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Monica [59]
3 years ago
15

Josh sold 465 bottles of apple cider and 605 bottles of wine. He made $10,050 altogether. If he sold 465 bottles of apple cider

and only 586 bottles of wine, he would just make $9,822. How much did 605 bottles of wine sell for? $______
Mathematics
2 answers:
kow [346]3 years ago
8 0

Answer:   605 bottles of wine sell for $5,445

Step-by-step explanation:

Hi, we have to write an equation for each case

465 a + 605w =10050

465a+586w=9882

Where “a” is the price of a bottle of apple cider, and “w”is the price of a bottle of wine.

Solving for a in the first equation

465 a + 605w =10050

465a =10050-605w

a= (10050-605w)/465

a= 21.61-1.30w

Replacing "a" in the second equation

465 ( 21.61-1.30w) +586w=9882

10048.65 -604.5w +586w=9882

-604.5w+586w =9882-10048.65

-18.5w=-166.65

w= -166.65/-18.5

w= 9  

Each bottle of wine sells for $9,  

605 bottles:

605x9 = $5,445

Anon25 [30]3 years ago
4 0

Answer:

465a + 605w = $10050

465a + 586w = 9822

465a + 605w = 10050

-465a - 586w = -9822

19w = 228

w= $12 each for wine

465a + 605(12)= 10050

465a + 7260= 10050

465a = 2790

a =$6 each apple cider

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Answer:

t=\frac{29.35-30}{\frac{1.365}{\sqrt{6}}}=-1.17  

p_v =P(t_{(5)}  

If we compare the p value and the significance level given \alpha=0.05 we see that p_v>\alpha so we can conclude that we have enough evidence to FAIL to reject the null hypothesis.  

We can say that at 5% of significance the true mean is not significantly less than 30 so then the claim that true average fuel efficiency is (at least) 30 mpg makes sense

Step-by-step explanation:

Data given and notation  

The mean and sample deviation can be calculated from the following formulas:

\bar X =\frac{\sum_{i=1}^n x_i}{n}

s=\sqrt{\frac{\sum_{i=1}^n (x_i -\bar X)}{n-1}}

\bar X=29.35 represent the sample mean  

s=1.365 represent the sample standard deviation  

n=6 sample size  

\mu_o =30 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 true mean is lower than 30 or no, the system of hypothesis are :  

Null hypothesis:\mu \geq 30  

Alternative hypothesis:\mu < 30  

Since we don't know the population deviation, 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{29.35-30}{\frac{1.365}{\sqrt{6}}}=-1.17  

P-value  

We need to calculate the degrees of freedom first given by:  

df=n-1=6-1=5  

Since is a one-side left tailed test the p value would given by:  

p_v =P(t_{(5)}  

Conclusion  

If we compare the p value and the significance level given \alpha=0.05 we see that p_v>\alpha so we can conclude that we have enough evidence to FAIL to reject the null hypothesis.  

We can say that at 5% of significance the true mean is not significantly less than 30 so then the claim that true average fuel efficiency is (at least) 30 mpg makes sense

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