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2 years ago

I need to write a decimal where the value of a digit is 1/10 as much as a digit in another place

Mathematics

1 answer:

KengaRu [80]2 years ago

6 0

1/10 is the same as saying 10 times less

Refer to the diagram below
The digit as the first decimal place is worth 1/10
The digit as the second decimal place is worth 1/100 which is 1/10 worth the digit as the first decimal place

Example of two decimal numbers:

2.56 and 2.68

The digit 6 in 2.56 is 6/100
The digit 6 in 2.68 is 6/10

The digit 6 in 2.56 is 1/10 as much as the digit 6 in 2.68

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2 years ago

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2 years ago

A certain type of thread is manufactured with a mean tensile strength of 78.3 kilograms and a standard deviation of 5.6 kilogram

azamat

Answer:

(a) The variance decreases.

(b) The variance increases.

Step-by-step explanation:

According to the Central Limit Theorem if we have a population with mean μ and standard deviation σ and we take appropriately huge random samples (n ≥ 30) from the population with replacement, then the distribution of the sample mean will be approximately normally distributed.

Then, the mean of the sample mean is given by,

$\mu_{\bar x}=\mu$

And the standard deviation of the sample mean is given by,

$\sigma_{\bar x}=\frac{\sigma}{\sqrt{n}}$

The standard deviation of sample mean is inversely proportional to the sample size, n.

So, if n increases then the standard deviation will decrease and vice-versa.

(a)

The sample size is increased from 64 to 196.

As mentioned above, if the sample size is increased then the standard deviation will decrease.

So, on increasing the value of n from 64 to 196, the standard deviation of the sample mean will decrease.

The standard deviation of the sample mean for n = 64 is:

$\sigma_{\bar x}=\frac{\sigma}{\sqrt{n}}=\frac{5.6}{\sqrt{64}}=0.7$

The standard deviation of the sample mean for n = 196 is:

$\sigma_{\bar x}=\frac{\sigma}{\sqrt{n}}=\frac{5.6}{\sqrt{196}}=0.4$

The standard deviation of the sample mean decreased from 0.7 to 0.4 when n is increased from 64 to 196.

Hence, the variance also decreases.

(b)

If the sample size is decreased then the standard deviation will increase.

So, on decreasing the value of n from 784 to 49, the standard deviation of the sample mean will increase.

The standard deviation of the sample mean for n = 784 is:

$\sigma_{\bar x}=\frac{\sigma}{\sqrt{n}}=\frac{5.6}{\sqrt{784}}=0.2$

The standard deviation of the sample mean for n = 49 is:

$\sigma_{\bar x}=\frac{\sigma}{\sqrt{n}}=\frac{5.6}{\sqrt{49}}=0.8$

The standard deviation of the sample mean increased from 0.2 to 0.8 when n is decreased from 784 to 49.

Hence, the variance also increases.

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2 years ago

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Answer:

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Step-by-step explanation:

I mean, it says it in the question.

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