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Gala2k [10]
3 years ago
9

PLSSS HURRY AND NO LINKS!!

Mathematics
1 answer:
matrenka [14]3 years ago
6 0

Answer:

Area of the hurricane = 7853 or 3.14  x  50 squared

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In rectangle ABCD, AB=8 units and BC=15 units. Find BD
BabaBlast [244]

Answer:


Step-by-step explanation:


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Simplify <br><img src="https://tex.z-dn.net/?f=%20%5Csqrt%7B324%7D%20" id="TexFormula1" title=" \sqrt{324} " alt=" \sqrt{324} "
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In a ∆ABC , angle A + Angle B = 125° and Angle B + Angle C = 150° . Find all the angles of ∆ABC.​
mel-nik [20]

\large\underline{\sf{Solution-}}

Given that,

<em>In triangle ABC</em>

\purple{\rm :\longmapsto\:\angle A + \angle B = 125 \degree \: -  -  - (1) }

\purple{\rm :\longmapsto\:\angle B + \angle C = 150 \degree \:  -  -  - (2)}

We know,

Sum of all interior angles of a triangle is supplementary.

\purple{\rm :\longmapsto\:\angle A + \angle B + \angle C = 180\degree }

<u>On adding equation (1) and (2), we get </u>

\purple{\rm :\longmapsto\:\angle A + \angle B + \angle B + \angle C = 125\degree  + 150 \degree \:}

\purple{\rm :\longmapsto\:\angle A + \angle B + \angle C + \angle B = 275\degree \:}

\purple{\rm :\longmapsto\:180\degree + \angle B = 275\degree \:}

\purple{\rm :\longmapsto\:\angle B = 275\degree - 180\degree  \:}

\purple{\rm :\longmapsto\:\angle B = 95\degree  \:}

On substituting the value in equation (1) and (2), we get

\purple{\rm :\longmapsto\:\angle A + 95\degree  = 125\degree }

\purple{\rm :\longmapsto\:\angle A =  125\degree - 95\degree  }

\purple{\rm :\longmapsto\:\angle A =  30\degree  }

Also, from equation (2), we get

\purple{\rm :\longmapsto\:95\degree  + \angle C = 150\degree }

\purple{\rm :\longmapsto\:\angle C = 150\degree  - 95\degree }

\purple{\rm :\longmapsto\:\angle C = 55\degree }

Hence,

\begin{gathered}\begin{gathered}\bf\: \rm\implies \:\begin{cases} &\sf{\angle A = 30\degree }  \\ \\ &\sf{\angle B = 95\degree } \\ \\ &\sf{\angle C = 55\degree } \end{cases}\end{gathered}\end{gathered}

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3 years ago
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Answer:

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The number of surface flaws in plastic panels used in the interior of automobiles has a Poisson distribution with a mean of 0.08
kvv77 [185]

Answer:

a) 44.93% probability that there are no surface flaws in an auto's interior

b) 0.03% probability that none of the 10 cars has any surface flaws

c) 0.44% probability that at most 1 car has any surface flaws

Step-by-step explanation:

To solve this question, we need to understand the Poisson and the binomial probability distributions.

Poisson distribution:

In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by the following formula:

P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}

In which

x is the number of sucesses

e = 2.71828 is the Euler number

\mu is the mean in the given interval.

Binomial distribution:

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

In which C_{n,x} is the number of different combinations of x objects from a set of n elements, given by the following formula.

C_{n,x} = \frac{n!}{x!(n-x)!}

And p is the probability of X happening.

Poisson distribution with a mean of 0.08 flaws per square foot of plastic panel. Assume an automobile interior contains 10 square feet of plastic panel.

So \mu = 10*0.08 = 0.8

(a) What is the probability that there are no surface flaws in an auto's interior?

Single car, so Poisson distribution. This is P(X = 0).

P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}

P(X = 0) = \frac{e^{-0.8}*(0.8)^{0}}{(0)!} = 0.4493

44.93% probability that there are no surface flaws in an auto's interior

(b) If 10 cars are sold to a rental company, what is the probability that none of the 10 cars has any surface flaws?

For each car, there is a p = 0.4493 probability of having no surface flaws. 10 cars, so n = 10. This is P(X = 10), binomial, since there are multiple cars and each of them has the same probability of not having a surface defect.

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

P(X = 10) = C_{10,10}.(0.4493)^{10}.(0.5507)^{0} = 0.0003

0.03% probability that none of the 10 cars has any surface flaws

(c) If 10 cars are sold to a rental company, what is the probability that at most 1 car has any surface flaws?

At least 9 cars without surface flaws. So

P(X \geq 9) = P(X = 9) + P(X = 10)

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

P(X = 9) = C_{10,9}.(0.4493)^{9}.(0.5507)^{1} = 0.0041

P(X = 10) = C_{10,10}.(0.4493)^{10}.(0.5507)^{0} = 0.0003

P(X \geq 9) = P(X = 9) + P(X = 10) = 0.0041 + 0.0003 = 0.0044

0.44% probability that at most 1 car has any surface flaws

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3 years ago
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