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nadezda [96]
2 years ago
11

What is the hundred thousand place in 201,569​

Mathematics
1 answer:
PolarNik [594]2 years ago
4 0
It would be the ( 2 )01,569. The answer would be two.
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He percent distribution of the number of grandchildren for a sample of 1904 grandparents is shown in the pie chart. find each pr
jekas [21]

down voteaccepted

P(X=1) = 0.09

P(X < 5) = 0.09 + 0.36

P(X >= 2) = .36 + .23 + .09 + .23

P(2 <= X <= 7) = .36 + .23

7 0
3 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 <em>μ</em> and standard deviation <em>σ</em> and we take appropriately huge random samples (<em>n</em> ≥ 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, <em>n</em>.

So, if <em>n</em> 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 <em>n</em> from 64 to 196, the standard deviation of the sample mean will decrease.

The standard deviation of the sample mean for <em>n</em> = 64 is:

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

The standard deviation of the sample mean for <em>n</em> = 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 <em>n</em> 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 <em>n</em> from 784 to 49, the standard deviation of the sample mean will increase.

The standard deviation of the sample mean for <em>n</em> = 784 is:

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

The standard deviation of the sample mean for <em>n</em> = 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 <em>n</em> is decreased from 784 to 49.

Hence, the variance also increases.

6 0
3 years ago
What are the degree and leading coefficient of the polynomial 4u^8 - 12 - 20u^2 - 2u^4
sasho [114]

Answer:

degree is 8 and leading coefficient is 4

Step-by-step explanation:

The degree is highest power or exponent in this case its 8.

The leading coefficient is 4 because that is the coefficient it has to do with the highest power.

Hope this helps

4 0
3 years ago
P(AB)
Anna71 [15]
The answer is 10/1
There are 10 boys and 1 of them is a sophomore
4 0
3 years ago
Need by 8:25AM please help me
olya-2409 [2.1K]

Step-by-step explanation:

42.75? 285/15=19, 285*0.19=42.75

4 0
3 years ago
Read 2 more answers
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