PLEASE HELP ME ASAP!
2 answers:
(For this, I'll be using the substitution method.)
So firstly, subtract 9 on each side of the first equation. Your equation will look like this:
Next, flip the signs around in the first equation. Your equation will look like this:
Next, substitute x in the second equation for (-2y+9). From there you can solve for y.
Firstly, foil 3(-2y+9).
Next, combine like terms:
Subtract 27 on each side:
Lastly, divide -12 on each side, and your answer should be
To solve for x, just substitute y for 1 in either equation. For this Ill be doing the first equation.
Multiply 2 and 1
Subtract 9 on each side
Lastly, flip the signs, and your answer should be
In short, the answer is (7,1).
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Answer:
Step-by-step explanation:
To find the population variance, we first need the mean:
so the mean is 7.2. To find the population variance (which is almost exactly the same as the sample variance except for a small difference in the denominators of the formula) we have to take each number minus the mean, and then square the difference. Add together all these squared numbers and then divide by the number of numbers. Like this:
Add together those numbers and divide them by 5:
So hmm let's take a peek at the cost first
so, they chartered the plane for 150 folks with a fixed cost of 250,000
now, incidental fees are 300 per person, if we use the quantity "x", for how many folks, then if "x" persons are booked, then incidental fees are 300x
so, more than likely an insurance agency is charging them 300x for coverage
anyway, thus the cost C(x) = 250,000 + 300x
now, the Revenue R(x), is simple is jut price * quantity
well, the price, thus far we know is 5000 for 80 folks, but it can be lowered by 30 to get one more person, thus increasing profits
so... let's see what the price say y(x) is
so.. now we know y(x) = -30x+7400
now, Revenue is just price * quantity
the price y(x) is -30x+7400, the quantity is "x"
that simply means R(x) = -30x²+7400x
now, for the profit P(x)
the profit is simple, that is just incoming revenue minus costs, whatever is left, is profit
so P(x) = R(x) - C(x)
P(x) = (7400x - 30x²) - (250,000+300x)
P(x) = -30x² + 7100x - 250,000
now, where does it get maximized? namely, where's the maximum for P(x)?
well
and as you can see, if you zero out the derivative, there's only 1 critical point, run a first-derivative test on it, to see if its a maximum
so, they chartered the plane for 150 folks with a fixed cost of 250,000
now, incidental fees are 300 per person, if we use the quantity "x", for how many folks, then if "x" persons are booked, then incidental fees are 300x
so, more than likely an insurance agency is charging them 300x for coverage
anyway, thus the cost C(x) = 250,000 + 300x
now, the Revenue R(x), is simple is jut price * quantity
well, the price, thus far we know is 5000 for 80 folks, but it can be lowered by 30 to get one more person, thus increasing profits
so... let's see what the price say y(x) is
so.. now we know y(x) = -30x+7400
now, Revenue is just price * quantity
the price y(x) is -30x+7400, the quantity is "x"
that simply means R(x) = -30x²+7400x
now, for the profit P(x)
the profit is simple, that is just incoming revenue minus costs, whatever is left, is profit
so P(x) = R(x) - C(x)
P(x) = (7400x - 30x²) - (250,000+300x)
P(x) = -30x² + 7100x - 250,000
now, where does it get maximized? namely, where's the maximum for P(x)?
well
and as you can see, if you zero out the derivative, there's only 1 critical point, run a first-derivative test on it, to see if its a maximum