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3 years ago

What is 251,723 in expanded form

Mathematics

2 answers:

mel-nik [20]3 years ago

6 0

Wow i havent done this since 4th grade, it feels so weird doing it again :)
200,000+ 50,000+ 1,000 + 700 + 20 + 3

inysia [295]3 years ago

4 0

Answer:

hope this helps :)

Step-by-step explanation:

200,000 +500,000+1,000+700+20+3

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In the seventh grade, Rachel read 15 books.

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Answer:

20%

Step-by-step explanation:

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A soccer team won only 10 of their first 30 games. How many of their remaining 20 games will they have to win to have a .500 rec

jeka94

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1 year ago

6x^2-2x+1=0 solve for x

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3 years ago

The fraction of defective integrated circuits produced in a photolithography process is being studied. A random sample of 300 ci

Olenka [21]

Answer:

The correct answer is

(0.0128, 0.0532)

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of $\pi$ , and a confidence interval $1-\alpha$ , we have the following confidence interval of proportions.

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

In which

Z is the zscore that has a pvalue of $1 - \frac{\alpha}{2}$

For this problem, we have that:

In a random sample of 300 circuits, 10 are defective. This means that $n = 300$ and $\pi = \frac{10}{300} = 0.033$

Calculate a 95% two-sided confidence interval on the fraction of defective circuits produced by this particular tool.

So $\alpha$ = 0.05, z is the value of Z that has a pvalue of $1 - \frac{0.05}{2} = 0.975$ , so $Z = 1.96$ .

The lower limit of this interval is:

$\pi - z\sqrt{\frac{\pi(1-\pi)}{300}} = 0.033 - 1.96\sqrt{\frac{0.033*0.967}{300}} = 0.0128$

The upper limit of this interval is:

$\pi + z\sqrt{\frac{\pi(1-\pi)}{300}} = 0.033 + 1.96\sqrt{\frac{0.033*0.967}{300}} = 0.0532$

The correct answer is

(0.0128, 0.0532)

4 0

2 years ago