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Marrrta [24]
3 years ago
14

Arrange the following and ascending and descending order first one is 1 by 12 1 by 23 1 by 5 1 by 7 1 by 50 1 by 9 1/17

Mathematics
1 answer:
Nonamiya [84]3 years ago
5 0

Answer:

Ascending order is given by

1/50<1/23<1/17<1/12<1/9<1/7<1/5

Descending order is given by

1/5>1/7>1/9>1/12>1/17>1/23>1/50

Step-by-step explanation:

We are given that

1/12,1/23,1/5,1/7,1/50,1/9,1/17

We have to arrange the numbers in ascending and descending order.

LCM of 12,23,5,7,50,9,17

23\times 5\times 12\times 7\times 50\times 17\times 9

=73899000

\frac{1}{12}\times \frac{6158250}{6158250}=\frac{6158250}{73899000}

\frac{1}{23}\times \frac{3213000}{3213000}=\frac{3213000}{73899000}

\frac{1}{5}\times \frac{14779800}{14779800}=\frac{14779800}{73899000}

\frac{1}{7}\times \frac{10557000}{10557000}=\frac{10557000}{73899000}

\frac{1}{50}\times \frac{1477980}{1477980}=\frac{1477980}{73899000}

\frac{1}{9}\times \frac{8211000}{8211000}=\frac{8211000}{73899000}

\frac{1}{17}\times \frac{4347000}{4347000}=\frac{4347000}{73899000}

14779800>10557000>8211000>6158250>4347000>3213000>1477980

Therefore,

Ascending order is given by

1/50<1/23<1/17<1/12<1/9<1/7<1/5

Descending order is given by

1/5>1/7>1/9>1/12>1/17>1/23>1/50

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The answer should be true
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A fourth grade class of 28 students is given a standardized math test. The mean score of the 12 boys is 25 with a standard devia
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Answer:

The standard error for the sampling distribution is 1.323.

Step-by-step explanation:

Let <em>X</em> = scores of girls and <em>Y</em> = scores of boys.

The information provided is:

\bar x=24\\s_{x}=4\\n_{x}=16\\\bar y=25\\s_{y}=3\\n_{y}=12

As the population standard deviations are not known, use a pooled standard deviation to estimate the standard error of the sampling distribution.

The formula of pooled standard deviation is:

S_{p}=\sqrt{\frac{s_{x}^{2}}{n_{x}^{2}}+\frac{s_{y}^{2}}{n_{y}^{2}}}

Compute the standard error for the sampling distribution as follows:

SE=\sqrt{\frac{s_{x}^{2}}{n_{x}^{2}}+\frac{s_{y}^{2}}{n_{y}^{2}}}=\sqrt{\frac{4^{2}}{16}+\frac{3^{2}}{12}}=\sqrt{1+\frac{9}{12}}=\sqrt{\frac{21}{12}}=1.323

Thus, the standard error for the sampling distribution is 1.323.

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3 years ago
According to a study of political​ prisoners, the mean duration of imprisonment for prisoners with chronic​ post-traumatic stres
Ray Of Light [21]

This question is incomplete, the complete question is;

According to a study of political​ prisoners, the mean duration of imprisonment for 33 prisoners with chronic​ post-traumatic stress disorder​ (PTSD) was 34.2 months. Assuming that σ = 40 ​months.

determine a 95​% confidence interval for the mean duration of​ imprisonment, ​μ , of all political prisoners with chronic PTSD. Interpret your answer in words.

Answer:

⇒ ( 20.6, 47.8 )

Therefore, 95% confidence interval for the mean duration of imprisonment μ of all political prisoners with PTSD is between 20.6 months and 47.8 months.

Step-by-step explanation:

Given the data in the question;

sample size; n = 33

standard deviation σ = 40 months

x' = 34.2 months

Now, at 95% confidence interval;

we know that z-value of 95% confidence interval is 1.96

so we substitute into the formula below;

⇒ x' ± Z( σ/√n )

⇒ 34.2 ± 1.96( 40/√33 )

⇒ 34.2 ± 13.647

so

we have

( 34.2 - 13.647 ), ( 34.2 + 13.647 )

⇒ ( 20.6, 47.8 )

Therefore, 95% confidence interval for the mean duration of imprisonment μ of all political prisoners with PTSD is between 20.6 months and 47.8 months.

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3 years ago
Suppose that two​ variables, X and​ Y, are negatively associated. Does this mean that​ above-average values of X will always be
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Answer:

If two​ variables, X and​ Y, are negatively associated, in a linear way, then the  above-average values of X will always be associated with​ below-average values of​ Y and vice-versa.

But not all points will fit the trend so the answer is NO, the above average values of X will not always be associated with below average values of Y.

Step-by-step explanation:

If two​ variables, X and​ Y, are negatively associated, in a linear way, then the  above-average values of X will always be associated with​ below-average values of​ Y and vice-versa.

But not all points will fit the trend so the answer is NO, the above average values of X will not always be associated with below average values of Y.

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a_sh-v [17]

Answer:

9 mile/ 1 hour

Step-by-step explanation:

Because the squirrel ran 9 miles in 1 hour, so u have to divide 9 miles by 1 hour.

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