Question in picture solve

kumpel [21]

If everything on inside of a line is 180->
(Line) --------/---------- everything on top equals 180 and there should be a number on either side. For instance 180-50 for the top would be 130, the wide angle's angle. I answer everything I like and come across so sorry I am four weeks late.

4 0

3 years ago

WILL GIVE BRAINLIEST Find the scale factor

Sloan [31]

Please I’m in 5th grade I tried but it’s too hard but please mark me Brainlyest

7 0

3 years ago

Read 2 more answers

Point E is located at (2, −3), and point F is located at (−2, −1). Find the point that is 3 over 4 the distance from point E to

irakobra [83]

It woud be very far.Explain how you would graph the line containing a slope of –1/5 that goes through the point (1,–4).

3 0

3 years ago

given that a test statistic in a left tailed test is z equals -1.25, use a 0.05 significance level to find the p-value and state

mr Goodwill [35]

Answer:

The p-value of the test is 0.106.

The null hypothesis will be accepted at 5% level of significance.

Step-by-step explanation:

The hypothesis test is left-tailed.

The test statistic value is: z = -1.25.

The significance level of the test is: α = 0.05.

The p-value of a left-tailed hypothesis test is:

$p-value=P(Z$

The p-value of the test is 0.106.

**Use the z-table for the probability.

Decision rule:

If the p-value is less than the significance level the null hypothesis will be rejected and if it is more than the significance level the null hypothesis will be accepted.

The p-value = 0.106 > α = 0.05.

The null hypothesis will be accepted at 5% level of significance.

5 0

4 years ago

A simple random sample of 60 households in city 1 is taken. In the sample, there are 45 households that decorate their houses wi

murzikaleks [220]

Answer:

The calculated value of z = - 0.197 falls in the critical region therefore we reject the null hypothesis and conclude that at the 5% significance level there is significant difference in population proportions of households that decorate their houses with lights for the holidays

Step-by-step explanation:

We formulate the null and alternative hypotheses as

H0: p1= p2 there is no difference in population proportions of households that decorate their houses with lights for the holidays

against Ha : p1≠ p2 (claim) ( two sided)

The significance level is set at ∝= 0.05

The critical value for two tailed test at alpha=0.05 is ± 1.96

or Z∝= 0.05/2= ± 1.96

The test statistic is

Z = p1-p2/√pq(1/n1 +1/n2)

p1= proportions of households decorating in city 1 = 45/60=0.75

p2= proportions of households decorating in city 2 = 40/50= 0.8

p = the common proportion on the assumption that the two proportion are same.

p = $\frac{n_1p_1 +n_2p_2}{n_1+n_2}$

Calculating

p =60 (0.75) + 50 (0.8) / 110

p= 45+ 40/110= 85/110 = 0.772

so q = 1-p= 1- 0.772= 0.227

Putting the values in the test statistic and calculating

z= 0.75- 0.8/ √0.772*0.227( 1/60 + 1/50)

z= -0.05/√ 0.175244 ( 110/300)

z= -0.05/0.25348

z= -0.197

The calculated value of z = - 0.197 falls in the critical region therefore we reject the null hypothesis and conclude that at the 5% significance level there is significant difference in population proportions of households that decorate their houses with lights for the holidays

4 0

3 years ago