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3 years ago

I need the answer ;(

Mathematics

2 answers:

antoniya [11.8K]3 years ago

6 0

Answer:

x=100

Step-by-step explanation:

Because they are vertical angles, AEG=FEB

You can see that FED+DEB=FEB

Therefore, 80+20=FEB=100

Since FEB=100 and FEB=AEG

AEG=100

Blizzard [7]3 years ago

5 0

Answer:

100°

Step-by-step explanation:

In this question, vertical angles are key. If you didn't know, vertical angles are basically opposites of each other. A pair of vertical angles are equal.

ACE = 20°

FED = 80°

FED and CEG are vertical angles (+the 20 from ACE), so your answer is 100°

hope this helps, brainliest is appreciated.

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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

Consider the following. C: counterclockwise around the triangle with vertices (0, 0), (1, 0), and (0, 1), starting at (0, 0)

horsena [70]

Answer:

$\mathbf{r_1 = (t,0) \implies t = 0 \ to \ 1}$

$\mathbf{r_2 = (2-t,t-1) \implies t = 1 \ to \ 2}$

$\mathbf{r_3 = (0,3-t) \implies t = 2 \ to \ 3}$

$\mathbf{\int \limits _{c} F \ dr =\dfrac{11 \sqrt{2}+11}{6}}$

Step-by-step explanation:

Given that:

C: counterclockwise around the triangle with vertices (0, 0), (1, 0), and (0, 1), starting at (0, 0)

a. Find a piecewise smooth parametrization of the path C.

r(t) = { 0

If C: counterclockwise around the triangle with vertices (0, 0), (1, 0), and (0, 1),

Then:

$C_1 = (0,0) \\ \\ C_2 = (1,0) \\ \\ C_3 = (0,1)$

Also:

$\mathtt{r_1 = (0,0) + t(1,0) = (t,0) }$

$\mathbf{r_1 = (t,0) \implies t = 0 \ to \ 1}$

$\mathtt{r_2 = (1,0) + t(-1,1) = (1- t,t) }$

$\mathbf{r_2 = (2-t,t-1) \implies t = 1 \ to \ 2}$

$\mathtt{r_3 = (0,1) + t(0,-1) = (0,1-t) }$

$\mathbf{r_3 = (0,3-t) \implies t = 2 \ to \ 3}$

b Evaluate :

Integral of (x+2y^1/2)ds

$\mathtt{\int \limits ^1_{c1} (x+ 2 \sqrt{y}) ds = \int \limits ^1_{0} \ (t + 0) \sqrt{1} } \\ \\ \mathtt{ \int \limits ^1_{c1} (x+ 2 \sqrt{y}) ds = \begin {pmatrix} \dfrac{t^2}{2} \end {pmatrix} }^1_0 \\ \\ \mathtt{\int \limits ^1_{c1} (x+ 2 \sqrt{y}) ds = \dfrac{1}{2}}$

$\mathtt{\int \limits _{c2} (x+ 2 \sqrt{y}) ds = \int \limits (x+2 \sqrt{y} \sqrt{(\dfrac{dx}{dt})^2 + (\dfrac{dy}{dt})^2 \ dt } }$

$\mathtt{\int \limits _{c2} (x+ 2 \sqrt{y}) ds = \int \limits 2- t + 2\sqrt{t-1} \ \sqrt{1+1} }$

$\mathtt{\int \limits _{c2} (x+ 2 \sqrt{y}) ds = \sqrt{2} \int \limits^2_1 2- t + 2\sqrt{t-1} \ dt }$

$\mathtt{\int \limits _{c2} (x+ 2 \sqrt{y}) ds = \sqrt{2} } \ \begin {pmatrix} 2t - \dfrac{t^2}{2}+ \dfrac{2(t-1)^{3/2}}{3} (2) \end {pmatrix} ^2_1}$

$\mathtt{\int \limits _{c2} (x+ 2 \sqrt{y}) ds = \sqrt{2} } \ \begin {pmatrix} 2 -\dfrac{1}{2} (4-1)+\dfrac{4}{3} (1)^{3/2} -0 \end {pmatrix} }$

$\mathtt{\int \limits _{c2} (x+ 2 \sqrt{y}) ds = \sqrt{2} } \ \begin {pmatrix} 2 -\dfrac{3}{2} + \dfrac{4}{3} \end {pmatrix} }$

$\mathtt{\int \limits _{c2} (x+ 2 \sqrt{y}) ds = \sqrt{2} } \ \begin {pmatrix} \dfrac{12-9+8}{6} \end {pmatrix} }$

$\mathtt{\int \limits _{c2} (x+ 2 \sqrt{y}) ds = \sqrt{2} } \ \begin {pmatrix} \dfrac{11}{6} \end {pmatrix} }$

$\mathtt{\int \limits _{c2} (x+ 2 \sqrt{y}) ds = \dfrac{ \sqrt{2} }{6} \ (11 )}$

$\mathtt{\int \limits _{c2} (x+ 2 \sqrt{y}) ds = \dfrac{ 11 \sqrt{2} }{6}}$

$\mathtt{\int \limits _{c3} (x+ 2 \sqrt{y}) ds = \int \limits ^3_2 0+2 \sqrt{3-t} \ \sqrt{0+1} }$

$\mathtt{\int \limits _{c3} (x+ 2 \sqrt{y}) ds = \int \limits ^3_2 2 \sqrt{3-t} \ dt}$

$\mathtt{\int \limits _{c3} (x+ 2 \sqrt{y}) ds = \int \limits^3_2 \begin {pmatrix} \dfrac{-2(3-t)^{3/2}}{3} (2) \end {pmatrix}^3_2 }$

$\mathtt{\int \limits _{c3} (x+ 2 \sqrt{y}) ds = -\dfrac{4}{3} [(0)-(1)]}$

$\mathtt{\int \limits _{c3} (x+ 2 \sqrt{y}) ds = -\dfrac{4}{3} [-(1)]}$

$\mathtt{\int \limits _{c3} (x+ 2 \sqrt{y}) ds = \dfrac{4}{3}}$

$\mathtt{\int \limits _{c} F \ dr =\dfrac{11 \sqrt{2}}{6}+\dfrac{1}{2}+ \dfrac{4}{3}}$

$\mathtt{\int \limits _{c} F \ dr =\dfrac{11 \sqrt{2}+3+8}{6}}$

$\mathbf{\int \limits _{c} F \ dr =\dfrac{11 \sqrt{2}+11}{6}}$

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