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SIZIF [17.4K]
2 years ago
14

Suppose a certain company sell reg keyboards for $82 and wireless $110 last week the store sold three times as many regular keyb

oards as wireless. If the total sales were 4628 how many was sold?
Mathematics
1 answer:
shutvik [7]2 years ago
3 0

Answer:

3,471 regular keyboards sold

1,157 wireless keyboards sold

Step-by-step explanation:

To represent this scenario use this system of equations:

x + y = 4628

x = 3y

x = regular keyboards

y = wireless keyboards

The first equation represents how many regular and wireless keyboards were sold to equal 4,628 units of keyboards.

The second equation represents how there are three times as many regular keyboards sold than wireless.

Solve this by plugging in x in the second equation to the first equation:

(3y) + y = 4628

4y = 4628                        Add the y's together to get 4y

y = 1,157

Therefore, 1,157 wireless keyboards were sold

To find how many regular keyboards were sold, plug y back into the second equation:

x = 3(1157)

x = 3,471

The final answer to this problem is that 3,471 regular keyboards were sold and 1,157 wireless keyboards were sold.

Hope this helps!! If you have any questions about the work please let me know!

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Answer:

s_p =\sqrt{\frac{(12 -1)(5100)^2 +(12-1)(5900)^2}{12 +12 -2}}=5514.526

t=\frac{(37900-39800)-0}{5514.526\sqrt{\frac{1}{12}+\frac{1}{12}}}}=-0.844  

p_v =2*P(t_{22}  

Comparing the p value with a significance level for example \alpha=0.1 we see that p_v>\alpha so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can conclude that the true means are not significantly different.  

Step-by-step explanation:

Data given and notation

\bar X_{A}=37900 represent the mean for A

\bar X_{B}=39800 represent the mean for B

s_{A}=5110 represent the sample standard deviation for A

s_{B}=5900 represent the sample standard deviation for B

n_{A}=12 sample size for the group A  

n_{B}=12 sample size for the group B

\alpha 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 means are equal or not, the system of  hypothesis would be:  

Null hypothesis:\mu_{A}-\mu_{B}= 0  

Alternative hypothesis:\mu_{A} - \mu_{B}\neq 0  

We don't have the population standard deviation's but we assume that the population deviation is equal for both populations, so we can apply a t test to compare means, and the statistic is given by:  

t=\frac{(\bar X_{A}-\bar X_{B})-\Delta}{s_p\sqrt{\frac{1}{n_{A}}+\frac{1}{n_{B}}}} (1)  

Where s_p represent the standard deviation pooled given by:

s_p =\sqrt{\frac{(n_A -1)s^2_A +(n_B -1)s^2_B}{n_A +n_B -2}}

s_p =\sqrt{\frac{(12 -1)(5100)^2 +(12-1)(5900)^2}{12 +12 -2}}=5514.526

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{(37900-39800)-0}{5514.526\sqrt{\frac{1}{12}+\frac{1}{12}}}}=-0.844  

P value  

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

df=n_A +n_{B}-2=12+12-2=22

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

p_v =2*P(t_{22}  

Comparing the p value with a significance level for example \alpha=0.1 we see that p_v>\alpha so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can conclude that the true means are not significantly different.  

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