Please help and you will get 5 points

Find the sum of the geometric series.<br> 8+6+9/2+27/8+...

Tema [17]

Answer: The answer is 58. Hope this helps!

3 0

3 years ago

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(4.3 x 10^-3)(2 x 10^-2) in standard form

notka56 [123]

Answer:

8.6•10^-5 or 0.000086

6 0

3 years ago

The amphitheater has two types of tickets available, reserved seats and lawn seats. The maximum capacity of the venue is 20,000

Tema [17]

Let the number of reserved tickets = x

Let the number of lawn seats = y

Constraint functions:

Maximum capacity means $x+y\leq 20000$

For concert to be held $x+y\geq 5000$

$lawn seats\leq reserved$ means $y\leq x$

Objective functions :

Maximum profit equation p = 65x +40y

Intersection points :

(10000,10000) (20000,0)(2500,2500)(5000,0)

p at (10000,10000) = 65(10000) + 40(10000) = $1050000

p at (20000,0) = 65(20000) + 40(0) = $1300000

p at (2500,2500) = 65(2500) + 40(2500) = $262500

p at (5000,0) = 65(5000) + 40(0) = $325000

Hence maximum profit occurs when all 20000 reserved seats are sold and the profit is $1300000

Please find attached the graph of it.

8 0

3 years ago

Write 77^1/10 in the form <br> n√77 where n is a whole number.

Sedaia [141]

Answer:

$\sqrt[10]{77}$

Step-by-step explanation:

$77^\frac{1}{10}$ is the same as $\sqrt[10]{77^1} ,$ or simply $\sqrt[10]{77}$

4 0

2 years ago

A small regional carrier accepted 23 reservations for a particular flight with 20 seats. 14 reservations went to regular custome

MrRissso [65]

Answer:

- The probability that overbooking occurs means that all 8 non-regular customers arrived for the flight. Each of them has a 56% probability of arriving and they arrive independently so we get that

P(8 arrive) = (0.56)^8 = 0.00967

- Let's do part c before part b. For this, we want an exact booking, which means that exactly 7 of the 8 non-regular customers arrive for the flight. Suppose we align these 8 people in a row. Take the scenario that the 1st person didn't arrive and the remaining 7 did. That odds of that happening would be (1-.56)*(.56)^7.

Now take the scenario that the second person didn't arrive and the remaining 7 did. The odds would be

(0.56)(1-0.56)(0.56)^6 = (1-.56)*(.56)^7. You can run through every scenario that way and see that each time the odds are the same. There are a total of 8 different scenarios since we can choose 1 person (the non-arriver) from 8 people in eight different ways (combination).

So the overall probability of an exact booking would be [(1-.56)*(.56)^7] * 8 = 0.06079

- The probability that the flight has one or more empty seats is the same as the probability that the flight is NOT exactly booked NOR is it overbooked. Formally,

P(at least 1 empty seat) = 1 - P(-1 or 0 empty seats)

= 1 - P(overbooked) - P(exactly booked)

= 1 - 0.00967 - 0.06079

= 0.9295.

Note that, the chance of being both overbooked and exactly booked is zero, so we don't have to worry about that.

Hope that helps!

Have a great day :P

7 0

3 years ago