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True [87]
2 years ago
15

State is a given functions are inverses. NO LINKS!! Part 2​

Mathematics
2 answers:
nadya68 [22]2 years ago
8 0

Step-by-step explanation:

5.

g(x)=-2x+1

let y=2x+1

Interchanging role of x & y

x=2y+1

y=½x-½

g-¹(x)=½(x-1)

not

equal to f(x)=-x+1

<u>Given function are not function of each other</u> .

6.

g(x)=-x

let

y=-x

Interchanging role of x & y

x=-y

y=-x

g-¹(x)=-x

not

equal to f(x)=3+⅓x

<u>Given function are not function of each other .</u>

nikklg [1K]2 years ago
8 0

9514 1404 393

Answer:

  none of the function pairs are inverses

Step-by-step explanation:

Inverse <u>linear</u> functions will meet <em>all</em> of several requirements:

  • each is a reflection of the other about the line y=x
  • the solution of the simultaneous equations in on the line y=x
  • in standard form, the coefficients are swapped (only)
  • the product of the slopes is 1

_____

For these problems, it is probably easiest to look at the last requirement. Each function is given in slope-intercept form, so the slope is simply the x-coefficient.

5) The product of slopes is (-2)(-1) = 2 ≠ 1 . . . NOT inverses

6) The product of slopes is (-1)(1/3) = -1/3 ≠ 1 . . . NOT inverses

7) The product of slopes is (-4/3)(-1/2) = 2/3 ≠ 1 . . . NOT inverses

8) The product of slopes is (3/2)(-5/2) = -15/4 ≠ 1 . . . NOT inverses

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Step-by-step explanation:

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A professor would like to test the hypothesis that the average number of minutes that a student needs to complete a statistics e
professor190 [17]

Answer:

\chi^2 =\frac{15-1}{25} 16 =8.96

The degrees of freedom are given by:

df = n-1 = 15-1=14

The p value for this case would be given by:

p_v =P(\chi^2

Since the p value is higher than the significance level we have enough evidence to FAIL to reject the null hypothesis and we can conclude that the true deviation is not ignificantly lower than 5 minutes

Step-by-step explanation:

Information given

n=15 represent the sample size

\alpha=0.05 represent the confidence level  

s^2 =16 represent the sample variance

\sigma^2_0 =25 represent the value that we want to  verify

System of hypothesis

We want to test if the true deviation for this case is lesss than 5minutes, so the system of hypothesis would be:

Null Hypothesis: \sigma^2 \geq 25

Alternative hypothesis: \sigma^2

The statistic is given by:

\chi^2 =\frac{n-1}{\sigma^2_0} s^2

And replacing we got:

\chi^2 =\frac{15-1}{25} 16 =8.96

The degrees of freedom are given by:

df = n-1 = 15-1=14

The p value for this case would be given by:

p_v =P(\chi^2

Since the p value is higher than the significance level we have enough evidence to FAIL to reject the null hypothesis and we can conclude that the true deviation is not ignificantly lower than 5 minutes

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Answer:

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