Simon can jump a distance that is 2 inches longer than 3 feet. Travis said he can jump farther than Simon because he can jump 38
2 answers:
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Answer:
2.28% of tests has scores over 90.
Step-by-step explanation:
Problems of normally distributed samples are solved using the z-score formula.
In a set with mean and standard deviation
, the zscore of a measure X is given by:
The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.
In this problem, we have that:
What proportion of tests has scores over 90?
This proportion is 1 subtracted by the pvalue of Z when X = 90. So
has a pvalue of 0.9772.
So 1-0.9772 = 0.0228 = 2.28% of tests has scores over 90.
Answer:
(A) Set A is linearly independent and spans . Set is a basis for
.
Step-by-Step Explanation
<u>Definition (Linear Independence)</u>
A set of vectors is said to be linearly independent if at least one of the vectors can be written as a linear combination of the others. The identity matrix is linearly independent.
<u>Definition (Span of a Set of Vectors)</u>
The Span of a set of vectors is the set of all linear combinations of the vectors.
<u>Definition (A Basis of a Subspace).</u>
A subset B of a vector space V is called a basis if: (1)B is linearly independent, and; (2) B is a spanning set of V.
Given the set of vectors , we are to decide which of the given statements is true:
In Matrix , the circled numbers are the pivots. There are 3 pivots in this case. By the theorem that The Row Rank=Column Rank of a Matrix, the column rank of A is 3. Thus there are 3 linearly independent columns of A and one linearly dependent column.
has a dimension of 3, thus any 3 linearly independent vectors will span it. We conclude thus that the columns of A spans
.
Therefore Set A is linearly independent and spans . Thus it is basis for
.