Evaluate the expression for c = 3 and d = -1 cd2

Which of these tables represents a non-linear function?

olchik [2.2K]

Answer:

we conclude that the table ''A 2-column table with 4 rows. Column 1 is labeled x with entries 17, 18, 19, 20. Column 2 is labeled y with entries 16, 17, 19, 20'' represents a non-linear function.

Step-by-step explanation:

The nature of whether a function can be linear or non-linear, we must check how the x and y values of the table change. If the first difference between y-values remains the same (constant first difference), then the table will represent a linear function, otherwise not.

Given the x and y entries of the first table:

x y

17 20

18 19

19 18

20 17

From the table, it is clear that as x constantly increases by 1 unit, the y-values are also changing constantly by 1 unit. The first difference between y-values remains the same.

i.e. 19-20=-1, 18-19=-1, 17-18=-1

Thus, this table represents a linear function.

Given the x and y entries of the second table:

x y

17 -16

18 -17

19 -18

20 -19

From the table, it is clear that as x constantly increases by 1 unit, the y-values are also changing constantly by 1 unit. The first difference between y-values remains the same.

i.e. -17-(-16)=-1, -18-(-17)=-1, -19-(-18)=-1

Thus, this table represents a linear function.

Given the x and y entries of the second table:

x y

17 16

18 17

19 19

20 20

From the table, it is clear that as x constantly increases by 1 unit, but the y-values are not changing constantly. The first difference between y-values does not remain the same.

i.e. 17- 16 = 1, 19 - 17 = 2, 20 - 19 = 1

Thus, this table does not represent a linear function. Hence, it is a non-linear function.

Given the x and y entries of the second table:

x y

17 -20

18 -19

19 -18

20 -17

From the table, it is clear that as x constantly increases by 1 unit, the y-values are also changing constantly by 1 unit. The first difference between y-values remains the same.

i.e. -19-(-20)=1, -18-(-19)=1, -17-(-18)=1

Thus, this table represents a linear function.

Therefore, we conclude that the table ''A 2-column table with 4 rows. Column 1 is labeled x with entries 17, 18, 19, 20. Column 2 is labeled y with entries 16, 17, 19, 20'' represents a non-linear function.

4 0

2 years ago

What are the answers

algol [13]

Unable to see the paper clearly

6 0

3 years ago

A certain geneticist is interested in the proportion of males and females in the population who have a minor blood disorder. In

lord [1]

Answer:

95% confidence interval for the difference between the proportions of males and females who have the blood disorder is [-0.064 , 0.014].

Step-by-step explanation:

We are given that a certain geneticist is interested in the proportion of males and females in the population who have a minor blood disorder.

A random sample of 1000 males, 250 are found to be afflicted, whereas 275 of 1000 females tested appear to have the disorder.

Firstly, the pivotal quantity for 95% confidence interval for the difference between population proportion is given by;

P.Q. = $\frac{(\hat p_1-\hat p_2)-(p_1-p_2)}{\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+ \frac{\hat p_2(1-\hat p_2)}{n_2}} }$ ~ N(0,1)

where, $\hat p_1$ = sample proportion of males having blood disorder= $\frac{250}{1000}$ = 0.25

$\hat p_2$ = sample proportion of females having blood disorder = $\frac{275}{1000}$ = 0.275

$n_1$ = sample of males = 1000

$n_2$ = sample of females = 1000

$p_1$ = population proportion of males having blood disorder

$p_2$ = population proportion of females having blood disorder

Here for constructing 95% confidence interval we have used Two-sample z proportion statistics.

So, 95% confidence interval for the difference between the population proportions, ( $p_1-p_2$ ) is ;

P(-1.96 < N(0,1) < 1.96) = 0.95 {As the critical value of z at 2.5% level

of significance are -1.96 & 1.96}

P(-1.96 < $\frac{(\hat p_1-\hat p_2)-(p_1-p_2)}{\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+ \frac{\hat p_2(1-\hat p_2)}{n_2}} }$ < 1.96) = 0.95

P( $-1.96 \times {\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+ \frac{\hat p_2(1-\hat p_2)}{n_2}} }$ < ${(\hat p_1-\hat p_2)-(p_1-p_2)}$ < $1.96 \times {\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+ \frac{\hat p_2(1-\hat p_2)}{n_2}} }$ ) = 0.95

P( $(\hat p_1-\hat p_2)-1.96 \times {\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+ \frac{\hat p_2(1-\hat p_2)}{n_2}} }$ < ( $p_1-p_2$ ) < $(\hat p_1-\hat p_2)+1.96 \times {\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+ \frac{\hat p_2(1-\hat p_2)}{n_2}} }$ ) = 0.95

95% confidence interval for ( $p_1-p_2$ ) =

[ $(\hat p_1-\hat p_2)-1.96 \times {\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+ \frac{\hat p_2(1-\hat p_2)}{n_2}} }$ , $(\hat p_1-\hat p_2)+1.96 \times {\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+ \frac{\hat p_2(1-\hat p_2)}{n_2}} }$ ]

= [ $(0.25-0.275)-1.96 \times {\sqrt{\frac{0.25(1-0.25)}{1000}+ \frac{0.275(1-0.275)}{1000}} }$ , $(0.25-0.275)+1.96 \times {\sqrt{\frac{0.25(1-0.25)}{1000}+ \frac{0.275(1-0.275)}{1000}} }$ ]

= [-0.064 , 0.014]

Therefore, 95% confidence interval for the difference between the proportions of males and females who have the blood disorder is [-0.064 , 0.014].

8 0

3 years ago

If a = 3 and b = 2, what is the value of the expression 11 + a - b ?

Vadim26 [7]

Substitute the values of a and b into the expression:
11 + 3 - 2
Add first because of order of operations (perform addition and subtraction in the order of occurrence from left to right):
14 - 2
Subtract (same justification as above just with subtraction instead):
12
Hope this helps!

4 0

3 years ago

What’s the answer to this equation?

Tresset [83]

Answer:

Part A) The graph in the attached figure

Part B) see the explanation

Step-by-step explanation:

Part A) Graph the function

we have the quadratic function

$y=3x^2-12x-36$

This is a vertical parabola open upward

The vertex is a minimum

using a graphing tool

The graph in the attached figure

Part B) What are the values of a, b and c?

we know that

The values of a and b represent the x-intercepts of the quadratic equation

The x-intercepts are

(-2,0) and (6,0)

so

$a=-2\\b=6$

Find the value of c

we know that

The x-coordinate of the vertex in a vertical parabola is equal to the midpoint of the roots

so

The value of c is equal to

$c=\frac{a+b}{2}$

substitute the given values

$c=\frac{-2+6}{2}=2$

see the attached figure

7 0

3 years ago