What negative number equals to the number 10

x Almost-equals 2.2, y Almost-equals 2.4

Step-by-step explanation:

a) H0: μ = 10

Ha: μ ≠ 10

This is a two tailed test

n = 13

Since α = 0.01, the critical value is determined from the t distribution table. Recall that this is a two tailed test. Therefore, we would find the critical value corresponding to 1 - α/2 and reject the null hypothesis if the absolute value of the test statistic is greater than the value of t 1 - α/2 from the table.

1 - α/2 = 1 - 0.01/2 = 1 - 0.005 = 0.995

The critical value is 3.012

The rejection region is area > 3.012

b) Ha: μ > 10

This is a right tailed test

n = 23

α = 0.1

We would reject the null hypothesis if the test statistic is greater than the table value of 1 - α

1 - α = 1 - 0.1 = 0.9

The critical value is 1.319

The rejection region is area > 1.319

c) Ha: μ > 10

This is a right tailed test

n = 99

α = 0.05

We would reject the null hypothesis if the test statistic is greater than the table value of 1 - α

1 - α = 1 - 0.05 = 0.95

The critical value is 1.66

The rejection region is area > 1.66

d) Ha: μ < 10

This is a left tailed test

n = 11

α = 0.1

We would reject the null hypothesis if the test statistic is lesser than the table value of 1 - α

1 - α = 1 - 0.1 = 0.9

The critical value is 1.363

The rejection region is area < 1.363

e) H0: μ = 10

Ha: μ ≠ 10

This is a two tailed test

n = 20

Since α = 0.05, we would find the critical value corresponding to 1 - α/2 and reject the null hypothesis if the absolute value of the test statistic is greater than the value of t 1 - α/2 from the table.

1 - α/2 = 1 - 0.05/2 = 1 - 0.025 = 0.975

The critical value is 2.086

The rejection region is area > 2.086

f) Ha: μ < 10

This is a left tailed test

n = 77

α = 0.01

We would reject the null hypothesis if the test statistic is lesser than the table value of 1 - α

1 - α = 1 - 0.01 = 0.99

The critical value is 2.376

The rejection region is area < 2.376

The last is the correct option

"all real values except x not-equals 7 and the x for which f (x) not-equals negative 3"

Step-by-step explanation:

Domain and Range of Functions

Given the function f(x), the domain of f is the set of all the values that x can take such f(x) exists. The range of f is the set of all the values that f takes.

We have a problem where we have to find the domain of a composite function. Let's recall that being f and g real functions, then

is the composite function of f and g.

We know the domain of f is the set of all real values except 7, and the domain of g is the set of all real values except –3.

Since f is the innermost function, the domain of the composite function is directly restricted by the domain of f. So, x cannot be 7.

Now, g takes f as its independent variable, and we know the domain of g excludes -3. It can be found that f(x) cannot be -3 because it will cause g not to exist.

Thus, the domain of is

All real numbers except x=7 and those where f(x)=-3

The last is the correct option

Step-by-step explanation:

Domain of a function means the set of values of the variable for which the given function exists.

(g circle f) (x) means g{f(x)}.

g{f(x)} will exists for all the real values of x except x = 7 and that particular x for which f(x) = -3.

Hence the domain of g{f(x)} will be all real values except x = 7 and the x for which f(x) = -3

Step-by-step explanation:

Domain of a function means the set of values of the variable for which the given function exists.

(g circle f) (x) means g{f(x)}.

g{f(x)} will exists for all the real values of x except x = 7 and that particular x for which f(x) = -3.

Hence the domain of g{f(x)} will be all real values except x = 7 and the x for which f(x) = -3

x Almost-equals 2.2, y Almost-equals 2.4

The last is the correct option

"all real values except x not-equals 7 and the x for which f (x) not-equals negative 3"

Step-by-step explanation:

Domain and Range of Functions

Given the function f(x), the domain of f is the set of all the values that x can take such f(x) exists. The range of f is the set of all the values that f takes.

We have a problem where we have to find the domain of a composite function. Let's recall that being f and g real functions, then

is the composite function of f and g.

We know the domain of f is the set of all real values except 7, and the domain of g is the set of all real values except –3.

Since f is the innermost function, the domain of the composite function is directly restricted by the domain of f. So, x cannot be 7.

Now, g takes f as its independent variable, and we know the domain of g excludes -3. It can be found that f(x) cannot be -3 because it will cause g not to exist.

