The length of a segment is the distance between its endpoints.

- AB and CD are not congruent
- AB does not bisect CD
- CD does not bisect AB
<u>(a) Length of AB</u>
We have:


The length of AB is calculated using the following distance formula

So, we have:


Simplify

<u>(b) Are AB and CD congruent</u>
First, we calculate the length of CD using:

Where:


So, we have:



By comparison

Hence, AB and CD are not congruent
<u>(c) AB bisects CD or not?</u>
If AB bisects CD, then:

The above equation is not true, because:

Hence, AB does not bisect CD
<u>(d) CD bisects AB or not?</u>
If CD bisects AB, then:

The above equation is not true, because:

Hence, CD does not bisect AB
Read more about lengths and bisections at:
brainly.com/question/20837270
Answer: Their weekly pay would be the same if xx equals $1,600
Step-by-step explanation: The first and most important step is to identify what the question requires, and that is, what is the value of the unknown in the equation of their weekly incomes that would make their pay to be the same? Their weekly pay as per individual is given as follows;
Khloe = 245 + 0.095x ———(1)
Emma = 285 + 0.07x ———(2)
Simply put, we need to find the value of x when equation (1) equals equation (2)
245 + 0.095x = 285 + 0.07x
Collect like terms and we now have
0.095x - 0.07x = 285 - 245
0.025x = 40
Divide both sides of the equation by 0.025
x = 1600
Therefore their weekly pay would be at the same level, if x equals $1600
It's 3 because 4×3=12 add 2 and it equals 14
Answer:
<em>The 95.7% confidence interval for the expected amount of garbage per bin for all bins in the city</em>
(48.937 , 50.863)
Step-by-step explanation:
<u><em>Explanation:-</em></u>
Given data random sample of 46 bins, the sample mean amount was 49.9 pounds and the sample standard deviation was 3.641
<em>The sample size 'n' =46</em>
<em>mean of the sample x⁻ = 49.9</em>
<em>Standard deviation of the sample S = 3.641</em>
<u>Confidence intervals:</u><em>-</em>
<em>The 95.7% confidence interval for the expected amount of garbage per bin for all bins in the city</em>
<em></em>
<em></em>
<em>Degrees of freedom = n-1 = 46-1 =45</em>
<em>The tabulated value t₀.₉₆ = 1.794 ( from t-table)</em>
<em></em>
<em></em>
(49.9 -0.9630 , 49.9+0.9630)
(48.937 , 50.863)
<u>Conclusion:</u>-
<em>The 95.7% confidence interval for the expected amount of garbage per bin for all bins in the city</em>
(48.937 , 50.863)
The water will rise 10,000m in 10000000