Please help, Brainly and 20 PTS to whoever gets it right

Which table represents a direct variation function? A table with 6 columns and 2 rows. The first row, x, has the entries, negati

IceJOKER [234]

Answer:

The correct option is;

A table with 6 columns and 2 rows. The first row, x, has entries, negative 3, negative 1, 2, 5, 10. The second row, y, has entries, negative 7.5, negative 2.5, 5.0, 12.5, 25

Please find attached the graphs of the table data

Step-by-step explanation:

Each of the given table data of in the tables are analysed to find direct variation;

Table 1

x, -3, -1, 2, 5, 10

y, -4.5, -3.0, -1.5, 0.0, 1.5

-4.5/-3 = 1.5 ≠ -3.0/-1 = 3

No direct variation

Table 2

x, -5.5, -4.5, -3.5, -2.5, -1.5

y, 10, 8, 6, 4, 2

10/(-5.5) = -20/11 ≠ 8/(-4.5) = -16/9

However, 10/(-5.5 + 0.5) = -2 = 8/(-4.5 + 0.5) = -2

Adjusted direct variation

Table 3

x, -5.5, -5.5, -5.5, -5.5, -5.5

y, -3, -1, 2, 5 , 10

-3/(-5.5) ≠ -1/-5.5

No direct variation

Table 4

x, -3, -1, 2, 5, 10

y, -7.5, -2.5, 5.0 , 12.5, 25

-7.5/-3 = 2.5 = -2.5/(-1) = 5.0/2 = 12.5/5 =25/10

Direct variation exists

8 0

3 years ago

Read 2 more answers

Estimate 73% of 120 <br><br> Thanks if you help!

xxMikexx [17]

Answer:

73/100 (120)

answer:-87.6

3 0

3 years ago

Read 2 more answers

Mathematical x4 - x2 ≥ 2x

kondaur [170]

X4-x2≥ 2x~~~2x≥ 2x True Because " _ " is the same thing with "=" ; and because 2x=2x then 2x≥ 2x ~~~~~~Have a nice day!!! ★★

4 0

3 years ago

What's bigger 3/4 or 9/12

pashok25 [27]

Change the common denominator so 3/4 ---> 9/12
3/4 and 9/12 are equal

Hope this helps!!!

3 0

3 years ago

Read 2 more answers

If a test of H subscript 0 colon space mu subscript D equals 0 space v s. space H subscript a colon space space mu subscript D g

lara [203]

Answer:

$p_v =P(t_{(n-1)}>t_{calculated}) =0.0601$

The p value on this case is given by the problem.

If we compare the p value with a significance level assumed $\alpha=0.05$ , we see that $p_v > \alpha$ and we can conclude that we FAIL to reject the null hypothesis that the difference mean between after and before is less or equal than 0.

Step-by-step explanation:

A paired t-test is used to compare two population means where you have two samples in which observations in one sample can be paired with observations in the other sample. For example if we have Before-and-after observations (This problem) we can use it.

Let put some notation

x=test value before , y = test value after

The system of hypothesis for this case are:

Null hypothesis: $\mu_y- \mu_x \leq 0$

Alternative hypothesis: $\mu_y -\mu_x >0$

The first step is calculate the difference $d_i=y_i-x_i$

The second step is calculate the mean difference

$\bar d= \frac{\sum_{i=1}^n d_i}{n}$

The third step would be calculate the standard deviation for the differences, and we got:

$s_d =\frac{\sum_{i=1}^n (d_i -\bar d)^2}{n-1}$

The 4 step is calculate the statistic given by :

$t=\frac{\bar d -0}{\frac{s_d}{\sqrt{n}}}=t_{calculated}$

The next step is calculate the degrees of freedom given by:

$df=n-1$

Now we can calculate the p value, since we have a right tailed test the p value is given by:

$p_v =P(t_{(n-1)}>t_{calculated}) =0.0601$

The p value on this case is given by the problem.

If we compare the p value with a significance level assumed $\alpha=0.05$ , we see that $p_v > \alpha$ and we can conclude that we FAIL to reject the null hypothesis that the difference mean between after and before is less or equal than 0.

4 0

3 years ago