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Artyom0805 [142]
1 year ago
14

I'm trying to evaluate and solve this quadratic function and I get so lost on word problems I will include a picture

Mathematics
1 answer:
SashulF [63]1 year ago
4 0

Given:

The Revenue function R(x):

R(x)\text{ = 550x}

The cost function C(x) :

C(x)=1240+2x^2

Solution

1. Profit is the difference between the revenue and the cost. Hence, the Profit function P(x) is:

\begin{gathered} P(x)\text{ = R(x) - C(x)} \\ =550x-(1240+2x^2) \\ =550x-1240-2x^2 \end{gathered}

2. The vertical intercept of P(x) can be obtained by substituting x =0 into P(x)

\begin{gathered} P(x)=550x-1240-2x^2 \\ P(0)\text{ = 0 -1240 -0} \\ P(0)\text{ = -1240} \end{gathered}

Ordered Pair : (0, -1240)

If the company sells no items, they would lose $ 1240

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What equation can be used to create the sequence -3, 2, 7, 12,
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y=x+5

Step-by-step explanation:

You can tell by looking at the sequence that each number is five more than the previous. With that in mind, you know that, to find the next number in the sequence, you have to add five to the one before it.

Making x the previous number and y the next number in the sequence, you get the equation y=x+5

You can plug pairs from the sequence to check:

2= -3+5

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12= 7+5

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What is the slope of the line shown on the graph?
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Step-by-step explanation:

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What’s 6 x 5 * divide* 10? I need answers plz
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3 years ago
Read 2 more answers
A researcher is interested in finding a 95% confidence interval for the mean number minutes students are concentrating on their
Sever21 [200]

Answer:

A. Normal

B. Between 40.08 minutes and 43.92 minutes.

C. About 95 percent of these confidence intervals will contain the true population mean number of minutes of concentration and about 5 percent will not contain the true population mean number of minutes of concentration.

Step-by-step explanation:

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean \mu and standard deviation \sigma, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean \mu and standard deviation s = \frac{\sigma}{\sqrt{n}}.

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

For a proportion p in a sample of size n, the sampling distribution of the sample proportion will be approximately normal with mean \mu = p and standard deviation s = \sqrt{\frac{p(1-p)}{n}}

x% confidence interval:

A confidence interval is built from a sample, has bounds a and b, and has a confidence level of x%. It means that we are x% confident that the population mean is between a and b.

Question A:

By the Central Limit Theorem, a normal distribution.

Question B:

We have that to find our \alpha level, that is the subtraction of 1 by the confidence interval divided by 2. So:

\alpha = \frac{1 - 0.95}{2} = 0.025

Now, we have to find z in the Ztable as such z has a pvalue of 1 - \alpha.

That is z with a pvalue of 1 - 0.025 = 0.975, so Z = 1.96.

Now, find the margin of error M as such

M = z\frac{\sigma}{\sqrt{n}}

In which \sigma is the standard deviation of the population and n is the size of the sample.

M = 1.96\frac{12}{\sqrt{150}} = 1.92

The lower end of the interval is the sample mean subtracted by M. So it is 42 - 1.92 = 40.08 minutes

The upper end of the interval is the sample mean added to M. So it is 42 + 1.92 = 43.92 minutes

Between 40.08 minutes and 43.92 minutes.

Question C:

x% confidence interval -> x% will contain the true population mean, (100-x)% wont.

So, 95% confidence interval:

About 95 percent of these confidence intervals will contain the true population mean number of minutes of concentration and about 5 percent will not contain the true population mean number of minutes of concentration.

3 0
3 years ago
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