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Zielflug [23.3K]
3 years ago
14

On a coordinate plane, a dashed straight line has a negative slope and goes through (0, 2) and (4, 0). Everything below and to t

he left of the line is shaded. Which point is a solution to the linear inequality y < Negative one-halfx + 2? (2, 3) (2, 1) (3, –2) (–1, 3)

Mathematics
1 answer:
yanalaym [24]3 years ago
4 0

Answer:

Option C.

Step-by-step explanation:

A dashed straight line has a negative slope and goes through (0, 2) and (4, 0).

The given inequality is

y

We need find the point which is a solution to the given linear inequality.

Check the given inequality for point (2, 3).

3

3    

This statement is false. Option 1 is incorrect.

Check the given inequality for point (2, 1).

1

1

This statement is false. Option 2 is incorrect.

Check the given inequality for point (3, -2).

-2

-2

This statement is false. Option 3 is correct.

Check the given inequality for point (-1,3).

3

3

This statement is false. Option 4 is incorrect.

Therefore, the correct option is C.

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Which of the following is a solution of x2 + 6x = −18? x = 3 − 3i x = −3 + 3i x = −6 + 3i x = 6 − 3i
Marina86 [1]

Step 1 ) Move all terms right side of the equation.

\displaystyle\ x^{2} + 6x = -18

\displaystyle\ x^{2} + 6x + 18 = -18 + 18

\displaystyle\ x^{2} +6x + 18 = 0

Step 2 ) Apply quadratic formula. (Note: There are 2 solutions)

\displaystyle\ x^{2} +6x + 18 = 0

\displaystyle\ x = \frac{-b\sqrt{b^{2}-4ac}}{2a}, \displaystyle\ x = \frac{-b-\sqrt{b^{2}-4ac}}{2a}

\displaystyle\ x = \frac{-6+\sqrt{6^{2}-4 * 18}}{2} , \displaystyle\ x = \frac{-6-\sqrt{6^{2}-4*18}}{2}

\displaystyle\ x = \frac{-6+6i}{2} ,

\displaystyle\ x = \frac{-6-6i}{2}

Step 3 ) Simplify.

\displaystyle\ x = \frac{-6+6i}{2} ,

\displaystyle\ x = \frac{-6-6i}{2}

\displaystyle\ x = -3+ 3i,

\displaystyle\ x = -3 - 3i

Since the options only provide one of the answer we found, the answer is...

\displaystyle\ x = -3 - 3i

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- <em>Marlon Nunez</em>

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zzz [600]

Answer:

t=\frac{\bar d -0}{\frac{s_d}{\sqrt{n}}}=\frac{0.983 -0}{\frac{1.685}{\sqrt{12}}}=2.021

df=n-1=12-1=11

p_v =P(t_{(11)}>2.021) =0.0342

If we compare the the p value with the significance level provided \alpha=0.1, we see that p_v < \alpha, so then we can reject the null hypothesis. and there is a significant increase in the miles per gallon from 2017 to 2019 at 10% of significance.

Step-by-step explanation:

Previous concepts

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 2017 , y = test value 2019

x: 28.7 32.1 29.6 30.5 31.9 30.9 32.3 33.1 29.6 30.8 31.1 31.6

y: 31.1 32.4 31.3 33.5 31.7 32.0 31.8 29.9 31.0 32.8 32.7 33.8

Solution to the problem

The system of hypothesis for this case are:

Null hypothesis: \mu_y- \mu_x \leq 0

Alternative hypothesis: \mu_y -\mu_x >0

Because if we have an improvement we expect that the values for 2019 would be higher compared with the values for 2017

The first step is calculate the difference d_i=y_i-x_i and we obtain this:

d: 2.4, 0.3, 1.7,3,-0.2, 1.1, -0.5, -3.2, 1.4, 2, 1.6, 2.2

The second step is calculate the mean difference  

\bar d= \frac{\sum_{i=1}^n d_i}{n}= \frac{11.8}{12}=0.983

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} =1.685

We assume that the true difference follows a normal distribution. The 4th step is calculate the statistic given by :

t=\frac{\bar d -0}{\frac{s_d}{\sqrt{n}}}=\frac{0.983 -0}{\frac{1.685}{\sqrt{12}}}=2.021

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

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If we compare the the p value with the significance level provided \alpha=0.1, we see that p_v < \alpha, so then we can reject the null hypothesis. and there is a significant increase in the miles per gallon from 2017 to 2019 at 10% of significance.

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