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AnnZ [28]
2 years ago
15

2 cities have nearly the same north-south line 90° W. The latitude of the first city is 23°N, the latitude of the second city is

36° N. Approximate the distance between the cities if the average radius of earth is 6400 km. The cities are approximately
Mathematics
1 answer:
Ivan2 years ago
4 0

Answer:

The cities are approximately 1452 km apart

Step-by-step explanation:

The two cities are on the same Longitude 90°W therefore the distance between them is on a great circle

Latitude of the first city = 23°N

Latitude of the second city = 36° N

Because their latitude are on the same polar axis, we subtract to get the angular difference

Angular Difference = 36 -23 =13°

Distance= \frac{\alpha}{360} X 2\pi R

where \alpha = Angular Difference, R= radius of the earth

Distance= \frac{13}{360} X 2\pi X 6400=1452.11km

The cities are approximately 1452 km apart

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konstantin123 [22]

Product means multiplication.

Given w= weight of a trout.

So, we need to convert "product of 16 and weight of trout: w " into an algebraic expression.

So, product of 16 and w can be written as 16*w or simply 16w.

Hence, B is the correct choice.

8 0
1 year ago
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Sheila has $47 to buy graphing pads for her math class. Each pad costs $8,
bija089 [108]

Answer:

5 tablets can be bought

Step-by-step explanation:

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8 0
2 years ago
In 1898 L. J. Bortkiewicz published a book entitled The Law of Small Numbers. He used data collected over 20 years to show that
attashe74 [19]

Answer:

(a) The probability of more than one death in a corps in a year is 0.1252.

(b) The probability of no deaths in a corps over 7 years is 0.0130.

Step-by-step explanation:

Let <em>X</em> = number of soldiers killed by horse kicks in 1 year.

The random variable X\sim Poisson(\lambda = 0.62).

The probability function of a Poisson distribution is:

P(X=x)=\frac{e^{-\lambda}\lambda^{x}}{x!};\ x=0,1,2,...

(a)

Compute the probability of more than one death in a corps in a year as follows:

P (X > 1) = 1 - P (X ≤ 1)

             = 1 - P (X = 0) - P (X = 1)

             =1-\frac{e^{-0.62}(0.62)^{0}}{0!}-\frac{e^{-0.62}(0.62)^{1}}{1!}\\=1-0.54335-0.33144\\=0.12521\\\approx0.1252

Thus, the probability of more than one death in a corps in a year is 0.1252.

(b)

The average deaths over 7 year period is: \lambda=7\times0.62=4.34

Compute the probability of no deaths in a corps over 7 years as follows:

P(X=0)=\frac{e^{-4.34}(4.34)^{0}}{0!}=0.01304\approx0.0130

Thus, the probability of no deaths in a corps over 7 years is 0.0130.

6 0
1 year ago
In the SuperLottery, three balls are drawn (at random) from ten white balls numbered from $1$ to $10$, and one SuperBall is draw
Mice21 [21]

Answer:

1/3

Step-by-step explanation:

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Now going to the white balls, you need atleast 2 numbers picked as the numbers drawn. Let's say we pick numbers 1-3. In order to win, the numbers 1 and 2, 2 and 3, 1 and 3, or 1 2 and 3 must be drawn. You can calculate the probability for each of these cases. In the case only 2 are drawn, the probability for each is 0.075. Since there are three cases that this could happen, you multiply this probability by 3, becoming 1/4. However, there is also the chance that you get all three. The probability of this is 3/10*2/9*1/8.

.225 + .1 + 0.008333  = 33.333...% or 1/3

4 0
2 years ago
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Area of a rectangle:
l \times w
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