In kite UVWX, mZXUV = 84º, and mZWVX = 68°. What is mZVWX?
w
H 44°
F 22°
G 42°
J 45°

your answer will be H 44°

Step-by-step explanation:

ΔWVX

∠WVX=68°

∠WVX=∠WXV(two sides are equal)

now in ΔVWX-

sum of angles=360

60+60+VWX=180

∠VWX=180-136

=44°

hope it helps

have a great day!!

The calculated z = 2.85 falls in the critical region z > 1.96 so we accept the null hypothesis that there is no difference between the population proportions for the two years.

Step-by-step explanation:

Let p1 be the sample of 1000 adults chosen in 2012 and

p2 be the sample of 1100 adults chosen in 2010

Part a:

The hypothesis that can be used to test for a significant difference between the population proportions for the two years is:

H0: p1-p2= 0           against the claim      Ha: p1-p2≠0

Part b:

The sample proportion indicating that their financial security was more that fair in 2012 is

p1^= 410/1000= 0.41

In 2010 is:

p2^= 385/1100= 0.35

Part c:

p^c= 410+385/1000+1100= 0.3785

q^c= 1-p^c= 1-0.3785= 0.6124

The critical region is z > ±1.96

The test statistic is

z= p1^- p2^ / sqrt ( p^cq^c( 1/n1+ 1/n2)

z= 0.41-0.35 /sqrt( 0.3785*0.6124(1/1000 +1/1100)

z=  0.06 / 0.021036

z= 2.85

Conclusion

The calculated z = 2.85 falls in the critical region z > 1.96 so we accept the null hypothesis that there is no difference between the population proportions for the two years.

P- value :  0.00466

The result is significant( accept the null hypothesis) for  value less than 0.05

Part d:

                                                     2012                         2010

Sample                                       1000                           1100

Proportions                               0.41                             0.35

Standard Error               √0.41( 0.59)/1000        √0.35( 0.65)/1100

                                               = 0.0155                0.01438

Standard Error for the difference = √0.0155² + 0.01438² =0.0211

p1-p2 ± z* standard error for the difference

0.41-0.35 ± 1.96 *( 0.0211)

0.06 ± 0.041356

-0.018644,   0.101356

The 95 % confidence interval estimate is (-0.018644,   0.10135