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AveGali [126]
3 years ago
15

A small company manufacturers a certain item and sells it online. The company has business model where the cost C, in dollars, t

o make x is given by the equation C=20/3x+50 and the revenue R, in dollars, made by selling x items is given by the equation R=10x. The break even point is the point where the cost and the revenue equations intersect.
Mathematics
1 answer:
Mars2501 [29]3 years ago
8 0

Answer:

Step-by-step explanation:

We know that the break even point is the point where the cost and the revenue equations intersect.  So the break even point in this situation is:

C= R

<=> 20/3x+50 = 10x (with x > 0 and x is a whole number)

<=> 10x^{2} -50x -20 = 0

<=> x ≈ 5

If the company sells more than 5 items, it will have benefit

If the company sells less than 5 items, it will lose

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t=\frac{(810-770)-0}{\sqrt{\frac{67^2}{13}+\frac{56^2}{19}}}}=1.771  

df=n_1 +n_2 -2=13+19-2=30  

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

p_v =P(t_{30}>1.771)=0.0434  

Comparing the p value with the significance level \alpha=0.05 we see that p_v so we can conclude that we have enough evidence to reject the null hypothesis, and the mean for group 1 is significantly higher than the mean for the group 2

Step-by-step explanation:

Data given

\bar X_{1}=810 represent the mean for sample 1  

\bar X_{2}=770 represent the mean for sample 2  

s_{1}=67 represent the sample standard deviation for 1  

s_{2}=56 represent the sample standard deviation for 2  

n_{1}=13 sample size for the group 2  

n_{2}=19 sample size for the group 2  

\alpha=0.05 Significance level provided

t would represent the statistic (variable of interest)  

Concepts and formulas to use  

We need to conduct a hypothesis in order to check if the mean for category 1 is higher than the mean for category 2, the system of hypothesis would be:  

Null hypothesis:\mu_{1}-\mu_{2}\leq 0  

Alternative hypothesis:\mu_{1} - \mu_{2}> 0  

We don't have the population standard deviation's, so for this case is better apply a t test to compare means, and the statistic is given by:  

t=\frac{(\bar X_{1}-\bar X_{2})-\Delta}{\sqrt{\frac{\sigma^2_{1}}{n_{1}}+\frac{\sigma^2_{2}}{n_{2}}}} (1)  

And the degrees of freedom are given by df=n_1 +n_2 -2=13+19-2=30  

t-test: Is used to compare group means. Is one of the most common tests and is used to determine whether the means of two groups are equal to each other.  

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

t=\frac{(810-770)-0}{\sqrt{\frac{67^2}{13}+\frac{56^2}{19}}}}=1.771  

P value  

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

p_v =P(t_{30}>1.771)=0.0434  

Comparing the p value with the significance level \alpha=0.05 we see that p_v so we can conclude that we have enough evidence to reject the null hypothesis, and the mean for group 1 is significantly higher than the mean for the group 2

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