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3 years ago

What is the value of 4 in the number 2743 in a fraction

Mathematics

2 answers:

stellarik [79]3 years ago

8 0

Answer:

40 is the answer as the sum below shows the .2743 workings equal to .2743/1 as fraction that equates to 2743/10000

Step-by-step explanation:

To find a percentage with the given number you would always need to use the .2743/1 .2743/1 x 10 or 2.743/10 or 27.43/100 etc.... When in doubt always put 1 below if we want a new fraction., we only use a 1 above to balance up a relation to another fraction or its meaning. .2743/1 your given number/Find fraction. Value of 4 therefore can only be seen as a completed fraction.

Contact [7]3 years ago

5 0

Step-by-step explanation:

Since, 4 is at ten's place in the number 2743.

Therefore, the value of 4 in the number 2743 = 4 * 10 = 40

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Arte-miy333 [17]

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3 years ago

Assume a simple random sample of 10 BMIs with a standard deviation of 1.186 is selected from a normally distributed population o

kirza4 [7]

Answer:

a) H0: $\sigma = 1.34$

H1: $\sigma \neq 1.34$

b) $df = n-1= 10-1=9$

And the critical values with $\alpha/2=0.005$ on each tail are:

$\chi_{\alpha/2}= 1.735, \chi_{1-\alpha/2}= 23.589$

c) $t=(10-1) [\frac{1.186}{1.34}]^2 =7.05$

d) For this case since the critical value is not higher or lower than the critical values we have enough evidence to FAIL to reject the null hypothesis and we can conclude that the true deviation is not significantly different from 1.34

Step-by-step explanation:

Information provided

n = 10 sample size

s= 1.186 the sample deviation

$\sigma_o =1.34$ the value that we want to test

$p_v$ represent the p value for the test

t represent the statistic (chi square test)

$\alpha=0.01$ significance level

Part a

On this case we want to test if the true deviation is 1,34 or no, so the system of hypothesis are:

H0: $\sigma = 1.34$

H1: $\sigma \neq 1.34$

The statistic is given by:

$t=(n-1) [\frac{s}{\sigma_o}]^2$

Part b

The degrees of freedom are given by:

$df = n-1= 10-1=9$

And the critical values with $\alpha/2=0.005$ on each tail are:

$\chi_{\alpha/2}= 1.735, \chi_{1-\alpha/2}= 23.589$

Part c

Replacing the info we got:

$t=(10-1) [\frac{1.186}{1.34}]^2 =7.05$

Part d

For this case since the critical value is not higher or lower than the critical values we have enough evidence to FAIL to reject the null hypothesis and we can conclude that the true deviation is not significantly different from 1.34

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2 years ago