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2 years ago

How are triangles different in spherical geometry as opposed to euclidean geometry?

Mathematics

1 answer:

Ivan2 years ago

5 0

Euclidean geometry, is simply plane and solid geometry. It is named after the Greek mathematician, Euclid, when he proposed his five postulates which serve as basis of drawing plane and solid figures. So, in a nutshell, a triangle in Euclidean geometry is a two-dimensional figure composed of three sides and whose interior angles sum up to 180°. A triangle in spherical geometry, on the other hand, is a triangle formed by three arcs. Thus, it is three-dimensional, and the interior angles sum up to more than 180°. The difference is shown in the attached picture.

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Two row boats start at the same location, and start traveling apart along straight lines which meet at an angle of π3. Boat A is

alina1380 [7]

Answer:

5√3 miles per hour

Step-by-step explanation:

Throughout their journey, the component of velocity of rowboat B to the east is the same as that of rowboat A:

(10 mph)(cos(π/3)) = 5 mph

Meanwhile, the component of velocity of rowboat B to the north is ...

(10 mph)(sin(π/3)) = 5√3 mph

Since rowboat A is always traveling in a direction that is at right angles to the direction between the boats, it contributes nothing to their relative speed. Since rowboat B is always directly north of rowboat A, its speed to the north is their relative speed.

The distance between the rowboats is increasing at 5√3 mph ≈ 8.66 mph.

5 0

2 years ago

Which transformations can be used to map a triangle with vertices A (2, 2), B (4, 1), C (4, 5) to A’ (–2, –2), B’ (–1, –4), C’ (

Aliun [14]

A is the the answer for you

6 0

2 years ago

On average, water flows over a particular water fall at a rate of 2.05 x 105 cubic feet per second. One cubic foot of water weig

iren [92.7K]

Answer:

5.024 × 10^8 tons /day

Step-by-step explanation:

This question has to deal with conversion

We are told in the question that:

On average, water flows over a particular water fall at a rate of 2.05 x 10^5 cubic feet per second.

Water flow rate = 2.05 × 10^5 ft³/s

From the question,

One cubic foot of water weighs 62.4 lb.

1 ft³ = 62.4 lb

1 ton = 2200 lb

We are to calculate the flow rate in tons per day

Hence,

1 ft³ = 62.4 lb

2.05 × 10^5 ft³ =

Cross Multiply

2.05 × 10^5 × 62.4

= 12792000lb

Hence the water flow rate = 12792000lb/s

From the question,

2200 lb = 1 ton

12792000 Ib =

Cross Multiply

12792000lb × 1 ton/2200 lb

= 5814.5454545 tons

Hence the water flow rate = 5814.5454545 tons/second

We want the flow rate to be in tons/day

1 ton / second = 86400 tons / day

5814.5454545 tons/second = x tons/ day

Cross Multiply

x = 5814.5454545 tons/second × 86400 tons /day/ 1 ton / second

= 502376727.27 tons/day

= 5.0237672727 × 10^8 tons/day

Approximately ≈ 5.024 × 10^8 tons /day

Therefore, the rate of water flow in tons of water per day is 5.024 × 10^8 tons of water /day

7 0

2 years ago

A 6 ft board is cut into two pieces, on twice as long as the other. How long are the pieces?

disa [49]

Answer:One is 2 feet and one is 4 feet

Step-by-step explanation:

8 0

2 years ago

Rocky Mountain National Park is a popular park for outdoor recreation activities in Colorado. According to U.S. National Park Se

Ugo [173]

Answer:

a) 0.6628 = 66.28% probability that at least 85 visitors had a recorded entry through the Beaver Meadows park entrance

b) 0.5141 = 51.41% probability that at least 80 but less than 90 visitors had a recorded entry through the Beaver Meadows park entrance

c) 0.5596 = 55.96% probability that fewer than 12 visitors had a recorded entry through the Grand Lake park entrance.

d) 0.9978 = 99.78% probability that more than 55 visitors have no recorded point of entry

Step-by-step explanation:

Binomial probability distribution

Probability of exactly x sucesses on n repeated trials, with p probability.

Can be approximated to a normal distribution, using the expected value and the standard deviation.

The expected value of the binomial distribution is:

$E(X) = np$

The standard deviation of the binomial distribution is:

$\sqrt{V(X)} = \sqrt{np(1-p)}$

Normal probability distribution

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

When we are approximating a binomial distribution to a normal one, we have that $\mu = E(X)$ , $\sigma = \sqrt{V(X)}$ .

In this problem, we have that:

$n = 175$

(a) What is the probability that at least 85 visitors had a recorded entry through the Beaver Meadows park entrance?

46.7% of visitors to Rocky Mountain National Park in 2018 entered through the Beaver Meadows. This means that $p = 0.467$ . So

$\mu = E(X) = np = 175*0.467 = 81.725$

$\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{175*0.467*0.533} = 6.6$

This probability, using continuity correction, is $P(X \geq 85 - 0.5) = P(X \geq 84.5)$ , which is 1 subtracted by the pvalue of Z when X = 84.5. So

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{84.5 - 81.725}{6.6}$

$Z = 0.42$

$Z = 0.42$ has a pvalue of 0.6628.

66.28% probability that at least 85 visitors had a recorded entry through the Beaver Meadows park entrance.

(b) What is the probability that at least 80 but less than 90 visitors had a recorded entry through the Beaver Meadows park entrance?

Using continuity correction, this is $P(80 - 0.5 \leq X < 90 - 0.5) = P(79.5 \leq X \leq 89.5)$ , which is the pvalue of Z when X = 89.5 subtracted by the pvalue of Z when X = 79.5. So

X = 89.5

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{89.5 - 81.725}{6.6}$

$Z = 1.18$

$Z = 1.18$ has a pvalue of 0.8810.

X = 79.5

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{79.5 - 81.725}{6.6}$

$Z = -0.34$

$Z = -0.34$ has a pvalue of 0.3669.

0.8810 - 0.3669 = 0.5141

51.41% probability that at least 80 but less than 90 visitors had a recorded entry through the Beaver Meadows park entrance

(c) What is the probability that fewer than 12 visitors had a recorded entry through the Grand Lake park entrance?

6.3% of visitors entered through the Grand Lake park entrance, which means that $p = 0.063$

$\mu = E(X) = np = 175*0.063 = 11.025$

$\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{175*0.063*0.937} = 3.2141$

This probability, using continuity correction, is $P(X < 12 - 0.5) = P(X < 11.5)$ , which is the pvalue of Z when X = 11.5. So

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{11.5 - 11.025}{3.2141}$

$Z = 0.15$

$Z = 0.15$ has a pvalue of 0.5596.

55.96% probability that fewer than 12 visitors had a recorded entry through the Grand Lake park entrance.

(d) What is the probability that more than 55 visitors have no recorded point of entry?

22.7% of visitors had no recorded point of entry to the park. This means that $p = 0.227$

$\mu = E(X) = np = 175*0.227 = 39.725$

$\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{175*0.227*0.773} = 5.54$

Using continuity correction, this probability is $P(X \leq 55 + 0.5) = P(X \leq 55.5)$ , which is the pvalue of Z when X = 55.5. So

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{55.5 - 39.725}{5.54}$

$Z = 2.85$

$Z = 2.85$ has a pvalue of 0.9978

0.9978 = 99.78% probability that more than 55 visitors have no recorded point of entry

8 0

2 years ago