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ycow [4]
2 years ago
14

A store specializing in mountain bikes is to open in one of two malls. If the first mall is selected, the store anticipates a ye

arly profit of $1,425,000 if successful, and a yearly loss of $475,000 otherwise. The probability of success is .5. If the second mall is selected, it is estimated that the yearly profit will be $950,000 if successful; otherwise, the annual loss will be $285,000. The probability of success at the second mall is .34
What is the expected profit for the second mall?
Mathematics
1 answer:
morpeh [17]2 years ago
7 0

Answer:

134900

Step-by-step explanation:

In this case, we are only asked to have the account for the second shopping center, now we know that the data to take into account is if it is successful it would be 950,000 gains but if not, the losses will be 285,000. With the success rate it is 0.34, that is, the failure rate would be 1 - 0.34 = 0.66.

Then the gain of would be the weighted average of these values, like this:

G = 0.34 * 950000 + 0.66 * (- 285000)

G = 134,900

Approximately 134900

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What is the area of the triangle? ​
vladimir2022 [97]

Answer:

10

Step-by-step explanation:

1/2×(-2-(-7))×(6-(-4))

5 0
3 years ago
5. Show that the following points are collinear. a) (1, 2), (4, 5), (8,9) ​
Irina-Kira [14]

Label the points A,B,C

  • A = (1,2)
  • B = (4,5)
  • C = (8,9)

Let's find the distance from A to B, aka find the length of segment AB.

We use the distance formula.

A = (x_1,y_1) = (1,2) \text{ and } B = (x_2, y_2) = (4,5)\\\\d = \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2}\\\\d = \sqrt{(1-4)^2 + (2-5)^2}\\\\d = \sqrt{(-3)^2 + (-3)^2}\\\\d = \sqrt{9 + 9}\\\\d = \sqrt{18}\\\\d = \sqrt{9*2}\\\\d = \sqrt{9}*\sqrt{2}\\\\d = 3\sqrt{2}\\\\

Segment AB is exactly 3\sqrt{2} units long.

Now let's find the distance from B to C

B = (x_1,y_1) = (4,5) \text{ and } C = (x_2, y_2) = (8,9)\\\\d = \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2}\\\\d = \sqrt{(4-8)^2 + (5-9)^2}\\\\d = \sqrt{(-4)^2 + (-4)^2}\\\\d = \sqrt{16 + 16}\\\\d = \sqrt{32}\\\\d = \sqrt{16*2}\\\\d = \sqrt{16}*\sqrt{2}\\\\d = 4\sqrt{2}\\\\

Segment BC is exactly 4\sqrt{2} units long.

Adding these segments gives

AB+BC = 3\sqrt{2}+4\sqrt{2} = 7\sqrt{2}

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

Now if A,B,C are collinear then AB+BC should get the length of AC.

AB+BC = AC

Let's calculate the distance from A to C

A = (x_1,y_1) = (1,2) \text{ and } C = (x_2, y_2) = (8,9)\\\\d = \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2}\\\\d = \sqrt{(1-8)^2 + (2-9)^2}\\\\d = \sqrt{(-7)^2 + (-7)^2}\\\\d = \sqrt{49 + 49}\\\\d = \sqrt{98}\\\\d = \sqrt{49*2}\\\\d = \sqrt{49}*\sqrt{2}\\\\d = 7\sqrt{2}\\\\

AC is exactly 7\sqrt{2} units long.

Therefore, we've shown that AB+BC = AC is a true equation.

This proves that A,B,C are collinear.

For more information, check out the segment addition postulate.

7 0
2 years ago
HW HELP ASAP PLZZZZZ
emmasim [6.3K]

Answer:

3

Step-by-step explanation:

I think its 3 because when we add 2 to x the graphic must be continue two step to right

6 0
2 years ago
Please help me asap!
Artemon [7]

Answer:   1. ∠A= 80.75°  2. 41.79 cm^2

Step-by-step explanation:

Since, According to the sines low,

\frac{sin A}{CB} = \frac{sin C}{AB}

Here, CB= 4.1 cm,  AB = 3.3 and ∠ C = 52.6°

\frac{sin A}{4.1} = \frac{sin 52.6}{3.3}

⇒ sin A = 4.1\times\frac{sin 52.6}{3.3}

⇒  sin A =0.98699998308

⇒ A = 80.75°

2. Since, the area of the given figure = Area of the rectangle having dimension 8.3 × 4.2 + Area of the half square of radius 2.1

=34.86 + 6.93

= 41.79 square cm

4 0
2 years ago
Which is the perimeter of the triangle below?<br>13/24 foot<br>22/24 foot<br>28/24 foot​
tensa zangetsu [6.8K]

Answer:

c.  22/24

Step-by-step explanation:

perimeter = 3/8 + 3/8 + 1/6

perimeter = 22/24 or 11/12

3 0
3 years ago
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