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Finger [1]
3 years ago
11

PLEASE HELP! I WILL MARK AS BRAINLIEST!! I REALLY NEED SOMEONES HELP!

Mathematics
1 answer:
Andrej [43]3 years ago
4 0

Since this is subtraction, everything must be turned negative in the second polynomial.

(6x^2 - 5x) - (2x^2 + 3x - 2)

6x^2 - 5x - 2x^2 - 3x - (-2)

6x^2 - 5x - 2x^2 - 3x + 2

Now, reorder the terms to make it easier.

6x^2 - 2x^2 - 5x - 3x + 2

Now, just combine like terms.

4x^2 - 5x - 3x + 2

4x^2 - 8x + 2

There’s the answer!

Please consider marking this answer as Brainliest to help me advance.

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In the past, the average age of employees of a large corporation has been 40 years. Recently, the company has been hiring older
Viktor [21]

Answer:

p_v =P(t_{(63)}>2.5)=0.0075  

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 fail reject the null hypothesis, so we can conclude that the mean age is significantly higher than 45 years at 5% of significance.  

Step-by-step explanation:

1) Data given and notation  

\bar X=45 represent the mean height for the sample  

s=16 represent the sample standard deviation for the sample  

n=64 sample size  

\mu_o =40 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 age is higher than 40 years, the system of hypothesis would be:  

Null hypothesis:\mu \leq 40  

Alternative hypothesis:\mu > 40  

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{45-40}{\frac{16}{\sqrt{64}}}=2.5    

P-value

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

df=n-1=64-1=63  

Since is a one right tailed test the p value would be:  

p_v =P(t_{(63)}>2.5)=0.0075  

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 fail reject the null hypothesis, so we can conclude that the mean age is significantly higher than 45 years at 5% of significance.  

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