All categories

3 years ago

Which of the following inequalities is best represented by this graph?

Mathematics

1 answer:

Nataly [62]3 years ago

5 0

For this case, the first thing to do is find the equation of the line correctly and then see which of the inequalities represents the shaded region.
We have then:
5x - y = 2
For x = 0
5 (0) - y = 2
-y = 2
y = -2
For y = 0
5x - 0 = 2
5x = 2
x = 2/5 = 0.4
Therefore the line is:
5x - y = 2
Then, we see that for the point (0, 0)
5x - y ≥ 2
5 (0) - (0) ≥ 2
0 ≥ 2
The inequality is not met, which is correct because this point is not part of the shaded region.
Answer:
The inequality that best is represented by the graph if:
5x - y ≥ 2

You might be interested in

Simplify cos theta csc theta

mihalych1998 [28]

Answer:

cot theta

Step-by-step explanation:

t = theta

csc t = 1/ sin t

cos t . csc t = cos t / sin t = cot t

5 0

4 years ago

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

Pls help zoom in if u can’t see

ale4655 [162]

Answer:

Step-by-step explanation:

because heat transfer from hot to cold but it does not transfer to something hotter or the same exact temp

5 0

3 years ago

Read 2 more answers

The perimeter of a trapezoid is 39a – 7. Three sides have the following lengths: 9a, 5a + 1, and 17a – 6. What is the length of

Airida [17]

We know the perimeter is the total sum of all sides of the shape, we can use this to form an equation: the perimeter=1st side + 2nd side + 3rd side + 4th side 39a-7=9a+(5a+1)+(17a-6)+4th side.
Collect like terms to get 39a-7=31a-5+4th side
If we move everything but the 4th side over to the left we can solve
C. 8a-2=4th side is the correct answer

3 0

3 years ago

Read 2 more answers

A car traveling at 22.4 m/s skids to a stop in 2.55 s. Determine the skidding distance of

Marta_Voda [28]

Answer:

57.12 m

Step-by-step explanation:

velocity= distance/time

distance= velocity×time

=22.4×2.55

=57.12 m

7 0

3 years ago