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Inga [223]
3 years ago
13

Which operation can be used with the augmented matrix below?

Mathematics
1 answer:
ycow [4]3 years ago
5 0

Answer:

Multiply row 2 by-1 and add it to row 3.

Step-by-step explanation:

The given augmented matrix is \left[\begin{array}{ccc}3&-21&15\\15&8&15\\-2&-1&3\end{array}\right]

The permissible row operations are:

1. Switching rows

2. Multiplying a row by a nonzero constant.

3. Adding/Subtracting two rows

 Therefore the correct option is: Multiply row 2 by-1 and add it to row 3.

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Pedro has  3 pounds and 10 ounces of fruit salad
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Complete the equation below to show a way to write 5/38 as a fraction
vaieri [72.5K]
5/38 is 0.13 because you have to divide them
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Gerald walks 1 km in 15 min. Express as a unit rate and explain step by step please
Simora [160]
The rate is 1km/15min

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To make it a unit rate, we need to make the denominator 1.

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5 0
3 years ago
Consider the following data set. The data were actually collected in pairs, and each row represents a pair.
Anna11 [10]

Answer:

Step-by-step explanation:

Hello!

a.

Using the data sets you have to analyze them as if they are two independent samples.

The hypotheses are:

H₀: μ₁=μ₂

H₁: μ₁≠μ₂

α: 0.05

Assuming that both data sets are from a normal distribution and both population variances, although unknown, are equal, the statistic to use is:

t=\frac{(X[bar]_1-X[bar]_2)-(Mu_1-Mu_2)}{Sa\sqrt{\frac{1}{n_1} +\frac{1}{n_2} } } ~~t_{n_1+n_2-2}

Sa^2= \frac{(n_1-1)S^2_1+(n_2-1)S^2_2}{n_1+n_2-2}

Sa^2= \frac{9*5.38+9*3.65}{10+10-2} =4.51

Sa= 2.12

t_{H_0}= \frac{(49.69-50.52)-0}{2.12*\sqrt{\frac{1}{10} +\frac{1}{10} } } = -0.875

The p-value for the two-tailed test is:

P(t_{18}≤-0.875) + P(t_{18}≥0.875)= P(t_{18}≤-0.875) + (1 - P(t_{18}≤0.875))= 0.1965 + ( 1 - 0.8035)= 0.393

Since there is no significant level I determined it at 5%, comparing it to the p-value, the hypothesis test is not significant. Meaning that the difference between the population means of both groups is equal to zero.

b.

Considering the given data as a paired sample, you have to determine the variable difference to conduct the test:

Xd: the difference between X₁ and X₂

Before the sample mean and sample standard deviation you have to calculate the difference between the observations of group 1 and group 2. (2nd attachment)

Mean X[bar]d= -0.83

Standard deviation Sd= 1.28

The hypotheses for the paired sample test are:

H₀: μd=0

H₁: μd≠0

α: 0.05

t= \frac{X[bar]d-Mud}{\frac{Sd}{\sqrt{n} } } ~~t_{n-1}

t_{H_0}= \frac{-0.83-0}{\frac{1.28}{ \sqrt{10} }} = -2.05

The p-value for the two-tailed test is:

P(t_{9}≤-2.05) + P(t_{9}≥2.05)= P(t_{9}≤-2.05) + (1 - P(t_{9}≤2.05))= 0.0353 + (1 - 0.9647)= 0.0706

Comparing the p-value with the significance level, the hypothesis test is not significant. Meaning that the population mean of the difference between group 1 and group 2 is equal to cero. There is no difference between the two groups.

c.

The value of the statistic in "a" is greater than the value of the statistic obtained in "b". Since there are two samples used an "a" the degrees of freedom of the test are the double as the ones used in "b". The p-value obtained in "b" is around half of the p-value of "a".

The main difference is that in "a" you compared two different samples but in "b" you compared two dependent samples when analyzing paired data the "effect of the individual" is removed from the equation.

I hope this helps!

3 0
3 years ago
A translation is
Eduardwww [97]

Answer:

The answer is B..

Step-by-step explanation:

A translation is a transformation in which every point moves the same direction and distance.

5 0
3 years ago
Read 2 more answers
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