For two triangles to be congruent by AAS:
1- Two angles in the first triangle must be equal to two angles in the second triangle
2- A non included side in the first triangle is equal to a non included side in the second triangle
Now, let's check our options. We will find that:
For the two triangles UTV and ABC:
angle T = angle A
angle V = angle C
TU (non-included between angles T & V) = AB (non-included between angles A & C)
Therefore, we can conclude that:
Triangles ABC and UTV are congruent by AAS
2x+6 = 20
take 6 from both sides
2x = 14
divide by 2 on both sides
x = 7
Using the normal distribution, it is found that there is a 0.0436 = 4.36% probability that a randomly selected caterpillar will have a length longer than (greater than) 4.0 centimeters.
<h3>Normal Probability Distribution</h3>
The z-score of a measure X of a normally distributed variable with mean
and standard deviation
is given by:

- The z-score measures how many standard deviations the measure is above or below the mean.
- Looking at the z-score table, the p-value associated with this z-score is found, which is the percentile of X.
In this problem, the mean and the standard deviation are given, respectively, by:
.
The probability that a randomly selected caterpillar will have a length longer than (greater than) 4.0 centimeters is <u>one subtracted by the p-value of Z when X = 4</u>, hence:


Z = 1.71
Z = 1.71 has a p-value of 0.9564.
1 - 0.9564 = 0.0436.
0.0436 = 4.36% probability that a randomly selected caterpillar will have a length longer than (greater than) 4.0 centimeters.
More can be learned about the normal distribution at brainly.com/question/24663213
#SPJ1
Over 3 up 6, or better know as 3/6
Answer:
Total cost is $205328
Step-by-step explanation:
Given data:
cost C(q) = 0.1 q^3 - 0.5 q^2 + 500 q + 200
current units q = 4 ( 4000 units )
The current level of production is 4000 ( 4 )units, and manufacturer is planning to upgrade this to 4100 ( 4.1 ) units
C'(q) = 3 * 0.1 q^2 - 2 * 0.5 q + 500
C'(q) = 0.3 q2 - q + 500
C(4.1) - C(4) ≈ C’(4) x Δq
≈ C’(4) x 0.1
≈ ( 0.3 * 4^2 - 4 + 500 ) x 0.1
≈ ( 0.3 * 16 - 4 + 500 ) x 0.1
≈ 500.8 x 0.1 ≈ 50.08
total cost = 4100 x 50.08 = $205328