The correct option that shows the population of the research carried out by Matthew is;
<u><em>Option D; all the students attending the college</em></u>
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- We are told that Matthew wants to estimate the mean height of students attending his college.
- Now, let us say the total number of students in his college is x. If he decides to select 100 students randomly to record their height, it means this 100 students is a sample out of the total number of students which is x that is the population.
In conclusion, we can say that the population of this research by Matthew is the total number of students that attend the college.
Read more at; brainly.com/question/6028584
Answer:
a) False
b) False
c) True
d) False
e) False
Step-by-step explanation:
a. A single vector by itself is linearly dependent. False
If v = 0 then the only scalar c such that cv = 0 is c = 0. Hence, 1vl is linearly independent. A set consisting of a single vector v is linearly dependent if and only if v = 0. Therefore, only a single zero vector is linearly dependent, while any set consisting of a single nonzero vector is linearly independent.
b. If H= Span{b1,....bp}, then {b1,...bp} is a basis for H. False
A sets forms a basis for vector space, only if it is linearly independent and spans the space. The fact that it is a spanning set alone is not sufficient enough to form a basis.
c. The columns of an invertible n × n matrix form a basis for Rⁿ. True
If a matrix is invertible, then its columns are linearly independent and every row has a pivot element. The columns, can therefore, form a basis for Rⁿ.
d. In some cases, the linear dependence relations among the columns of a matrix can be affected by certain elementary row operations on the matrix. False
Row operations can not affect linear dependence among the columns of a matrix.
e. A basis is a spanning set that is as large as possible. False
A basis is not a large spanning set. A basis is the smallest spanning set.
5+2=7 which means there will be 7 zeros so it is 10000000 times greater than
The inequality x < 9.7 tells us x needs to smaller than 9.7.
A number like 5 will be fine. Because 5 < 9.
But someone like 10 won't be fine.
Because 9 < 10.
Even 9.7 won't be suffice.
Because 9.7 < 9.7 does not make sense. If it was 9.7 <= 9.7 then it would be fine.
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.
