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hoa [83]
3 years ago
10

HELP ME PLEASE PLEASE I NEED HELP

Mathematics
1 answer:
ryzh [129]3 years ago
7 0

Answer:

I guess the answer is 3 million on a number line

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A study tested whether significant social activities outside the house in young children affected their probability of later dev
Alona [7]

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

3 0
3 years ago
4 (2x - 8) + 5 (x + 4)
ASHA 777 [7]

Answer:

???

Step-by-step explanation:

6 0
3 years ago
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Write as a product 4x^4+8x^3-2x^2
TiliK225 [7]

Answer:

764x

Step-by-step explanation:

4^4=256

8^3=512

2^2=4

256-4+512=764

764x

3 0
3 years ago
What is the image point of (-1, 1) after a translation right 2 units and<br> down 1 unit?
hram777 [196]
1,0 should be right :)
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What is the estimate value of 976 divide by 0.188
RUDIKE [14]

Answer:

Estimating is used to help predict the answer to a calculation. We use estimation to make calculations more manageable and to give us an idea of the answer. Estimation is used in the construction industry, for example, when buying materials and planning labour.

6 0
3 years ago
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