$

$sin \angle AOB = \dfrac{Perpendicular}{Hypotenuse}$

$\Rightarrow sin \angle AOB = \dfrac{\frac{1}{4}}{\frac{5}{12}}\\\Rightarrow sin \angle AOB = \dfrac{3}{5}$

$cos\angle AOB = \dfrac{Base}{Hypotenuse}$

$\Rightarrow cos \angle AOB = \dfrac{\frac{1}{3}}{\frac{5}{12}}\\\Rightarrow cos\angle AOB = \dfrac{4}{5}$

$tan\angle AOB = \dfrac{Perpendicular}{Base}$

$\Rightarrow tan\angle AOB = \dfrac{3}{4}$

$cosec \angle AOB = \dfrac{Hypotenuse}{Perpendicular}$

$\Rightarrow cosec\angle AOB = \dfrac{5}{3}$

$sec\angle AOB = \dfrac{Hypotenuse}{Base}$

$\Rightarrow sec\angle AOB = \dfrac{5}{4}$

$cot\angle AOB = \dfrac{Base}{Perpendicular}$

$\Rightarrow cot\angle AOB = \dfrac{4}{3}$

3 0

2 years ago

Find all the complex square roots of W= 100 (cos 60 + I sin 60). Write the roots in polar form with theta in degrees.

serg [7]

Answer:

2√3+2i

hope it helps

please mark Brainliest

5 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

2 years ago