Answer:
a: given
b: definition of congruent
c: definition of bisect
Step-by-step explanation:
just took the assignment
Answer:
C. y = 3/5x + 13/5
Step-by-step explanation:
Answer:

Step-by-step explanation:
The formula to find the area of a circle is

We know that

Therefore the area is

Answer:
Given the mean = 205 cm and standard deviation as 7.8cm
a. To calculate the probability that an individual distance is greater than 218.4 cm, we subtract the probability of the distance given (i.e 218.4 cm) from the mean (i.e 205 cm) divided by the standard deviation (i.e 7.8cm) from 1. Therefore, we have 1- P(Z
). Using the Z distribution table we have 1-0.9573. Therefore P(X >218.4)= 0.0427.
b. To calculate the probability that mean of 15 (i.e n=15) randomly selected distances is greater than 202.8, we subtract the probability of the distance given (i.e 202.8cm) from the mean (i.e 205 cm) divided by the standard deviation (i.e 7.8cm) divided by the square root of mean (i.e n= 15) from 1. Therefore, we have 1- P(Z
). Using the Z distribution table we have 1-0.1378. Therefore P(X >202.8)= 0.8622.
c. This will also apply to a normally distributed data even if it is not up to the sample size of 30 since the sample distribution is not a skewed one.
Step-by-step explanation:
Given the mean = 205 cm and standard deviation as 7.8cm
a. To calculate the probability that an individual distance is greater than 218.4 cm, we subtract the probability of the distance given (i.e 218.4 cm) from the mean (i.e 205 cm) divided by the standard deviation (i.e 7.8cm) from 1. Therefore, we have 1- P(Z
). Using the Z distribution table we have 1-0.9573. Therefore P(X >218.4)= 0.0427.
b. To calculate the probability that mean of 15 (i.e n=15) randomly selected distances is greater than 202.8, we subtract the probability of the distance given (i.e 202.8cm) from the mean (i.e 205 cm) divided by the standard deviation (i.e 7.8cm) divided by the square root of mean (i.e n= 15) from 1. Therefore, we have 1- P(Z
). Using the Z distribution table we have 1-0.1378. Therefore P(X >202.8)= 0.8622.
c. This will also apply to a normally distributed data even if it is not up to the sample size of 30 since the sample distribution is not a skewed one.
1. Let x = #
2. "Five less than six times a number" can be written as: 6x - 5
3. "is at least" is the same as "greater than or equal to" and can be written as: >=
4. "nine subtracted from two times that number" can be written as: 2x - 9
5. the equation is: 6x - 5 >= 2x - 9
6. solve for x by grouping the x variable terms on one side and the constants on the other side:
6x - 5 >= 2x - 9
-2x +5 -2X +5
4x >= -4
7. divide each side by 4, and you get: x = -1