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2 years ago

The Grade 5, 6, and 7 classes

Mathematics

1 answer:

Llana [10]2 years ago

8 0

Answer:

<em>x</em> = 1/5

Step-by-step explanation:

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3 years ago

TWO QUESTIONS PLS HELP ASAP

ASHA 777 [7]

Answer: 4. (-1,-1) 3. (3,-2)

4)

Set the equations equal to each other.

4x+3=-x-2

Subtract 3 from both sides

4x=-x-5

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5x=-5

Divide both sides by 5

x=-1

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3)

Do the same thing for this one and set them equal to each other

-2x+4=-1/3x-1

Add 1 to both sides

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3 years ago

A random sample of 864 births in a state included 426 boys. Construct a 95% confidence interval estimate of the proportion of bo

oksian1 [2.3K]

Using the z-distribution, it is found that the 95% confidence interval is (0.46, 0.526), and it does not provide strong evidence against that belief.

<h3>What is a confidence interval of proportions?</h3>

A confidence interval of proportions is given by:

$\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}$

In which:

$\pi$ is the sample proportion.

z is the critical value.

n is the sample size.

In this problem, we have a 95% confidence level, hence $\alpha = 0.95$ , z is the value of Z that has a p-value of $\frac{1+0.95}{2} = 0.975$ , so the critical value is z = 1.96.

We have that a random sample of 864 births in a state included 426 boys, hence the parameters are given by:

$n = 864, \pi = \frac{426}{864} = 0.493$

Then, the bounds of the interval are given by:

$\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.493 - 1.96\sqrt{\frac{0.493(0.507)}{864}} = 0.46$

$\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.493 + 1.96\sqrt{\frac{0.493(0.507)}{864}} = 0.526$

The 95% confidence interval estimate of the proportion of boys in all births is (0.46, 0.526). Since the interval contains 0.506, it does not provide strong evidence against that belief.

More can be learned about the z-distribution at brainly.com/question/25890103

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2 years ago