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2 years ago

Find the additive inverse of 6-3i.

1 answer:

5 0

Hey there!

In order to find the additive inverse of any number, regardless if it is real or not, you can simply multiply the number by -1 or negate it.

This should look like this:
-(6-3i)
=-6+3i

Therefore, the additive inverse of 6-3i is -6+3i.

**Note: Keep in mind, when the additive inverse of a number is added to the number, the number equals 0:
(6-3i)+(-6+3i)=0

Hope this helps and have a marvelous day! :)

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A liquid dietary product implies in its advertising that use of the product for one month results in an average weight loss of a

BigorU [14]

Answer:

Following are the responses to the given question:

Step-by-step explanation:

Please find the table in the attached file.

mean and standard deviation difference: $\bar{d}=\frac{\Sigma d}{n} =\frac{-4-6-.......-4-4}{8}=-4.125 \\\\S_d=\sqrt{\frac{\Sigma (d-\bar{d})^2 }{n-1}}=\sqrt{\frac{(-4 + 4.125)^2 +.......+(-4 +4.125)^2 }{8-1}}= 1.246$

For point a:

hypotheses are:

$H_0 : \mu_d \geq -3\\\\H_a : \mu_d < -3\\\\$

degree of freedom:

$df=n-1=8-1=7$

From t table, at $\alpha = 0.05$ , reject null hypothesis if $t$ .

test statistic:

$t=\frac{\bar{d}-\mu_d }{\frac{s_d}{\sqrt{d}}}=\frac{ -4.125- (-3)}{\frac{1.246}{ \sqrt{8}}} =-2.55$

because the $t=-2.553$ , removing the null assumption. Data promotes a food product manufacturer's assertion with a likelihood of Type 1 error of 0.05.

For point b:

From t table, at $\alpha =0.01$ , removing the null hypothesis if $t$ .

because $t=-2.553 >-2.908$ , fail to removing the null hypothesis.

The data do not help the foodstuff producer's point with the likelihood of a .01-type mistake.

For point c:

Hypotheses are:

$H_0: \mu_d \geq -5\\\\H_a: \mu_d < -5$

Degree of freedom:

$df=n-1=8-1=7$

From t table, at $\alpha =0.05$ , removing the null hypothesis if $t$ .

test statistic: $t=\frac{\bar{d}-\mu_d}{\frac{s_d}{\sqrt{n}}} =\frac{-4.125-(-5)}{\frac{1.246}{\sqrt{8}}}=1.986$

Since $t-1.986 >-1.895$ , The null hypothesis fails to reject. The results do not support the packaged food producer's claim with a Type 1 error probability of 0,05.

From t table, at $\alpha= 0.01$ , reject null hypothesis if $t$ .

Since $t=1.986>-2.998$ , fail to reject null hypothesis.

Data do not support the claim of the producer of the dietary product with the probability of Type 1 error of .01.

5 0

2 years ago

Simplify the expression 3x(x – 12x) + 3x2 – 2(x – 2)2. Which statements are true about the process and simplified product? Check

olasank [31]

We have the expression:
3x(x-12x) + 3x^2 - 2(x-2)^2

First, we will expand the power 2 bracket as follows:
3x(x-12x) + 3x^2 - 2(x^2 - 4x +4)

Then, we will get rid of the brackets as follows:
3x^2 - 36x^2 + 3x^2 - 2x^2 + 8x - 8

Now, we will gather the like terms and add them as follows:
-32 x^2 + 8x - 8

We can take the 8 as a common factor:
8 ( -4x^2 + x -1)

3 0

2 years ago

On a coordinate plane, a piecewise function has 3 lines. The graph shows cleaning time in hours on the x-axis and total cost in

sveta [45]

Answer:

C. (2, 6] hour jobs cost $100

Step-by-step explanation:

Let's consider each of these statements in view of the graph:

A cleaning time of 2 hours will cost $100. -- The closed circle at (2, 50) tells you the cost of a 2-hour job is $50, not $100.
A cleaning time of 6 hours will cost $150. -- The closed circle at (6, 100) tells you the cost of a 6-hour job is $100, not $150.
Cost is a fixed rate of $100 for jobs requiring more than 2 hours, up to a maximum of 6 hours. -- The line between the open circle at (2, 100) and the closed circle at (6, 100) tells you this is TRUE.
Cost is a fixed rate of $200 for jobs that require at least 6 hours. -- "At least 6 hours" means "greater than or equal to 6 hours." The closed circle at (6, 100) means a 6-hour job is $100, not $200.

3 0

2 years ago

Which properties are present in a table that represents an exponential function in the form y-b* when b > 1?

Oksana_A [137]

Answer:

<u>Properties that are present are </u>

Property I

Property IV

Step-by-step explanation:

The function given is $y=b^x$ where b > 1

Let's take a function, for example, $y=2^x$

Let's check the conditions:

I. As the x-values increase, the y-values increase.

Let's put some values:

y = 2 ^ 1

y = 2

and

y = 2 ^ 2

y = 4

So this is TRUE.

II. The point (1,0) exists in the table.

Let's put 1 into x and see if it gives us 0

y = 2 ^ 1

y = 2

So this is FALSE.

III. As the x-value increase, the y-value decrease.

We have already seen that as x increase, y also increase in part I.

So this is FALSE.

IV. as the x value decrease the y values decrease approaching a singular value.

THe exponential function of this form NEVER goes to 0 and is NEVER negative. So as x decreases, y also decrease and approached a value (that is 0) but never becomes 0.

This is TRUE.

Option I and Option IV are true.

7 0

2 years ago

Read 2 more answers

An expression is given: 2 open parentheses square root of k minus 1 close parenthesis plus square root of 8. If on adding negati

neonofarm [45]

Answer:

Possible value of k is √2

Step-by-step explanation:

The information given are;

The expression, 2·(√k - 1) + √8 to which may be added -6·√2 to obtain a rational number, we therefore have;

2·(√k - 1) + √8 - 6·√2 = R

Therefore, simplifying gives;

2·√k - 2 + 2·√2 - 6·√2 = 2·√k - 2 - 4·√2 = R

2·√k - 2 - 4·√2 + 2= R + 2 = R

2·√k - 2+ 2 - 4·√2 = R

2·√k + 0 - 4·√2 = 2·√k - 4·√2 = 2·(√k - 2·√2) = R

(√k - 2·√2) = R/2 = R

Therefore, √2 is a factor of √k such that √k - 2·√2 = R

Which gives k = x·√2, where x = a rational number

When x = 1, k = √2.

Therefore, a possible value of k is √2

3 0

2 years ago