Solve the system of equations.What is x and y −2x+15y=−24 2x+9y=24
2 answers:
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Answer:
Step-by-step explanation:
The summary of the given statistics data include:
sample size n = 400
sample mean = 6.86
standard deviation = 4.37
Level of significance ∝ = 0.01
Population Mean = 6.00
Assume that a simple random sample has been selected. Identify the null and alternative hypotheses, test statistic, P-value, and state the final conclusion that addresses the original claim.
To start with the hypothesis;
The null and the alternative hypothesis can be computed as :
The test statistics for this two tailed test can be computed as:
z = 3.936
degree of freedom = n - 1
degree of freedom = 400 - 1
degree of freedom = 399
At the level of significance ∝ = 0.01
P -value = 2 × (z < 3.936) since it is a two tailed test
P -value = 2 × ( 1 - P(z ≤ 3.936)
P -value = 2 × ( 1 -0.9999)
P -value = 2 × ( 0.0001)
P -value = 0.0002
Since the P-value is less than level of significance , we reject at level of significance 0.01
Conclusion: There is sufficient evidence to conclude that the original claim that the mean of the population of earthquake depths is 5.00 km.
Answer:
We conclude that the proportion of dropouts has changed from the historical value of 0.081.
Step-by-step explanation:
We are given that in 2009, the high school dropout rate was 8.1%. A polling company recently took a survey of 1000 people between the ages of 16 and 24 and found 6.5% of them are high school dropouts.
The polling company would like to determine whether the proportion of dropouts has changed from the historical value of 0.081.
<em>Let p = proportion of school dropouts rate</em>
SO, <u>Null Hypothesis,</u> : p = 0.081 {means that the proportion of dropouts has not changed from the historical value of 0.081}
<u>Alternate Hypothesis</u>, : p
0.081 {means that the proportion of dropouts has changed from the historical value of 0.081}
The test statistics that will be used here is <u>One-sample z proportion statistics</u>;
T.S. = ~ N(0,1)
where, = sample proportion of high school dropout rate = 6.5%
n = sample of people = 1000
So, <u><em>test statistics</em></u> =
= -2.05
<u>Also, P-value is given by the following formula;</u>
P-value = P(Z < -2.05) = 1 - P(Z 2.05)
= 1 - 0.97982 = <u>0.0202</u> or 2.02%
<em>Now at 5% significance level, the z table gives critical values between -1.96 and 1.96 for two-tailed test. Since our test statistics does not lies within the range of critical values of z so we have sufficient evidence to reject our null hypothesis as it will fall in the rejection region.</em>
Therefore, we conclude that the proportion of dropouts has changed from the historical value of 0.081.