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Vitek1552 [10]
3 years ago
12

An artist spends a total of $150 on the supplies needed to create the paintings, p, and drawings, d, he plans to sell. Choose th

e expression that would represent the profit the artist will make if each painting sells for $75 and each drawings for $35.
(A) 75p + 35d - 150
(B) 35p + 75d - 150
(C) 75p - 35d - 150
(D) 35 + p + 75 + d + 150 ​
Mathematics
2 answers:
Andru [333]3 years ago
8 0
<h3>Answer: Choice A) 75p + 35d - 150</h3>

===================================================

Work Shown:

p = number of paintings sold

75p = amount made from paintings alone ($75 per painting)

d = number of drawings sold

35d = amount made from drawings only ($35 per drawing)

R = revenue = amount of money coming in

R = 75p+35d

C = cost = amount of money going out

C = 150

P = profit

P = R - C

P = ( R ) - ( C )

P = ( 75p+35d ) - ( 150 )

P = 75p + 35d - 150

Llana [10]3 years ago
3 0

Answer:

The answer is both (A) and (B)

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Answer:

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A line passes through the point (-3, -3) and has a slope of 1/2. What is the equation of the line? A y= 1/2x + 3/2 B y= 1/2x -3/
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Answer:

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Step-by-step explanation:

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While researching the cost of school lunches per week across the state, you use a sample size of 45 weekly lunch prices. The sta
Drupady [299]

We assume the lunch prices we observe are drawn from a normal distribution with true mean \mu and standard deviation 0.68 in dollars.


We average n=45 samples to get \bar{x}.


The standard deviation of the average (an experiment where we collect 45 samples and average them) is the square root of n times smaller than than the standard deviation of the individual samples. We'll write


\sigma = 0.68 / \sqrt{45} = 0.101


Our goal is to come up with a confidence interval (a,b) that we can be 90% sure contains \mu.


Our interval takes the form of ( \bar{x} - z \sigma, \bar{x} + z \sigma ) as \bar{x} is our best guess at the middle of the interval. We have to find the z that gives us 90% of the area of the bell in the "middle".


Since we're given the standard deviation of the true distribution we don't need a t distribution or anything like that. n=45 is big enough (more than 30 or so) that we can substitute the normal distribution for the t distribution anyway.


Usually the questioner is nice enough to ask for a 95% confidence interval, which by the 68-95-99.7 rule is plus or minus two sigma. Here it's a bit less; we have to look it up.


With the right table or computer we find z that corresponds to a probability p=.90 the integral of the unit normal from -z to z. Unfortunately these tables come in various flavors and we have to convert the probability to suit. Sometimes that's a one sided probability from zero to z. That would be an area aka probability of 0.45 from 0 to z (the "body") or a probability of 0.05 from z to infinity (the "tail"). Often the table is the integral of the bell from -infinity to positive z, so we'd have to find p=0.95 in that table. We know that the answer would be z=2 if our original p had been 95% so we expect a number a bit less than 2, a smaller number of standard deviations to include a bit less of the probability.


We find z=1.65 in the typical table has p=.95 from -infinity to z. So our 90% confidence interval is


( \bar{x} - 1.65 (.101),  \bar{x} + 1.65 (.101) )


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\pm 1.65(.101) = \pm 0.167 dollars


That's around plus or minus 17 cents.




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