It would be (A) , the point 9 (38.9) only needs another point to be a whole tenth .
Answer: The correct congruence statement is .
Explanation:
It is given that A triangle Q D J. The base D J is horizontal and side Q D is vertical. Another triangle M C W is made. The base C W and side M C are neither horizontal nor vertical. Triangle M C W is to the right of triangle Q D J.
The sides Q D and M C are labeled with a single tick mark. The sides D J and C W are labeled with a double tick mark. The sides Q J and M W are labeled with a triple tick mark.
Draw two triangles according to the given information.
From the figure it is noticed that
So by SSS rule of congruence we can say that .
The test statistic value lies to the right of the critical value. So we have sufficient evidence to reject the null hypothesis.
<h3>What are null hypotheses and alternative hypotheses?</h3>
In null hypotheses, there is no relationship between the two phenomenons under the assumption or it is not associated with the group. And in alternative hypotheses, there is a relationship between the two chosen unknowns.
Students who take a statistics course are given a pre-test on the concepts and skills for the first chapter of a statistics course.
Then they are given a post-test once the professor has concluded lecturing on the material.
Pre-test and post-test scores for 4 students in an elementary statistics class are given below.
Then we have
Then the null hypotheses and alternative hypotheses will be
H₀:
Hₐ:
Then the test statistic will be
Then
The test statistic value is given by
Since this is a right-tailed test, so the critical value is given by
Since the test statistic value lies to the right of the critical value. So we have sufficient evidence to reject the null hypothesis.
Hence, we can conclude that that is test scores have improved.
More about the null hypotheses and alternative hypotheses link is given below.
brainly.com/question/9504281
#SPJ1
Step-by-step explanation:
<em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em>
<em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em>here's</em><em> your</em><em> solution</em>
<em> </em><em>=</em><em>></em><em> </em><em>in </em><em>first </em><em>figure</em><em> </em><em>,</em><em> </em><em>base </em><em>=</em><em> </em><em>5</em><em>.</em><em>5</em><em>,</em><em> </em><em>perpendicular</em><em> </em><em>=</em><em>7</em><em>.</em><em>8</em>
<em>=</em><em>></em><em> </em><em>h^</em><em>2</em><em> </em><em>=</em><em> </em><em>5</em><em>.</em><em>5</em><em>^</em><em>2</em><em> </em><em>+</em><em> </em><em>7</em><em>.</em><em>8</em><em>^</em><em>2</em>
<em>=</em><em>></em><em> </em><em>h^</em><em>2</em><em> </em><em>=</em><em> </em><em>3</em><em>0</em><em>.</em><em>2</em><em>5</em><em> </em><em>+</em><em> </em><em>6</em><em>0</em><em>.</em><em>8</em><em>4</em><em> </em>
<em>=</em><em>></em><em>h^</em><em>2</em><em> </em><em>=</em><em> </em><em>9</em><em>1</em><em>.</em><em>0</em><em>9</em>
<em>=</em><em>></em><em> </em><em>h </em><em>=</em><em> </em><em>√</em><em>9</em><em>1</em><em>.</em><em>0</em><em>9</em>
<em>=</em><em>></em><em> </em><em>h </em><em>=</em><em> </em><em>9</em><em>.</em><em>5</em>
<em> </em><em> </em><em>Both </em><em>figure</em><em> </em><em>are </em><em>congruent</em>
<em>enc </em><em>we </em><em>will </em><em>get </em><em>a </em><em>rectangle</em><em> </em><em>by </em><em>add </em><em>both </em>
<em>hope</em><em> it</em><em> helps</em>