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3 years ago

7

Một khối kim loại có thể tích 3,2cm3 và cân nặng 22,4g. Hỏi một khối kim loại cùng chất có thể tích là 4,5cm3 cân nặng bao nhiêu

gam ?

1 answer:

Ymorist [56]3 years ago

6 0

Answer:

Một cm3 của khối kim loại đó cân nặng số gam là:

22,4:3,2=7(g)

Một khối kim loại cùng chất có thể tích là 4,5cm3 cân nặng số gam là:

7x4,5=31,5(g)

Step-by-step explanation:

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3 years ago

A newspaper carrier has $1.80 in change. He has two more quarters than dimes but three times as many nickels as quarters. How ma

Mazyrski [523]

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3 years ago

Darren wants to estimate the mean age in a population of trees. He'll sample nnn trees and build a 90\%90%90, percent confidence

Alchen [17]

Using the z-distribution, as we have the standard deviation for the population, it is found that the smallest sample size required to obtain the desired margin of error is of 77.

<h3>What is a z-distribution confidence interval?</h3>

The confidence interval is:

$\overline{x} \pm z\frac{\sigma}{\sqrt{n}}$

In which:

$\overline{x}$ is the sample mean.

z is the critical value.

n is the sample size.

$\sigma$ is the standard deviation for the population.

The margin of error is given by:

$M = z\frac{\sigma}{\sqrt{n}}$

In this problem, we have that the parameters are given as follows:

$M = 3, z = 1.645, \sigma = 16$ .

Hence, solving for n, we find the sample size.

$M = z\frac{\sigma}{\sqrt{n}}$

$3 = 1.645\frac{16}{\sqrt{n}}$

$3\sqrt{n} = 1.645 \times 16$

$\sqrt{n} = \frac{1.645 \times 16}{3}$

$(\sqrt{n})^2 = \left(\frac{1.645 \times 16}{3}\right)^2$

$n = 76.97$

Rounding up, the smallest sample size required to obtain the desired margin of error is of 77.

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

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3 years ago