What is an equation of the line that passes through the points (4, -1) and

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

You can jog around the block twice and the park once in 10 min. You can jog around the block twice and the park 3 times in 22 mi

stich3 [128]

B = block p = park

2b + p = 10 min 2 runs around the block takes 10 minutes
2b + 3p = 22 min 2 runs around the block plus 3 runs around the park takes 22 minutes

To solve, I will use substitution. First, I will divide the first equation by 2 to get b by itself and subtract p from both sides to get b by itself. Then, substitute 5 - 1/2p in for b in the second equation because they have equal values. Then, subtract 10 from both sides to eliminate it. Divide by 2 to both sides to get 'p' by itself.

2b + p = 10 min ---------→b + 1/2p = 5 ----------→ b = 5 - 1/2p

2(5 - 1/2p) + 3p = 22 ---------→ 10 -p + 3p = 22 --------→ 10 + 2p = 22

10 + 3p = 22 ----------→ 2p = 12

2p = 12 -------------→ p = 6

It takes 6 minutes to jog around the park

Hope this helps!

6 0

3 years ago

which expression is equivalent to 2m(3/2m+1) +3 (5/3m-2) ? A .3m^2+5m-1 B. 3/4m^2 +23/9m-6 C. 3m^2+7m-6 D. 3/4m^2+5/9m-1 please

horrorfan [7]

Answer:

66

Step-by-step explanation:

5 0

4 years ago

Read 2 more answers

10%of what is 14 i need to find the whole number

zlopas [31]

What makes a band a band?

3 0

3 years ago

A story sells boxes of juice in equal-size packs. Ava bought 18 boxes, Katie bought 36 boxes, and Alexa bought 45 boxes. What is

Bas_tet [7]

Answer:

9

Step-by-step explanation:

Find the factors of each number

18 - 1, 2, 3, 6, 9, 18

36 - 1, 2, 3, 4, 6, 9, 12, 18, 36

45 - 1, 5, 9, 45

The highest common factor is 9. So there is 9 boxes in a pack.

6 0

3 years ago