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Contact [7]
2 years ago
6

Juanita wants to perform row operations on the augmented matrix for the system below (See pictures). Which matrix should Juanita

use to perform the operations?

Mathematics
1 answer:
Artist 52 [7]2 years ago
5 0

Answer:

The second option

Step-by-step explanation:

The given system of equation is

x+2y=3

-x+y+z=2

y-2z=-3

The augment matrix is obtained by combining the coefficient matrix with the constant matrix to obtain;

\left[\begin{array}{cccc}1&2&0&|3\\-1&1&1&|2\\0&1&-2&|-3\end{array}\right]

Note that the absence of z, in the first equation means its coefficient is zero. The same thing applies to x in the last equation.

The correct choice is the second option.

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A bridge is rated to a capacity of 100 British tons. What is the maximum weight the bridge can support in kilograms? (Round to t
S_A_V [24]

Maximum weight the bridge can support in kilograms is 101696

Step-by-step explanation:

  • Step 1: Given capacity of bridge = 100 British tons. Find how many kilograms are equivalent to 1 British ton.

1 British ton = 2240 pounds

1 pound = 0.454 kg

⇒ 1 British ton = 2240 × 0.454 kg = 1016.96 kg

  • Step 2: Find how many kilograms are in 100 British tons.

⇒ 100 × 1016.96 = 101696

3 0
2 years ago
PLZ HELP! Oliver is 8 feet above the surface of the water. There is a school of fish 10 feet below the surface. A ledge with som
qwelly [4]

Answer:

26 feet, Oliver will have to travel 26 feet deep.

Step-by-step explanation:

8 feet added to 18 feet will be your answer.

8 0
2 years ago
Compute E(X) for the following random variable X : X=Number of tosses until all 10 numbers are seen (including the last toss) by
julsineya [31]

Answer:

For 10 tosses we have that E(X)=10

Therefore E(i)= 1/10 +2/10 +3/10....10/10

This implies that 40/10=E(i)

Therefore E(10) =40/10

= 4.

8 0
1 year ago
14) Rounded to the nearest whole number, the square
Katen [24]

Answer:

3934

Step-by-step explanation:

Round 15,479,652 and find square root.

4 0
1 year ago
We're testing the hypothesis that the average boy walks at 18 months of age (H0: p = 18). We assume that the ages at which boys
marusya05 [52]

Answer:

II. This finding is significant for a two-tailed test at .01.

III. This finding is significant for a one-tailed test at .01.

d. II and III only

Step-by-step explanation:

1) Data given and notation    

\bar X=19.2 represent the battery life sample mean    

\sigma=2.5 represent the population standard deviation    

n=25 sample size    

\mu_o =18 represent the value that we want to test    

\alpha represent the significance level for the hypothesis test.    

t would represent the statistic (variable of interest)    

p_v represent the p value for the test (variable of interest)    

2) State the null and alternative hypotheses.    

We need to conduct a hypothesis in order to check if the mean battery life is equal to 18 or not for parta I and II:    

Null hypothesis:\mu = 18    

Alternative hypothesis:\mu \neq 18    

And for part III we have a one tailed test with the following hypothesis:

Null hypothesis:\mu \leq 18    

Alternative hypothesis:\mu > 18  

Since we know the population deviation, is better apply a z test to compare the actual mean to the reference value, and the statistic is given by:    

z=\frac{\bar X-\mu_o}{\frac{\sigma}{\sqrt{n}}} (1)    

z-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".    

3) Calculate the statistic    

We can replace in formula (1) the info given like this:    

z=\frac{19.2-18}{\frac{2.5}{\sqrt{25}}}=2.4    

4) P-value    

First we need to calculate the degrees of freedom given by:  

df=n-1=25-1=24  

Since is a two tailed test for parts I and II, the p value would be:    

p_v =2*P(t_{(24)}>2.4)=0.0245

And for part III since we have a one right tailed test the p value is:

p_v =P(t_{(24)}>2.4)=0.0122

5) Conclusion    

I. This finding is significant for a two-tailed test at .05.

Since the p_v. We reject the null hypothesis so we don't have a significant result. FALSE

II. This finding is significant for a two-tailed test at .01.

Since the p_v >\alpha. We FAIL to reject the null hypothesis so we have a significant result. TRUE.

III. This finding is significant for a one-tailed test at .01.

Since the p_v >\alpha. We FAIL to reject the null hypothesis so we have a significant result. TRUE.

So then the correct options is:

d. II and III only

6 0
2 years ago
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