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3 years ago

Use rounding or compatible numbers to estimate the sum 271 + 425

Mathematics

2 answers:

Setler [38]3 years ago

6 0

I’m not sure how rounded you have to get but 271 would be 270 and 425 would be 430. 270+430=700

zysi [14]3 years ago

4 0

Answer:

700

Step-by-step explanation:

For this one I'll do estimates. 271 estimated would drop down to 270. And 425 would bump up to 430. Adding 270 and 400 you would get a round number of 700!

Hope this helped.

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4 years ago

A study tested whether significant social activities outside the house in young children affected their probability of later dev

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Answer:

Step-by-step explanation:

From the question we are told that

The first sample size is $n_1 = 1000$

The second sample size is $n_2 = 6000$

The number that had significant outside activity in the sample with ALL is $k_1 = 700$

The number that had significant outside activity in the sample without ALL is $k_2 = 5000$

Considering question a

The percentage of children with ALL have significant social activity outside the home when younger is mathematically represented as

$\^ p_1 = \frac{700}{1000} * 100$

=> $\^ p_ 1 = 0.7 = 70\%$

Considering question b

The percentage of children without ALL have significant social activity outside the home when younger is mathematically represented as

$\^ p_2 = \frac{5000}{6000}$

=> $\^ p_ 2 = 0.83$

Generally the sample odds ratio for significant social activity outside the home when younger, comparing the groups with and without ALL is mathematically represented as

$r = \frac{\* p _1}{ \^ p_2 }$

=> $r = \frac{0.7}{ 0.83 }$

=> $r = 0.141$

Considering question c

From the question we are told the confidence level is 95% , hence the level of significance is

$\alpha = (100 - 95 ) \%$

=> $\alpha = 0.05$

Generally from the normal distribution table the critical value of $\frac{\alpha }{2}$ is

$Z_{\frac{\alpha }{2} } = 1.96$

Generally the lower limit of the 95% confidence interval for the population odds ratio for significant social activity outside the home when younger, comparing the groups with and without ALL is mathematically represented as

$a = e^{ln ( r ) - Z_{\frac{\alpha }{2}} \sqrt{ [ \frac{1}{ k_1 } ] + [ \frac{1}{ c_1 } ] + [\frac{1}{k_2} ] + [\frac{1}{ c_2 } ] } }$

Here $c_1 \ and \ c_2$ are the non-significant values i.e people that did not play outside when they were young in both samples

The values are

$c_1 = 1000 - 700 = 300$

and $c_2 = 6000 - 5000$

=> $c_2 = 1000$

=> $a = e^{ln ( 0.141 ) - 1.96 \sqrt{ [ \frac{1}{ 700 } ] + [ \frac{1}{ 1000} ] + [\frac{1}{5000} ] + [\frac{1}{ 300 } ] } }$

=> $a = 0.1212$

Generally the upper limit of the 95% confidence interval for the population odds ratio for significant social activity outside the home when younger, comparing the groups with and without ALL is mathematically represented as

$b = e^{ln ( 0.141 ) + 1.96 \sqrt{ [ \frac{1}{ 700 } ] + [ \frac{1}{ 1000} ] + [\frac{1}{5000} ] + [\frac{1}{ 300 } ] } }$

$b = 0.1640$

Generally the 95% confidence interval for the population odds ratio for significant social activity outside the home when younger, comparing the groups with and without ALL is

$95\% CI = [ 0.1212 , 0.1640 ]$

Generally looking and the confidence interval obtained we see that it is less that 1 hence this means that there is a greater odd of developing ALL in groups with insignificant social activity compared to groups with significant social activity

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