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Gre4nikov [31]
2 years ago
9

(05.06 MC)

Mathematics
1 answer:
irina [24]2 years ago
6 0

The point (-2, 5) is not included in the solution area

<h3>The graph of the inequalities</h3>

The system of inequalities is given as:

y > 5x + 5

y > -2x+1

See attachment for the graph.

Since the inequalities use the greater than symbol, then the lines of the inequalities would be a dotted line and the upper part would be shaded

<h3>The solution area</h3>

The point is given as:

(-2, 5)

The point (-2, 5) is not in the shaded area of the system of inequalities

Hence, the point is not included in the solution area

Mathematically, we have:

5 > 5 * -2 + 5 ⇒ 5 > -5 --- true

5 > -2 * -2 +1 ⇒ 5 > 5 --- false

Since both inequalities are not true, then the point is justified

Read more about system of inequalities at:

brainly.com/question/19526736

#SPJ1

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t=\frac{201-190}{\frac{10}{\sqrt{25}}}=5.5    

p_v =P(t_{(24)}>5.5)=0.00000589  

If we compare the p value and the significance level given \alpha=0.05 we see that p_v so we can conclude that we have enough evidence to reject the null hypothesis, and we can conclude that the true mean is higher than 190

Step-by-step explanation:

Data given and notation  

\bar X=201 represent the mean

s=10 represent the sample standard deviation for the sample  

n=25 sample size  

\mu_o =190 represent the value that we want to test

\alpha=0.05 represent the significance level for the hypothesis test.  

t would represent the statistic (variable of interest)  

p_v represent the p value for the test (variable of interest)  

State the null and alternative hypotheses.  

We need to conduct a hypothesis in order to check if the mean is higher than 190, the system of hypothesis would be:  

Null hypothesis:\mu \leq 190  

Alternative hypothesis:\mu > 190  

If we analyze the size for the sample is < 30 and we don't know the population deviation so is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:  

t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}  (1)  

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".  

Calculate the statistic

We can replace in formula (1) the info given like this:  

t=\frac{201-190}{\frac{10}{\sqrt{25}}}=5.5    

P-value

The first step is calculate the degrees of freedom, on this case:  

df=n-1=25-1=24  

Since is a one side test the p value would be:  

p_v =P(t_{(24)}>5.5)=0.00000589  

Conclusion  

If we compare the p value and the significance level given \alpha=0.05 we see that p_v so we can conclude that we have enough evidence to reject the null hypothesis, and we can conclude that the true mean is higher than 190

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3 years ago
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