Answer:
<em>Answer is option d</em><em> </em>
<em>Answer is </em><em>given below with explanations</em><em>. </em>
Step-by-step explanation:
We can prove that the two triangles are similar.
We can prove this using AA criterion of similarity.
In triangle DNC and triangle QSC
Vertically opposite angles are equal.
Then Angle QCS = Angle DCN
Two parallel lines cut by a transversal line make the alternate angles are equal.
Then Angle NDC = Angle CQS
By AA criterion of similarity
TRIANGLE DNC ~ TRIANGLE QSC
<em>HAVE A NICE DAY</em><em>!</em>
<em>THANKS FOR GIVING ME THE OPPORTUNITY</em><em> </em><em>TO ANSWER YOUR QUESTION</em><em>. </em>
Answer:
False
Step-by-step explanation:
2x + y = 0
Try (1, -2)
2(1) + (-2) = 0
2 + (-2) = 0
0 = 0
The solution works on the first equation.
-x + 2y = 5
-(1) + 2(-2) = 5
-1 - 4 = 5
-5 = 5
The solution does not work in the second equation.
Answer: False
Given:
30-hour review course average a score of 620 on that exam.
70-hour review course average a score of 749.
To find:
The linear equation which fits this data, and use this equation to predict an average score for persons taking a 57-hour review course.
Solution:
Let x be the number of hours of review course and y be the average score on that exam.
30-hour review course average a score of 620 on that exam. So, the linear function passes through the point (30,620).
70-hour review course average a score of 749. So, the linear function passes through the point (70,749).
The linear function passes through the points (30,620) and (70,749). So, the linear equation is:
Adding 620 on both sides, we get
We need to find the y-value for 
.
Therefore, the required linear equation for the given situation is and the average score for persons taking a 57-hour review course is 707.1.
Answer: it is the 4th graph
Step-by-step explanation:
(-3,25)(0,20)(3,16)(6,12.8)
Answer:
We accept the null hypothesis that the breaking strength mean is less and equal to 1750 pounds and has not increased.
Step-by-step explanation:
The null and alternative hypotheses are stated as
H0: u ≥ 1750 i.e the mean is less and equal to 1750
against the claim
Ha: u > 1750 ( one tailed test) the mean is greater than 1750
Sample mean = x`= 1754
Population mean = u = 1750
Population deviation= σ = 65 pounds
Sample size= n = 100
Applying the Z test
z= x`- u / σ/ √n
z= 1754- 1750 / 65/ √100
z= 4/6.5
z= 0.6154
The significance level alpha = 0.1
The z - value at 0.1 for one tailed test is ± 1.28
The critical value is z > z∝.
so
0.6154 is < 1.28
We accept the null hypothesis that the breaking strength mean is less and equal to 1750 pounds and has not increased.
