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kvasek [131]
3 years ago
13

A manufacturer of widgets finds that the production cost, C, in dollars per unit is a function of the number of widgets produced

. The selling price, S, of each widget in dollars is a function of the production cost per unit. C(x)=-0.1x^2+100 S(C)=1.4C
Mathematics
2 answers:
adell [148]3 years ago
8 0

Answer:

D. S(C(x))= –0.14x^2+140; $108.50

Step-by-step explanation:

Cause the others are wrong

Elis [28]3 years ago
3 0

Answer:

I guess that you want to find the profit:

We have two equations:

the cost equation:

C(x) = -0.1*x^2 + 100.

And the selling equation, that is a vertical stretch of the cost equation by a factor of 1.4:

S(x) = 1.4*C(x) = 1.4*( -0.1*x^2 + 100.) = -0.14*x^2 + 140

Now, whit those two equations we can find the profit equation, that is defined as the difference between the selling price, and the cost:

P(x) = S(x) - C(x) = 1.4*C(x) - C(x) = (1.4 - 1)*C(x) = 0.4*C(x).

Then the profit is 0.4 times the initial cost.

P(x) = 0.4*( -0.1*x^2 + 100.) = -0.04*x^2 + 40

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Answer

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Read 2 more answers
According to an​ airline, flights on a certain route are on time 80 ​% of the time. suppose 10 flights are randomly selected and
natulia [17]
<span>(a) This is a binomial experiment since there are only two possible results for each data point: a flight is either on time (p = 80% = 0.8) or late (q = 1 - p = 1 - 0.8 = 0.2).
​(b) Using the formula:</span><span>
P(r out of n) = (nCr)(p^r)(q^(n-r)), where n = 10 flights, r = the number of flights that arrive on time:
P(7/10) = (10C7)(0.8)^7 (0.2)^(10 - 7) = 0.2013
Therefore, there is a 0.2013 chance that exactly 7 of 10 flights will arrive on time.
​(c) Fewer than 7 flights are on time means that we must add up the probabilities for P(0/10) up to P(6/10).
Following the same formula (this can be done using a summation on a calculator, or using Excel, to make things faster):
P(0/10) + P(1/10) + ... + P(6/10) = 0.1209
This means that there is a 0.1209 chance that less than 7 flights will be on time.
​(d) The probability that at least 7 flights are on time is the exact opposite of part (c), where less than 7 flights are on time. So instead of calculating each formula from scratch, we can simply subtract the answer in part (c) from 1.
1 - 0.1209 = 0.8791.
So there is a 0.8791 chance that at least 7 flights arrive on time.
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0.0264 + 0.0881 + 0.2013 = 0.3158, so the probability that between 5 to 7 flights arrive on time is 0.3158.
</span>
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Snezhnost [94]
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