1/2y^2(6x+2y+12x-2y) is equivalent

tatuchka [14]

Correct question:

Evaluate and simplify the expression when a = 3 and b = 2,

given expression = 3b-2(1 - b)a - 2

Answer:

3b-2(1 - b)a - 2 at the given values is 10.

Step-by-step explanation:

Given;

3b-2(1 - b)a - 2

The expression is simplified as;

3b-2(1 - b) a - 2 = 3b - 2a(1 - b) - 2

= 3b - 2a + 2ab - 2

When a = 3 and b = 2; then, substitute these values in the simplified expression above;

3b - 2a + 2ab - 2 = 3(2) - 2(3) + 2(3 x 2) - 2

= 6 - 6 + 12 - 2

= 10

Therefore, 3b-2(1 - b)a - 2 at the given values is 10.

8 0

3 years ago

Which algebraic expression represents the difference of e squared and twenty-one? A. E2+21 b.21 times e2 c. E2 - 21 d. 21 divide

Digiron [165]

Answer: c. E2 - 21

Step-by-step explanation:

difference means to subtract

Now, e squared=E² and twenty-one=21

Therefore the difference of e squared and twenty-one is E2 - 21

6 0

3 years ago

5x + 13-14 or-4x + 6< 10

Blizzard [7]

Answer:

5x - 1 or x>−1

Step-by-step explanation:

Just combine like terms: 13,-14

Combine like terms: 6,10

Divide by -4:-4x/-4<4/-4

7 0

3 years ago

−6x + 14 < −28 OR 9x + 15 ≤ −12. solve for x.

sasho [114]

Answer:

x>7 or x ≤ -3

Step-by-step explanation:

Solving the 1st inequality

-6x +14 < -28 --------------- (Collect like terms)

-6x < -28 - 14

-6x < - 42 -------------------- (Divide both sides by -6)

Note: If you decide an inequality expression by a negative value, the inequality sign changes)

-6x/-6 > -42/-6

x > 7

Solving the 2nd inequality

9x + 15 ≤ −12 ----------- (Collect like terms)

9x ≤ −12 - 15

9x ≤ −27 ------------------(Divide both sides by 9)

9

9x/9 ≤ −27/9

x ≤ -3

Bring both results together, we get

x>7 or x ≤ -3

The final result is complex (i.e. can't be combined together).

4 0

4 years ago

within a kiambu county,students were randomly assigned to one of two mathematics teachers.Mrs.Elite and Mrs. Bright. After the a

DerKrebs [107]

Using the t-distribution, as we have the standard deviation for the samples, it is found that since the absolute value of the test statistic is greater than the critical value, we reject the claim that Mrs.Elite and Mrs. Bright are equally effective teachers.

<h3>What are the hypothesis tested?</h3>

At the null hypothesis, we test if they are equally effective teachers, that is, the subtraction of their means is 0, hence:

$H_0: \mu_1 - \mu_2 = 0$

At the alternative hypothesis, we test if they are not equally effective teachers, that is, the subtraction of their means is not 0, hence:

$H_1: \mu_1 - \mu_2 \neq 0$

<h3>What is the distribution of the differences?</h3>

For Mrs. Elite, we have that:

$\mu_1 = 78, \sigma_1 = 10, n_1 = 30, s_1 = \frac{10}{\sqrt{30}} = 1.82574$

For Mrs. Bright, we have that:

$\mu_2 = 85, \sigma_2 = 15, n_2 = 25, s_2 = \frac{15}{\sqrt{25}} = 3$

For the distribution of differences, we have that:

$\overline{x} = \mu_1 - \mu_2 = 78 - 85 = -7$

$s = \sqrt{s_1^2 + s_2^2} = \sqrt{1.82574^2 + 3^2} = 3.5119$

<h3>What is the test statistic?</h3>

The test statistic is given by:

$t = \frac{\overline{x} - \mu}{s}$

In which $\mu = 0$ is the value tested at the null hypothesis.

Hence:

$t = \frac{\overline{x} - \mu}{s}$

$t = \frac{-7 - 0}{3.5119}$

$t = -1.99$

Considering a<em> two-tailed test</em>, as we are testing if the mean is different of a value, with 30 + 25 - 2 = <em>53 df and a significance level of 0.1</em>, the critical value is of $|t^{\ast}| = 1.6741$ .

Since the absolute value of the test statistic is greater than the critical value, we reject the claim that Mrs.Elite and Mrs. Bright are equally effective teachers.

To learn more about the t-distribution, you can take a look at brainly.com/question/13873630

3 0

3 years ago