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pshichka [43]
3 years ago
13

How long will it take you to ski a distance of 36 miles at a speed of 3 miles per 30 minutes

Mathematics
1 answer:
vaieri [72.5K]3 years ago
7 0
36 ÷ 3 = 12
12 x 30 = 360 (mins)

The answer is that it would take you 360 minutes to ski a distance of 36 miles.
or you could say: t would take you 6 hours to ski a distance of 36 miles.
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The answer is 10

The equation would look like
5(4)-4(3)+2
And if you solve it, you’ll get 10 as your answer
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What is 94 million,23 in standard form
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Nine million, four hundred thousand, twenty three9

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Find a power series representation of e^x sin(x)
liberstina [14]
The Taylor series is defined by:
f(x) = \sum \frac{f^n (a)}{n!} (x-a)^n

Let a = 0.
Then its just a matter of finding derivatives and determining how many terms is needed for the series.

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Do this successively to n = 6.
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Plug in x=0 and sub into taylor series:
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3 years ago
A display order of numbers are called​
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Answer:

I think

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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

Seats available = 17 - 14 = 3

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
3 years ago
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