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3 years ago

What is the opposite of 1?

Mathematics

2 answers:

Verizon [17]3 years ago

7 0

The opposite of 1 is -1 :)

Vanyuwa [196]3 years ago

4 0

Answer:

-1

Step-by-step explanation:

You might be interested in

Find an equation for the line parallel to 4 y + 24 x = 32 and goes through the point ( 1 , − 4 ) . Write your answer in the form

Snezhnost [94]

Answer: y = - 6x + 2

Step-by-step explanation:

The equation of a straight line can be represented in the slope intercept form as

y = mx + c

Where

m = slope = (change in the value of y in the y axis) / (change in the value of x in the x axis)

The equation of the given line is

4y + 24x = 32

Rearranging it so that it will take the slope intercept form, it becomes

4y = - 24x + 32

Dividing through by 4, it becomes

y = - 6x + 8

Comparing with the slope intercept form, slope = - 6

If two lines are parallel, it means that they have the same slope. Therefore, the slope of the line passing through (1,-4) is - 6

To determine the intercept, we would substitute m = - 6, x = 1 and y = -4 into y = mx + c. It becomes

- 4 = - 6 × 1 + c = - 6 + c

c = - 4 + 6 = 2

The equation becomes

y = - 6x + 2

8 0

3 years ago

An article reported on the results of an experiment in which half of the individuals in a group of 60 postmenopausal overweight

slavikrds [6]

Answer:

Step-by-step explanation:

From the given information:

Let's first compute the null and alternative hypothesis

Null hypothesis:

$\mathbf{H_o: \mu_1-\mu_2=1}$

Alternative hypothesis:

$H_a :\mu_1 -\mu_2 > 1$

The number of samples is half in a group of 60

i.e

$n_1=n_2 = 30$

the sample mean for sample 1 $\overline x_1$ = 5.7

the standard deviation for sample 1 $s_1$ = 3.1

the sample mean for sample 2 $\overline x_2$ = 3.9

the standard deviation for sample 2 $s_2$ = 2.7

degree of freedom for this test can be computed by using the formula:

$df = \dfrac{\begin {pmatrix} \dfrac{s_1^2}{n_1} + \dfrac{s^2_2}{n_2} \end {pmatrix}^2 } {\dfrac{ (\dfrac{s_1^2}{n_1}^2)}{ n_1-1} + \dfrac{ (\dfrac{s_2^2}{n_2}^2)}{ n_2-1} }$

$df = \dfrac{\begin {pmatrix} \dfrac{3.1^2}{30} + \dfrac{2.7^2}{30} \end {pmatrix}^2 } {\dfrac{ \begin {pmatrix} \dfrac{3.1^2}{30} \end {pmatrix}^2}{ 30-1} + \dfrac{ \begin {pmatrix} \dfrac{2.7^2}{30} \end {pmatrix}^2}{ 30-1} }}$