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2 years ago

For a certain company, the cost for producing x items is 60x+300 and the revenue for selling x items is 100x−0.5x2 .

Mathematics

1 answer:

Step2247 [10]2 years ago

5 0

The equation for the profit from selling the items is P = 40x-0.5x²-300. The values of x are 10 and 70.

<h3>How to solve for the expression</h3>

We have the cost function in this question to be

Cx = 60x+300

Revenue = 100x−0.5x²

A. The revenue function is stated as the revenue - cost

= (100x−0.5x²)-60x+300

We have to open the equation

40x-0.5x²-300

B. when profit is 50 dollars

50 = 40x-0.5x²-300

We have to multiply both sides by 10

500 = 400x - 5x² - 3000

We have to collect like terms and form a quadratic equation

5x² - 400x - 3500 = 0

We have to solve the quadratic equation above using a quadratic calculator.

Hence

x = 10, 70

So the two values that would create a profit of 50 dollars is 10 and 70.

Read more on quadratic polynomials here:

brainly.com/question/25841119

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Answer:

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3 years ago

Read 2 more answers

The population of a species of rabbit triples every year. This

jok3333 [9.3K]

Answer:

"4" represents that there are 4 rabbit in the beginning.

Step-by-step explanation:

Given that the population of a species of rabbit triples every year. This

can be modeled by function $f\left(x\right)=4\left(3\right)^x$ and $f(5) = 972$ .

Now we need to find about what does the 4 represent in the given function.

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3 0

3 years ago

At a university, 60% of the 7,400 students are female. The student newspaper reports the results of a survey of a random sample

omeli [17]

Given Information:

Population mean = p = 60% = 0.60

Population size = N = 7400

Sample size = n = 50

Required Information:

Sample mean = μ = ?

standard deviation = σ = ?

Answer:

Sample mean = μ = 0.60

standard deviation = σ = 0.069

Step-by-step explanation:

We know from the central limit theorem, the sampling distribution is approximately normal as long as the expected number of successes and failures are equal or greater than 10

np ≥ 10

50*0.60 ≥ 10

30 ≥ 10 (satisfied)

n(1 - p) ≥ 10

50(1 - 0.60) ≥ 10

50(0.40) ≥ 10

20 ≥ 10 (satisfied)

The mean of the sampling distribution will be same as population mean that is

Sample mean = p = μ = 0.60

The standard deviation for this sampling distribution is given by

$\sigma = \sqrt{\frac{p(1-p)}{n} }$

Where p is the population mean that is proportion of female students and n is the sample size.

$\sigma = \sqrt{\frac{0.60(1-0.60)}{50} }\\\\\sigma = \sqrt{\frac{0.60(0.40)}{50} }\\\\\sigma = \sqrt{\frac{0.24}{50} }\\\\\sigma = \sqrt{0.0048} }\\\\\sigma = 0.069$