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3 years ago

FOR BRAINLIEST ANSWER ONLY: 2. 3. 4. 6.

Mathematics

1 answer:

nasty-shy [4]3 years ago

6 0

Answer:

2. x = 2 & y = 4

3. x = 4 & y = 2

Step-by-step explanation:

2. x + y = 6

2x + y = 8 (multiply the first equation by -1, so you can eliminate the ys)

- x - y = -6

2x + y = 8 (now add the variables together)

x = 2 (plug in x in one of the equations to find out y)

x + y = 6

(2) + y = 6

-2 -2

y = 4

3. 3x + y = 14

x = 2y (plug in x into the first equation and solve it for y)

3(2y) + y = 14

6y + y = 14

7y = 14

y = 2 (plug in y in one of the equations to find out x)

x = 2y

x = 2(2)

x = 4

4. One number (x) is 2 more (+2) than twice (times 2) as large as another. their sum is 17. Find the numbers.

2x + 2 = 17 (solve for x)

-2 -2

2x = 15

x = 7.5

6. 7 (4x + 1) - (x + 6) (start by distributing 7 into the first parenthesis)

(28x + 7) - (x + 6) (do the same to the other parenthesis by distributing -1)

(28x + 7) (-x - 6) (and now just combine like terms)

28x + 7 - x - 6

28x - x + 7 - 6

27x + 1

i hope this helped! if you have any question, pls let me know!

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Step-by-step explanation:

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B (2, -2)

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3 years ago

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viva [34]

Answer:

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P-value=0.042.

Step-by-step explanation:

The question is incomplete:

The data of the scores for each student is:

Before After

430 465

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520 535

360 410

440 425

500 505

425 450

470 480

515 520

430 430

450 460

495 500

540 530

We will generate a sample for the difference of scores (before - after) and test that sample.

The sample of the difference is [35 -10 15 50 -15 5 25 10 5 0 10 5 -10]

This sample, of size n=13, has a mean of 9.615 and a standard deviation of 18.423.

The claim is that the scores after the stats course are significantly higher than the scores before (the difference in the scores is higher than 0).

Then, the null and alternative hypothesis are:

$H_0: \mu=0\\\\H_a:\mu> 0$

The significance level is 0.01.

The estimated standard error of the mean is computed using the formula:

$s_M=\dfrac{s}{\sqrt{n}}=\dfrac{18.423}{\sqrt{13}}=5.11$

Then, we can calculate the t-statistic as:

$t=\dfrac{M-\mu}{s/\sqrt{n}}=\dfrac{9.615-0}{5.11}=\dfrac{9.615}{5.11}=1.882$

The degrees of freedom for this sample size are:

$df=n-1=13-1=12$

This test is a right-tailed test, with 12 degrees of freedom and t=1.882, so the P-value for this test is calculated as (using a t-table):

$P-value=P(t>1.882)=0.042$

As the P-value (0.042) is bigger than the significance level (0.01), the effect is not significant.

The null hypothesis failed to be rejected.

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