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2 months ago

How much profit was generated by the sales of gold label and black label combined?

Mathematics

2 answers:

lawyer [12.5K]2 months ago

8 0

The profit produced by combining sales of gold label and black label equals the total of their individual profits. From the table $\begin{array}{ccc}&\text{Number of bottles sold}&\text{Average profit (per bottle)}\\\text{Gold label}&1,000&\$2.75\end{array}$ you find the gold label's profit $1,000\cdot \$2.75=\$2,750.$ and similarly from another table $\begin{array}{ccc}&\text{Number of bottles sold}&\text{Average profit (per bottle)}\\\text{Black label}&2,000&\$1.50\end{array}$ find the black label's profit $2,000\cdot \$1.50=\$3,000.$ . Adding these yields $\$2,750+\$3,000=\$5,750.$ . Thus, option 5 is the correct choice.

PIT_PIT [12.4K]2 months ago

4 0

The sum of profits from gold label and black label sales amounts to $\boxed{{\mathbf{\$5750}}}$ . Here's the breakdown: From the provided data, calculate gold label profit as ${\text{profit earned by gold label}}={\text{ number of bottles sold }}\times{\text{ average profit}}$ based on the number of bottles sold and average profit per unit. Similarly, calculate black label profit using the 2000 bottles sold and average profit ${\text{profit earned by black label}}={\text{ number of bottles sold }}\times{\text{ average profit}}$ . Adding these profits gives $\begin{aligned}{\text{combined profit}}&={\text{profit earned by black label}}+{\text{profit earned by gold label}}\\&=\$3000+\$2750\\&=\$5750\\\end{aligned}$ , concluding that the total combined profit is $\$5750$ , which matches option five $\boxed{{\mathbf{\$5750}}}$ . Further relevant resources are linked for deeper understanding. This explanation is suited for middle school math covering profit and loss concepts, involving keywords such as profit, sales, combined totals, and average calculations.

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23 days ago

Construct an interval estimate for the given parameter using the given sample statistic and margin of error. For μ1-μ2, using x¯

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Answer: CI = (0, 8)

Step-by-step explanation: The confidence interval for the difference in means is given as

Lower limit

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Upper limit

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Where x1 - x2 = 8 and the margin of error = 8

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1 month ago

A random sample of 16 students selected from the student body of a large university had an average age of 25 years and a standar

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Answer:

The P-value ranges between 2.5% and 5% according to the t-table.

Step-by-step explanation:

A random sample of 16 students from a large university showed an average age of 25 years with a standard deviation of 2 years.

Let $\mu$ = true average age of all students at the university.

So, the Null Hypothesis, $H_0$ : $\mu \leq$ 24 years {indicating the average age is less than or equal to 24 years}

Alternate Hypothesis, $H_A$ : $\mu$ > 24 years {indicating the average age is significantly greater than 24 years}

Here we employ the One-sample t-test statistics as the population's standard deviation is unknown;

T.S. = $\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }$ ~ $t_n_-_1$

where, $\bar X$ = sample average age = 25 years

s = sample standard deviation = 2 years

n = sample size = 16

This gives us the test statistics = $\frac{25-24}{\frac{2}{\sqrt{16} } }$ ~ $t_1_5$

= 2

The value of the t-test statistics is 2.

Moreover, the P-value of the test-statistics can be found as follows;

P-value = P( $t_1_5$ > 2) = 0.034 {as per the t-table}

Thus, the P-value lies between 2.5% and 5% based on the t-table.

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