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3 years ago

Is this correct ? .......................

Mathematics

1 answer:

ladessa [460]3 years ago

5 0

To find the midpoint of the two points (x₁ ,y₁) and (x₂, y₂) we need to use the formula:

$\sf{Midpoint=(\frac{x_1 + x_2}{2}, \frac{y_1+y_2}{2})}$

So to find the midpoint of E (a,a) and F (3a, a), let's plug it in to the formula:

$\sf{Midpoint=(\frac{a + 3a}{2}, \frac{a+a}{2})}$

Simplify the numerator:

$\sf{Midpoint=(\frac{4a}{2}, \frac{2a}{2})}$

Simplifying the fractions more:

$\sf{Midpoint=(2a, a)}$

So the midpoint of EF is (2a,a).

Your answer of (a,a) would be wrong.

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A company that manufactures guitars has a fixed cost of $100,000. It costs $100 to produce each guitar and the selling price is

miv72 [106K]

a. Write the cost function:: C(x) = 100x + 100,000 where x is number of guitars

---------------------------------

b. Write the revenue function:: R(x) = 300x where x is number of guitars

-----------------------

c. Find the profit function.

Profit = Revenue - Cost

P(x) = R(x) - C(x)

P(x) = [ R(x) ] - [ C(x) ]

P(x) = [ 300x ] - [ 100x+100,000 ]

P(x) = 300x - 100x-100,000

P(x) = 200x - 100,000

The break even point is when the profit is 0 dollars. You don't lose any money. And you don't gain any money.

Solve 125x - 100,000 = 0

125x = 100,000

x = 800 (# of guitars made and sold)

7 0

3 years ago

Suppose you have two urns with poker chips in them. Urn I contains two red chips and four white chips. Urn II contains three red

Neporo4naja [7]

Answer:

Multiple answers

Step-by-step explanation:

The original urns have:

Urn 1 = 2 red + 4 white = 6 chips
Urn 2 = 3 red + 1 white = 4 chips

We take one chip from the first urn, so we have:

The probability of take a red one is : $\frac{1}{3}$ (2 red from 6 chips(2/6=1/2))

For a white one is: $\frac{2}{3}$ (4 white from 6 chips(4/6=(2/3))

Then we put this chip into the second urn:

We have two possible cases:

First if the chip we got from the first urn was white. The urn 2 now has 3 red + 2 whites = 5 chips
Second if the chip we got from the first urn was red. The urn two now has 4 red + 1 white = 5 chips

If we select a chip from the urn two:

In the first case the probability of taking a white one is of: $\frac{2}{5}$ = 40% ( 2 whites of 5 chips)
In the second case the probability of taking a white one is of: $\frac{1}{5}$ = 20% ( 1 whites of 5 chips)

This problem is a dependent event because the final result depends of the first chip we got from the urn 1.

For the fist case we multiply :

$\frac{4}{6}$ x $\frac{2}{5}$ = $\frac{4}{15}$ = 26.66% ( $\frac{4}{6}$ the probability of taking a white chip from the urn 1, $\frac{2}{5}$ the probability of taking a white chip from urn two)

For the second case we multiply:

$\frac{1}{3}$ x $\frac{1}{5}$ = $\frac{1}{30}$ = .06% ( $\frac{1}{3}$ the probability of taking a red chip from the urn 1, $\frac{1}{5}$ the probability of taking a white chip from the urn two)

8 0

3 years ago

A bag of gumballs contains 2 pink, 3 orange, 4 blue, 2 red, and 1 yellow

podryga [215]

Answer:

1/4

Step-by-step explanation:

there are 12 gumballs in total

1 yellow + 2 pink = 3 gumballs out of all 12

3 /12 is simplified to 1/4 as both numbers is divisible by 3

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2 years ago

Help I need it I’m almost done

sesenic [268]

Answer:

Step-by-step explanation:

3 0

3 years ago

Just looking for a boy bsf (less than 14) also im friking bored :|

Anettt [7]

Answer:

I am bored too

Step-by-step explanation:

Go to school, sit down, and you will come to find the answer

7 0

3 years ago