Thus, the domain of is

All real numbers except x=7 and those where f(x)=-3

The last is the correct option

Step-by-step explanation:

a) H0: μ = 10

Ha: μ ≠ 10

This is a two tailed test

n = 13

Since α = 0.01, the critical value is determined from the t distribution table. Recall that this is a two tailed test. Therefore, we would find the critical value corresponding to 1 - α/2 and reject the null hypothesis if the absolute value of the test statistic is greater than the value of t 1 - α/2 from the table.

1 - α/2 = 1 - 0.01/2 = 1 - 0.005 = 0.995

The critical value is 3.012

The rejection region is area > 3.012

b) Ha: μ > 10

This is a right tailed test

n = 23

α = 0.1

We would reject the null hypothesis if the test statistic is greater than the table value of 1 - α

1 - α = 1 - 0.1 = 0.9

The critical value is 1.319

The rejection region is area > 1.319

c) Ha: μ > 10

This is a right tailed test

n = 99

α = 0.05

We would reject the null hypothesis if the test statistic is greater than the table value of 1 - α

1 - α = 1 - 0.05 = 0.95

The critical value is 1.66

The rejection region is area > 1.66

d) Ha: μ < 10

This is a left tailed test

n = 11

α = 0.1

We would reject the null hypothesis if the test statistic is lesser than the table value of 1 - α

1 - α = 1 - 0.1 = 0.9

The critical value is 1.363

The rejection region is area < 1.363

e) H0: μ = 10

Ha: μ ≠ 10

This is a two tailed test

n = 20

Since α = 0.05, we would find the critical value corresponding to 1 - α/2 and reject the null hypothesis if the absolute value of the test statistic is greater than the value of t 1 - α/2 from the table.

1 - α/2 = 1 - 0.05/2 = 1 - 0.025 = 0.975

The critical value is 2.086

The rejection region is area > 2.086

f) Ha: μ < 10

This is a left tailed test

n = 77

α = 0.01

We would reject the null hypothesis if the test statistic is lesser than the table value of 1 - α

1 - α = 1 - 0.01 = 0.99

The critical value is 2.376

The rejection region is area < 2.376

A. Two slack variables are needed

B. S1 and s2 (option b)

C. Option d (5X1 + 8X2 + 10X3 + s1 = 173)

D. Option a (5X1 + 4X2 + 17X3 + s2 = 245)

E. Z is maximized at 173 when (X1, X2, X3) = (34.6, 0, 0)

Step-by-step explanation:

In a linear maximization problem like this, if we want to convert the inequality (constraint) into a linear equation, we add slack variables to the left hand side of each inequality.

Therefore we add s1 and s2 to the first and second inequality respectively.

Imputing the slack variables, we obtain as follows:

Subject to

Solution :

5X1 on the first equation is the pivot because the negative value of -5 is the highest and 173/5 is less than 245/5

We perform eq1 - eq2 on the first row and eq1 + eq3 on the third row to get X1 and clear the rest.

We

X1 = 173/5 = 34.6

X2 = 0 (inactive)

X3 = 0 (inactive)

s1 = 0 , s2 = -72/-1 = 72,

Z = 173/1 =173

Therefore the minimum value of Z is 173 when (X1, X2, X3) = (34.6, 0, 0)

A. Two slack variables are needed

B. S1 and s2 (option b)

C. Option d (5X1 + 8X2 + 10X3 + s1 = 173)

D. Option a (5X1 + 4X2 + 17X3 + s2 = 245)

E. Z is maximized at 173 when (X1, X2, X3) = (34.6, 0, 0)

Step-by-step explanation:

In a linear maximization problem like this, if we want to convert the inequality (constraint) into a linear equation, we add slack variables to the left hand side of each inequality.

Therefore we add s1 and s2 to the first and second inequality respectively.

Imputing the slack variables, we obtain as follows:

Subject to

Solution :

5X1 on the first equation is the pivot because the negative value of -5 is the highest and 173/5 is less than 245/5

We perform eq1 - eq2 on the first row and eq1 + eq3 on the third row to get X1 and clear the rest.

We

X1 = 173/5 = 34.6

X2 = 0 (inactive)

X3 = 0 (inactive)

s1 = 0 , s2 = -72/-1 = 72,

Z = 173/1 =173

Therefore the minimum value of Z is 173 when (X1, X2, X3) = (34.6, 0, 0)