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uranmaximum [27]
3 years ago
5

Complete the following proof.

Mathematics
1 answer:
Zanzabum3 years ago
7 0

Answer:

Step-by-step explanation:

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Solve the following system of linear equations {2x-7y=10 {5x -6y=2
Sliva [168]

2x-7y=10 = \frac{2}{7}

5x -6y=2 = \frac{5}{6}

5 0
3 years ago
Where are the minimum and maximum values for f(x) = -2 + 4 cos x on the interval (0,21]?
Dafna11 [192]

Answer:

Step-by-step explanation:

Maximum value is when cos x = 1

So it is -2 + 4(1) = 2.

Minimum value, when  cos x = -1:

= -2  + 4(-1) = -6.

4 0
3 years ago
Read 2 more answers
Need help finding the domain and range
pogonyaev

Here are a few things you'll need to know for this question:

  • Domain: <u>The list of x-values that are possible on a line.</u>
  • Range: <u>The list of y-values that are possible on a line.</u>
  • Interval Notation: <u>Shows the domain/range using the endpoints</u>. Brackets mean that the endpoint is included, parentheses mean the endpoint is excluded. Ex: (2,10]. 2 is excluded, 10 is included.
  • Closed Circles: <u>The endpoint is included.</u>
  • Open Circles: <u>The endpoint is excluded.</u>

So firstly, let's look at the domain. We see that there is a closed circle at x = -2 and an open circle at x = 5. Using what we know, <u>the interval notation of the domain is [-2,5).</u>

Next, let's look at the range. We see that there is a closed circle at y = -5 and an open circle at y = 2. Using what we know, <u>the interval notation of the range is [-5,2).</u>

3 0
3 years ago
What is the length of MN
Dafna1 [17]
Because this is an isosceles triangle, 3x = x + 10.  Thus, 2x + 10 and x = 5.

With x=5, MN has the length 5+10, or 15.
5 0
3 years ago
Read 2 more answers
"The municipal transportation authority determined that 58% of all drivers were speeding along a busy street. In an attempt to r
vredina [299]

Answer:

a) X=77 drivers

b) Power of the test = 0.404

c) Increasing the sample size.

Step-by-step explanation:

This is a hypothesis test of proportions. As the claim is that the speed monitors were effective in reducing the speeding, this is a left-tail test.

For a left-tail test at a 5% significance level, we have a critical value of z that is zc=-1.645. This value is the limit of the rejection region. That means that if the test statistic z is smaller than zc=-1.645, the null hypothesis is rejected.

The proportion that would have a test statistic equal to this critical value can be expressed as:

p_c=\pi+z_c\cdot\sigma_p

The standard error of the proportion is:

\sigma_p=\sqrt{\dfrac{\pi(1-\pi)}{n}}=\sqrt{\dfrac{0.58*0.42}{150}}\\\\\\ \sigma_p=\sqrt{0.001624}=0.04

Then, the proportion is:

p_c=\pi+z_c\cdot\sigma_p=0.58-1.645*0.04=0.58-0.0658=0.5142

This proportion, with a sample size of n=150, correspond to

x=n\cdot p=150\cdot0.5142=77.13\approx 77

The power of the test is the probability of correctly rejecting the null hypothesis.

The true proportion is 0.52, but we don't know at the time of the test, so the critical value to make a decision about rejecting the null hypothesis is still zc=-1.645 corresponding to a critical proportion of 0.51.

Then, we can say that the probability of rejecting the null hypothesis is still the probability of getting a sample of size n=150 with a proportion of 0.51 or smaller, but within a population with a proportion of 0.52.

The standard error has to be re-calculated for the new true proportion:

\sigma_p=\sqrt{\dfrac{\pi(1-\pi)}{n}}=\sqrt{\dfrac{0.52*0.48}{150}}\\\\\\ \sigma_p=\sqrt{0.001664}=0.041

Then, we calculate the z-value for this proportion with the true proportion:

z=\dfrac{p-\pi'}{\sigma_p}=\dfrac{0.51-0.52}{0.041}=\dfrac{-0.01}{0.041}=-0.244

The probability of getting a sample of size n=150 with a proportion of 0.51 or lower is:

P(p

Then, the power of the test is β=0.404.

The only variable left to change in the test in order to increase the power of the test is the sample size, as the significance level can not be changed (it is related to the probability of a Type I error).

It the sample size is increased, the standard error of the proprotion decreases. As the standard error tends to zero, the critical proportion tend to 0.58, as we can see in its equation:

\lim_{\sigma_p \to 0} p_c=\pi+ \lim_{\sigma_p \to 0}(z_c\cdot\sigma_p)=\pi=0.58

Then, if the critical proportion increases, the z-score increases, and also the probability of rejecting the null hypothesis.

5 0
3 years ago
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