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Lostsunrise [7]

3 years ago

15

Which of the following ordered pairs are solutions to the equation y=−2x+6? (Select all that apply.) Select all that apply: (0,6

) (1,4) (−2,−2) none of the above

1 answer:

yuradex [85]3 years ago

5 0

Answer:

(0,6)

(1,4)

Step-by-step explanation:

(0,6)

1. Plug in 0 for x and solve to see if 6 results for y

y = -2(0) + 6

0 + 6 = 6

y = 6

When 0 is plugged in for x, y equals 6 which makes (0,6) a solution.

(1,4)

1. Plug in 1 for x and solve to see if 4 results for y

y = -2(1) + 6

-2 + 6 = 4

y = 4

When 1 is plugged in for x, y equals 4 which makes (1,4) a solution.

(-2,-2)

1. Plug in -2 for x and solve to see if -2 results for y

y = -2(-2) + 6

4 + 6 = 10

10 ≠ -2

When -2 is plugged in for x, y does not equal -2 which means (-2,-2) is not a solution.

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Ilia_Sergeevich [38]

b

a and d should not be considered bc it is not logically correct. c is eliminated because that may not always be the case. therefore it should be b.

7 0

3 years ago

3. There are 6 places set at each table for the big outdoor party.

Ainat [17]

You do 6x4 , The amount of legs on the chairs is irrelevant information, but your answer will be 24 people
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8 0

3 years ago

I have no idea what im doing

omeli [17]

Answer:

x = - 6

Step-by-step explanation:

A hole is an input value that makes both numerator and denominator equal to zero.

<u>Let's factorize the numerator and denominator:</u>

3x² - 108 = 3(x² - 36) = 3(x + 6)(x - 6)
x² + 2x - 24 = x² + 6x - 4x - 24 = x(x + 6) - 4(x + 6) = (x - 4)(x + 6)

<u>We have same factor of x + 6:</u>

x + 6 = 0
x = - 6 is a hole

8 0

2 years ago

Solve the system of equations y=x-4, y=-3x

abruzzese [7]

Y=x-4
x=y+4
y=-3(y+4)
y=-3y-12
4y=-12
y=-3
x=1

5 0

3 years ago

Standard Error from a Formula and a Bootstrap Distribution Sample A has a count of 30 successes with and Sample B has a count of

tia_tia [17]

Answer:

Using a formula, the standard error is: 0.052

Using bootstrap, the standard error is: 0.050

Comparison:

The calculated standard error using the formula is greater than the standard error using bootstrap

Step-by-step explanation:

Given

Sample A Sample B

$x_A = 30$ $x_B = 50$

$n_A = 100$ $n_B =250$

Solving (a): Standard error using formula

First, calculate the proportion of A

$p_A = \frac{x_A}{n_A}$

$p_A = \frac{30}{100}$

$p_A = 0.30$

The proportion of B

$p_B = \frac{x_B}{n_B}$

$p_B = \frac{50}{250}$

$p_B = 0.20$

The standard error is:

$SE_{p_A-p_B} = \sqrt{\frac{p_A * (1 - p_A)}{n_A} + \frac{p_A * (1 - p_B)}{n_B}}$

$SE_{p_A-p_B} = \sqrt{\frac{0.30 * (1 - 0.30)}{100} + \frac{0.20* (1 - 0.20)}{250}}$

$SE_{p_A-p_B} = \sqrt{\frac{0.30 * 0.70}{100} + \frac{0.20* 0.80}{250}}$

$SE_{p_A-p_B} = \sqrt{\frac{0.21}{100} + \frac{0.16}{250}}$

$SE_{p_A-p_B} = \sqrt{0.0021+ 0.00064}$

$SE_{p_A-p_B} = \sqrt{0.00274}$

$SE_{p_A-p_B} = 0.052$

Solving (a): Standard error using bootstrapping.

Following the below steps.

Open Statkey
Under Randomization Hypothesis Tests, select Test for Difference in Proportions
Click on Edit data, enter the appropriate data
Click on ok to generate samples
Click on Generate 1000 samples ---- <em>see attachment for the generated data</em>

From the randomization sample, we have:

Sample A Sample B

$x_A = 23$ $x_B = 57$

$n_A = 100$ $n_B =250$

$p_A = 0.230$ $p_A = 0.228$

So, we have:

$SE_{p_A-p_B} = \sqrt{\frac{p_A * (1 - p_A)}{n_A} + \frac{p_A * (1 - p_B)}{n_B}}$

$SE_{p_A-p_B} = \sqrt{\frac{0.23 * (1 - 0.23)}{100} + \frac{0.228* (1 - 0.228)}{250}}$

$SE_{p_A-p_B} = \sqrt{\frac{0.1771}{100} + \frac{0.176016}{250}}$

$SE_{p_A-p_B} = \sqrt{0.001771 + 0.000704064}$

$SE_{p_A-p_B} = \sqrt{0.002475064}$

$SE_{p_A-p_B} = 0.050$

5 0

3 years ago