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Sedaia [141]
3 years ago
9

Translate the English phrase into an algebraic expression fifteen more than m

Mathematics
1 answer:
Vikki [24]3 years ago
5 0

Answer:

m + 15

Step-by-step explanation:

Start at the end and work backwards to the beginning of the sentence.  Most times these work like that.  I'll start with "m", then I know that 15 more than "m" is m + 15.  It would be the same thing to express that particular problem as 15 + m because addition is commutative.  But subtraction is not.

"15 less than m" is m - 15, wheras

"m less than 15" is 15 - m.  

This matters because if m = 10, for example, 10 - 15 = -5 where 15 - 10 = 5

5 and -5 are not the same answer.

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(3 + 4) 9 + 3 = (7)9 +3

= 63 + 3

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A university claims that the average cost of books per student, per semester is $300. A group of students believes that the actu
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Answer:

t=\frac{345-300}{\frac{200}{\sqrt{100}}}=2.25    

p_v =P(t_{(99)}>2.25)=0.0133  

Conclusion  

If we compare the p value and the significance level assumed \alpha=0.01 we see that p_v>\alpha so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, so we can conclude that the mean is higher than 300 at 1% of significance.  

Step-by-step explanation:

Data given and notation  

\bar X=345 represent the sample mean

s=200 represent the sample standard deviation

n=100 sample size  

\mu_o =300 represent the value that we want to test

\alpha 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 300, the system of hypothesis would be:  

Null hypothesis:\mu \leq 300  

Alternative hypothesis:\mu > 300  

If we analyze the size for the sample is > 30 but 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{345-300}{\frac{200}{\sqrt{100}}}=2.25    

P-value

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

df=n-1=100-1=99  

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

p_v =P(t_{(99)}>2.25)=0.0133  

Conclusion  

If we compare the p value and the significance level assumed \alpha=0.01 we see that p_v>\alpha so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, so we can conclude that the mean is higher than 300 at 1% of significance.  

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