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madam [21]
4 years ago
15

PLEASE ANSWER List: What are the three types of joints in the human body?

Physics
1 answer:
horsena [70]4 years ago
4 0
1.Synovial joint
2. pivot joint
3. Hinge joint
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If hot air rises then why is it cold in space please explain in detail thank you
Helen [10]
Hot air has a lower density than the surrounding atmosphere, therefore,it will rise.. The reason it is "cold" in the upper atmosphere is actually because of a lower air density.
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3 years ago
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Atmospheric pressure decreases as we go higher up the earth's surface. give reasons
julsineya [31]
Atmospheric pressure is molecular's heavy. So the total number of molecules decreases by going up the surface of the earth. Ok?
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3 years ago
Why alloys<br> are useful to someone with no science<br> knowledge<br><br> Will give Brainiest
Andreyy89

Answer:

The key idea behind the synthesis, processing and application of alloys consist in combining desire qualities from each mineral component in the form of a material.

Explanation:

The key idea behind the synthesis, processing and application of alloys consist in combining desire qualities from each mineral component in the form of a material. For example, steel is the combination of iron, which is ductile but prone to rusting in wet environments, and carbon, which offers high hardness and additional resistence to rusting in wet environments, but is fragile. Hence, steel is more resistant to rusting, ductile as iron and more resistant than iron.

6 0
3 years ago
Scientists treat the number of stars in a given volume of space as a Poisson random variable. The density of our galaxy in the v
vladimir1956 [14]

Answer:

P(X\ge 1) = 0.9502

Explanation:

Given

Density = 3 starts in 10 cubic light years.

Required

Determine the probability of 1 or more in 10 cubic light years

Since the number of stars follow a Poisson distribution, we make use of:

P(X=k) = f(x) = (\lambda T)^k\frac{ e^{-\lambda T}}{k!}

\lambda = density

\lambda = \frac{3}{10}

\lambda = 0.3

T = the light years

T = 10

Calculating P(X \ge 1)

In probability:

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

Calculating P(X=0)

Substitute 0 for k and the values for \lambda and T in

P(X=k) = f(x) = (\lambda T)^k\frac{ e^{-\lambda T}}{k!}

P(X=0) = (0.3* 10)^0 * \frac{ e^{-0.3 * 10}}{0!}

P(X=0) = (3)^0 * \frac{ e^{-0.3 * 10}}{1}

P(X=0) = (3)^0 *  e^{-0.3 * 10}

P(X=0) = 1 *  e^{-0.3 * 10}

P(X=0) = 1 *  e^{-3}

P(X=0) = e^{-3}

P(X=0) = 0.04979

Substitute 0.04979 for P(X=0) in P(X \ge 1) = 1 - P(X = 0)

P(X\ge 1) = 1 - 0.04979

P(X\ge 1) = 0.95021

P(X\ge 1) = 0.9502 ---  approximated

<em>Hence, the required probability is 0.9502</em>

3 0
3 years ago
34.9x46x809 Please helpp
Fed [463]
<h2>34.9×46×809</h2><h3>=1605.4×809 </h3><h3>=1298768.6</h3>

please mark this answer as brainlist

6 0
3 years ago
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