|y-2|+11>0; solve for y.<br>Algebra 1

One more question on this test guys. please answer as fast as possible. If you do,

GaryK [48]

1. The given rectangular equation is $x=2$ .

We substitute $x=r\cos \theta$ .

$r\cos \theta=2$

Divide through by $\cos \theta$

$r=\frac{2}{\cos \theta}$

$r=2}\sec \theta$

$\boxed{x=2\to r=2\sec \theta}$

2. The given rectangular equation is:

$x^2+y^2=36$

This is the same as:

$x^2+y^2=6^2$

We use the relation $r^2=x^2+y^2$

This implies that:

$r^2=6^2$

$\therefore r=6$

$\boxed{x^2+y^2=36\to r=6}$

3. The given rectangular equation is:

$x^2+y^2=2y$

This is the same as:

We use the relation $r^2=x^2+y^2$ and $y=r\sin \theta$

This implies that:

$r^2=2r\sin \theta$

Divide through by r

$r=2\sin \theta$

$\boxed{x^2+y^2=2y\to r=2\sin \theta}$

4. We have $x=\sqrt{3}y$

We substitute $y=r\sin \theta$ and $x=r\cos \theta$

$r\cos \theta=r\sin \theta\sqrt{3}$

This implies that;

$\tan \theta=\frac{\sqrt{3}}{3}$

$\theta=\frac{\pi}{6}$

$\boxed{x=\sqrt{3}y\to \theta=\frac{\pi}{6}}$

5. We have $x=y$

We substitute $y=r\sin \theta$ and $x=r\cos \theta$

$r\cos \theta=r\sin \theta$

This implies that;

$\tan \theta=1$

$\theta=\frac{\pi}{4}$

$\boxed{x=y\to \theta=\frac{\pi}{4}}$

6 0

3 years ago

True or False: A p is the probability that the results are not due to chance, the probability that the null hypothesis (H0) is f

professor190 [17]

Answer:

False

Step-by-step explanation:?

The hypothesis tests compare weather an event is meant to alter a population mean results, for example, a scientist experiment might have or not have a significant effect over the population results. The test aims to reject the null hypothesis, so what it really want to find out is if the alternative Hypothesis H1 is likely true. The null hypothesis is the probability that the results are not due to chance – if it’s rejected, then the results are due to chance.The level of significance , or so called p-value, is the probability that the null hypothesis (H0) happen , If p is very small then the null hypothesis is rejected - isn’t true- and the alternative Hypothesis is accepted. A higher P value implies a higher probability than results are not happening so that the H0 is accepted and H1 rejected. The null Hypothesis will normally will rejected when the level of significance are either lower than 0.05 or 0.01, the lower P value the higher the level of confidence that the results are due to chance.

Since the first part of the statement (A p is the probability that the results are not due to chance) is correct, and the second part is wrong (…the probability that the null hypothesis (H0) is false), the total statement is false. The correct statement would be as follows : A p is the probability that the results are not due to chance, the probability that the null hypothesis (H0) is true.

7 0

3 years ago

Pleaseeee pleaseee help mee with precal...One root of 2x2 + 8x + c = 0 is r1 = 1 + √5. Find r2.

igomit [66]

If a root is a+√b then another is a-√b because of quadatic formula

r2=1-√5

6 0

4 years ago

Read 2 more answers

The brand manager for a brand toothpaste must plan a campaign designed to increase brand recognition. He wants to first determin

bulgar [2K]

We need to find out how many adults must the brand manager survey in order to be 90% confident that his estimate is within five percentage points of the true population percentage.

From the given data we know that our confidence level is 90%. From Standard Normal Table we know that the critical level at 90% confidence level is 1.645. In other words, $Z_{Critical}=1.645$ .

We also know that E=5% or E=0.05

Also, since, $\hat{p}$ is not given, we will assume that $\hat{p}$ =0.5. This is because, the formula that we use will have $\hat{p}(1-\hat{p})$ in the expression and that will be maximum only when $\hat{p}$ =0.5. (For any other value of $\hat{p}$ , we will get a value less than 0.25. For example if, $\hat{p}$ is 0.4, then $1-\hat{p}=0.6$ and thus, $\hat{p}(1-\hat{p})=0.24$ .).

We will now use the formula

$n=(\frac{Z_{Critical}}{E})^2\hat{p}(1-\hat{p})$

We will now substitute all the data that we have and we will get

$n=(\frac{1.645}{0.05})^2\times0.5(1-0.5)$

$n=(32.9)^2\times0.25$

$n=270.6025$

which can approximated to n=271.

So, the brand manager needs a sample size of 271

3 0

3 years ago

Which rule can be used to describe the growth of the y vales of the function y=12^x over each interval from x=-1 to x=3?

valentinak56 [21]

The function is $f(x)=12^x$ .

This means that as we get from 1 to 2, we have the following values:

$f(1)=12^1=12$ , and $f(2)=12^2=12\cdot12$ .

Similarly, if we check x=3, we have $f(3)=12^3=12\cdot12\cdot12$ .

We can see that as x changes by 1 (with greater values), y increases by a factor of 12.

Answer: multiply by 12

5 0

3 years ago

Read 2 more answers

|y-2|+11>0; solve for y.Algebra 1