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IRINA_888 [86]
3 years ago
6

How do I use substitution to solve

Mathematics
2 answers:
nekit [7.7K]3 years ago
4 0
<span>The method of solving "by substitution" works by solving one of the equations (you choose which one) for one of the variables (you choose which one), and then plugging this back into the other equation, "substituting" for the chosen variable and solving for the other. Then you back-solve for the first variable.</span>
Olin [163]3 years ago
3 0
We can substitute y in the second equation with the first equation since y = y. Thesolution<span> of the linear </span>system<span> is (1, 6). You can </span>use<span> the </span>substitution<span> method even if both equations of the linear </span>system<span> are in standard form. Just begin by </span>solving<span> one of the equations for one of its variables.</span>
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Area of square/rectangle = length times width
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3 years ago
Temperature transducers of a certain type are shipped in batches of 50. A sample of 58 batches was selected, and the number of t
andrew11 [14]

Answer:

a) X        Freq.         Rel Freq.

__________________________

0           7              (7/58) = 0.121

1            10            (10/58)=0.172

2           13            (13/58) =0.224

3           14            (14/58)=0.241

4           6              (6/58)=0.103

5           3              (3/58)=0.052

6           3              (3/58)=0.052

7            1              (1/58) =0.017

8            1              (1/58)=0.017

_________________________

Total     58                 1.00

b) \frac{7+10+13+14+6}{58}= \frac{50}{58}=0.862

Step-by-step explanation:

Assuming this question: Temperature transducers of a certain type are shipped in batches of 50. A sample of 60 batches was selected, and the number of transducers in each batch not conforming to design specifications was determined, resulting in the following data:

2,1,2,4,0,1,3,2,0,5,3,3,1,3,2,4,7,0,2,3,

0,4,2,1,3,1,1,3,4,2,3,2,2,8,4,5,1,3,1,

5,0,2,3,2,1,0,6,4,2,1,6,0,3,3,3,6,2,3

(a) Determine frequencies and relative frequencies for the observed values of x = number of nonconforming transducers in a batch. (Round your relative frequencies to three decimal places.)

For this case first we order the dataset on increasing way and we got:

0 0 0 0 0 0 0

1 1 1 1 1 1 1 1 1 1

2 2 2 2  2 2 2 2 2 2 2 2 2

3 3 3 3 3 3 3 3 3 3 3 3  3 3

4 4 4 4 4 4

5 5 5

6 6 6

7

8

And we can construct the following table:

X        Freq.         Rel Freq.

__________________________

0           7              (7/58) = 0.121

1            10            (10/58)=0.172

2           13            (13/58) =0.224

3           14            (14/58)=0.241

4           6              (6/58)=0.103

5           3              (3/58)=0.052

6           3              (3/58)=0.052

7            1              (1/58) =0.017

8            1              (1/58)=0.017

_________________________

Total     58                 1.00

(b) What proportion of batches in the sample have at most four nonconforming transducers? (Round your answer to three decimal places.)

For this case this proportion would be:

\frac{7+10+13+14+6}{58}= \frac{50}{58}=0.862

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