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BARSIC [14]
1 year ago
7

Determine whether each statement is true or false. If a statement is true, give a reason or cite an appropriate statement from t

he text. If a statement is false, provide an example that shows the statement is not true in all cases or cite an appropriate statement from the text.
1. Adding a multiple of one column of a square matrix to another column changes only the sign of the determinant.

a. False, adding a multiple of one column to another does not change the value of the determinant.
b. False, adding a multiple of one column to another changes the sign of the determinant.
c. True, adding a multiple of one column to another changes only the sign of the determinant.
d. False, adding a multiple of one column to another changes the value of the determinant by the multiple of the minor.

2. Two matrices are column-equivalent when one matrix can be obtained by performing elementary column operations on the other.

a. False, row-equivalent matrices are matrices that can be obtained from each other by performing elementary column operations on the other.
b. True, column-equivalent matrices are matrices that can be obtained from each other by performing elementary row operations on the other.
c. False, row-equivalent matrices are matrices that can be obtained from each other by performing elementary row operations on the other.
d. True, column-equivalent matrices are matrices that can be obtained from each other by performing elementary column operations on the other.
Mathematics
1 answer:
Over [174]1 year ago
3 0

Answer:

1) a. False, adding a multiple of one column to another does not change the value of the determinant.

2) d. True, column-equivalent matrices are matrices that can be obtained from each other by performing elementary column operations on the other.

Step-by-step explanation:

1) If the multiple of one column of a matrix A is added to another to form matrix B then we get: |A| = |B|. Here, the value of the determinant does not change. The correct option is A

a. False, adding a multiple of one column to another does not change the value of the determinant.

2) Two matrices can be column-equivalent when one matrix is changed to the other using a sequence of elementary column operations. Correc option is d.

d. True, column-equivalent matrices are matrices that can be obtained from each other by performing elementary column operations on the other.

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Step-by-step explanation:

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23%

Step-by-step explanation:

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2 years ago
A newspaper article about an opinion poll says that "53% of Americans approve of the president's overall job performance." The p
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Answer:

<u>Part A:</u> The population are the American adult citizens, excluding the ones from Alaska and Hawaii. The population is the people which the sample is trying to represent, as a hole.

The sample is a portion of this population, and in this case is represented by a randomly selected amount of people whose response to the interview has been selected.

<u>Part B:</u> The sample here has been selected in two steps. The <u>first</u> step is the one that we must pay attention to: the numbered list of one million responses from around the US (excluding Alaska and Hawaii). Because these responses were obtained by an online survey, the sample looks like a convenience sampling, as it depends on the availability and willingness from participants to take part of the study (is not compulsory for everyone, so not everyone is going to response, then there are people that is not going to be represented by). The <u>second</u> step is the random selection of a part of the previous responses. This last part will ensure that, the individuals that took part of the group that was interviewed, are well represented in the results.

<u>Part C:</u> As it was mentioned, there is a selection bias, because the information from the sample comes from a specific group of people that has certain features that may not represent all American adults citizens. For <u>example, the opinion of  those people who do not use internet</u>, will not be considered (and they may be a large number of persons). This situations weaken the conclusions obtained in the study, as they are not representative of the hole population.

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Which shows one way to determine the factors of 4x3 + x2 – 8x – 2 by grouping?
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A prticular type of tennis racket comes in a midsize versionand an oversize version. sixty percent of all customers at acertain
svetlana [45]

Answer:

a) P(x≥6)=0.633

b) P(4≤x≤8)=0.8989 (one standard deviation from the mean).

c) P(x≤7)=0.8328

Step-by-step explanation:

a) We can model this a binomial experiment. The probability of success p is the proportion of customers that prefer the oversize version (p=0.60).

The number of trials is n=10, as they select 10 randomly customers.

We have to calculate the probability that at least 6 out of 10 prefer the oversize version.

This can be calculated using the binomial expression:

P(x\geq6)=\sum_{k=6}^{10}P(k)=P(6)+P(7)+P(8)+P(9)+P(10)\\\\\\P(x=6) = \binom{10}{6} p^{6}q^{4}=210*0.0467*0.0256=0.2508\\\\P(x=7) = \binom{10}{7} p^{7}q^{3}=120*0.028*0.064=0.215\\\\P(x=8) = \binom{10}{8} p^{8}q^{2}=45*0.0168*0.16=0.1209\\\\P(x=9) = \binom{10}{9} p^{9}q^{1}=10*0.0101*0.4=0.0403\\\\P(x=10) = \binom{10}{10} p^{10}q^{0}=1*0.006*1=0.006\\\\\\P(x\geq6)=0.2508+0.215+0.1209+0.0403+0.006=0.633

b) We first have to calculate the standard deviation from the mean of the binomial distribution. This is expressed as:

\sigma=\sqrt{np(1-p)}=\sqrt{10*0.6*0.4}=\sqrt{2.4}=1.55

The mean of this distribution is:

\mu=np=10*0.6=6

As this is a discrete distribution, we have to use integer values for the random variable. We will approximate both values for the bound of the interval.

LL=\mu-\sigma=6-1.55=4.45\approx4\\\\UL=\mu+\sigma=6+1.55=7.55\approx8

The probability of having between 4 and 8 customers choosing the oversize version is:

P(4\leq x\leq 8)=\sum_{k=4}^8P(k)=P(4)+P(5)+P(6)+P(7)+P(8)\\\\\\P(x=4) = \binom{10}{4} p^{4}q^{6}=210*0.1296*0.0041=0.1115\\\\P(x=5) = \binom{10}{5} p^{5}q^{5}=252*0.0778*0.0102=0.2007\\\\P(x=6) = \binom{10}{6} p^{6}q^{4}=210*0.0467*0.0256=0.2508\\\\P(x=7) = \binom{10}{7} p^{7}q^{3}=120*0.028*0.064=0.215\\\\P(x=8) = \binom{10}{8} p^{8}q^{2}=45*0.0168*0.16=0.1209\\\\\\P(4\leq x\leq 8)=0.1115+0.2007+0.2508+0.215+0.1209=0.8989

c. The probability that all of the next ten customers who want this racket can get the version they want from current stock means that at most 7 customers pick the oversize version.

Then, we have to calculate P(x≤7). We will, for simplicity, calculate this probability substracting P(x>7) from 1.

P(x\leq7)=1-\sum_{k=8}^{10}P(k)=1-(P(8)+P(9)+P(10))\\\\\\P(x=8) = \binom{10}{8} p^{8}q^{2}=45*0.0168*0.16=0.1209\\\\P(x=9) = \binom{10}{9} p^{9}q^{1}=10*0.0101*0.4=0.0403\\\\P(x=10) = \binom{10}{10} p^{10}q^{0}=1*0.006*1=0.006\\\\\\P(x\leq 7)=1-(0.1209+0.0403+0.006)=1-0.1672=0.8328

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