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4 years ago

What is the value of p? 10p+5 = 18p-6over 2

1 answer:

5 0

1) we have to subtract by the number 18p from each of your sides
Like: 10p + 5 - 18p
= 18p -3 -180
This gives us: -8p + 5 = -3
Nextly, we have to subtract by 5 on each of your sides
Like: -8p + 5 - 5
= -3 - 5
This gives us: -8p = -8
Now we have divide by the -8 on each of your sides:
Like: -8p ÷ -8
= -8 ÷ -8
Answer: p = 1

I hope it helped you:)

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Assume the world catch of fish in 1950 was 12 million tons and in 1955 it was 96 million tons.

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Solution:

Given that, the world catch of fish in 1950 was 12 million tons and in 1955 it was 96 million tons

The average rate of change is given by formula:

$\text{ Average rate of change } = \frac{\text{ change in value }}{\text{ number of years }}$

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Change in value = value in 1955 - value in 1950

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3 years ago

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3 years ago

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3 years ago

A small regional carrier accepted 19 reservations for a particular flight with 17 seats. 14 reservations went to regular custome

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Answer:

(a) The probability of overbooking is 0.2135.

(b) The probability that the flight has empty seats is 0.4625.

Step-by-step explanation:

Let the random variable X represent the number of passengers showing up for the flight.

It is provided that a small regional carrier accepted 19 reservations for a particular flight with 17 seats.

Of the 17 seats, 14 reservations went to regular customers who will arrive for the flight.

Number of reservations = 19

Regular customers = 14

Seats available = 17 - 14 = 3

Remaining reservations, n = 19 - 14 = 5

P (A remaining passenger will arrive), p = 0.52

The random variable X thus follows a Binomial distribution with parameters n = 5 and p = 0.52.

(1)

Compute the probability of overbooking as follows:

P (Overbooking occurs) = P(More than 3 shows up for the flight)

$=P(X>3)\\\\={5\choose 4}(0.52)^{4}(1-0.52)^{5-4}+{5\choose 5}(0.52)^{5}(1-0.52)^{5-5}\\\\=0.175478784+0.0380204032\\\\=0.2134991872\\\\\approx 0.2135$

Thus, the probability of overbooking is 0.2135.

(2)

Compute the probability that the flight has empty seats as follows:

P (The flight has empty seats) = P (Less than 3 shows up for the flight)

$=P(X$

Thus, the probability that the flight has empty seats is 0.4625.

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3 years ago