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soldier1979 [14.2K]
3 years ago
6

2. Which relation below is NOT a function?

Mathematics
2 answers:
Nadya [2.5K]3 years ago
8 0

Answer:

b

Step-by-step explanation:

The answer is B because different X values cannot have the same y values. For example in A the x values, -2 and 0, have the same y value 4. That's not possible.

Although C also has a similar problem ALL of the X values have the same y value meaning that the function is an undefined function which is still a type of function. Its basically a function where all the y values are the same.

rusak2 [61]3 years ago
7 0

C is the answer to that question i think

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Answer:

I'm pretty sure that'd be 8567

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Evaluate |x|+(-2x^2) for x=3<br> Answers<br> -9<br> -15<br> 21
Nikolay [14]

Hello from MrBillDoesMath!

Answer:

-15

Discussion:

We are asked to evaulate:

abs(3) + ( -2 (3)^2) =                         => as abs(3) = 3

3          + (-2*9)  =

3           - 18 =

-15    

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8 0
3 years ago
∆ABC is a right angled at B and D is a point on BC if AD = 18cm BD = 9cm and CD =4cm Find AC
Gemiola [76]

In triangle, ABD,

AD²= AB²+BD²

AB² = AD²-BD²

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AB = √243

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AC² = (√243)²+(13)²

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3 0
3 years ago
What does it mean reflection across ×=3
Advocard [28]
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3 years ago
Read 2 more answers
You are given the following information obtained from a random sample of 5 observations. 20 18 17 22 18 At 90% confidence, you w
Margaret [11]

Answer:

a

  The null hypothesis is  

         H_o  : \mu  =  21

The Alternative  hypothesis is  

           H_a  :  \mu<   21

b

     \sigma_{\= x} =   0.8944

c

   t = -2.236

d

  Yes the  mean population is  significantly less than 21.

Step-by-step explanation:

From the question we are given

           a set of  data  

                               20  18  17  22  18

       The confidence level is 90%

       The  sample  size  is  n =  5  

Generally the mean of the sample  is  mathematically evaluated as

        \= x  =  \frac{20 + 18 +  17 +  22 +  18}{5}

       \= x  =  19

The standard deviation is evaluated as

        \sigma =  \sqrt{ \frac{\sum (x_i - \= x)^2}{n} }

         \sigma =  \sqrt{ \frac{ ( 20- 19 )^2 + ( 18- 19 )^2 +( 17- 19 )^2 +( 22- 19 )^2 +( 18- 19 )^2 }{5} }

         \sigma = 2

Now the confidence level is given as  90 %  hence the level of significance can be evaluated as

         \alpha = 100 - 90

        \alpha = 10%

         \alpha =0.10

Now the null hypothesis is  

         H_o  : \mu  =  21

the Alternative  hypothesis is  

           H_a  :  \mu<   21

The  standard error of mean is mathematically evaluated as

         \sigma_{\= x} =   \frac{\sigma}{ \sqrt{n} }

substituting values

         \sigma_{\= x} =   \frac{2}{ \sqrt{5 } }

        \sigma_{\= x} =   0.8944

The test statistic is  evaluated as  

              t =  \frac{\= x - \mu }{ \frac{\sigma }{\sqrt{n} } }

substituting values

              t =  \frac{ 19  - 21 }{ 0.8944 }

              t = -2.236

The  critical value of the level of significance is  obtained from the critical value table for z values as  

                   z_{0.10} =  1.28

Looking at the obtained value we see that z_{0.10} is greater than the test statistics value so the null hypothesis is rejected

6 0
4 years ago
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