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notsponge [240]
3 years ago
14

Kevin caught some fish. Of them 4/9 were herring, and 2/9 salmon. What was the total amount of fish, amount of herring and amoun

t of salmon if there were 12 flounders?
Mathematics
1 answer:
julia-pushkina [17]3 years ago
6 0

There were 3 different fish caught, Salmon, Herring and Flounder.

4/9 were Herring, 2/9 were Salmon, 4/9 +2/9 = 6/9

This means 3/9 were Flounders ( 3/9 + 6/9 = 9/9 = 1)

3/9 can be reduced to 1/3.

1/3 of the fish were Flounders.

Divide the amount of flounders by the portion caught:

12 / 1/3 = 12 * 3/1 = 36 total fish.

Now you have total number of fish, multiply the total by each portion for each type of fish.

Total fish = 36

Herring = 4/9 x 36 = 16

Salmon = 2/9 x 36 = 8

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The perimeter of a rectangle is 140 feet. The length of the rectangle is 6 more than 2 times the width. What is the length of th
Alja [10]
The equations we get are
l = 6 + 2w (We get this from "The length of the rectangle is 6 more than 2 times the width.")
and
2l + 2w = 140 (We get this from the perimeter. Two times the length plus two times the width equals the perimeter of a quadrilateral.)

The first equation can be written as
l - 2w = 6

So now we have
2l + 2w = 140
l - 2w = 6
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Add the two equations.
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Divide by 3 on both sides.
l = 48\frac{2}{3}

Your answer is 48\frac{2}{3}.
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3 years ago
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Step-by-step explanation:

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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
Ludmilka [50]

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 <em>X</em> 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

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Remaining reservations, n = 19 - 14 = 5

P (A remaining passenger will arrive), <em>p</em> = 0.52

The random variable <em>X</em> thus follows a Binomial distribution with parameters <em>n</em> = 5 and <em>p</em> = 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.

4 0
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Answer:

Neither i think

Step-by-step explanation:

7 0
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