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3 years ago

Help pls I'm confused

Mathematics

2 answers:

Trava [24]3 years ago

4 0

Answer:

Sorry this took so long, hope this helps

also the "a^2" is meant to mean this " $a^{2}$ "

Step-by-step explanation:

1) a^2+b^2=c^2

4^2+3^2=c^2

16+9=c^2

25=c^2

$\sqrt{25}$ =c

5=c

2) a^2+b^2=c^2

15^2+8^2=c^2

225+64=c^2

289=c^2

$\sqrt{225$ =c

15=c

3)a^2+b^2=c^2

10^2+50^2=c^2

100+2500=c^2

2600=c^2

$\sqrt{2600}$ =c

50.99≈c

4) a= $\sqrt{c^2-b^2}$

a= $\sqrt{15^2-12^2$

a= $\sqrt{225-144}$

a= $\sqrt{81}$

a=9

SashulF [63]3 years ago

3 0

Answer:

Step-by-step explanation:

Pythagorean theorem

Is a^2 + b^2 = C^2

A can be the base

B can be the height

C is the hypotenuse which is the side opposite the right angle

Is that what you don’t understand or?

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3 years ago

A certain kind of sheet metal has, on average, 3 defects per 18 square feet. Assuming a Poisson distribution, find the probabili

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Answer:

75.8% probability that a 31 square foot metal sheet has at least 4 defects.

Step-by-step explanation:

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

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In which

x is the number of sucesses

e = 2.71828 is the Euler number

$\mu$ is the mean in the given time interval.

In this problem, we have that:

3 defects per 18 square feet.

So for 31 square feet, we have to solve a rule of three

3 defects - 18 square feet

x defects - 31 square feet

$18x = 3*31$

$x = \frac{31*3}{18}$

$x = 5.17$

So $\mu = 5.17$

Assuming a Poisson distribution, find the probability that a 31 square foot metal sheet has at least 4 defects.

Either it has three or less defects, or it has at least 4 defects. The sum of the probabilities of these events is decimal 1.

$P(X \leq 3) + P(X \geq 4) = 1$

We want $P(X \geq 4)$

$P(X \geq 4) = 1 - P(X \leq 3)$

In which

$P(X \leq 3) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)$

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

$P(X = 0) = \frac{e^{-5.17}*(5.17)^{0}}{(0)!} = 0.0057$

$P(X = 1) = \frac{e^{-5.17}*(5.17)^{1}}{(1)!} = 0.0294$

$P(X = 2) = \frac{e^{-5.17}*(5.17)^{2}}{(2)!} = 0.0760$

$P(X = 3) = \frac{e^{-5.17}*(5.17)^{3}}{(3)!} = 0.1309$

$P(X \leq 3) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) = 0.0057 + 0.0294 + 0.0760 + 0.1309 = 0.242$

Finally

$P(X \geq 4) = 1 - P(X \leq 3) = 1 - 0.242 = 0.758$

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4 years ago

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Answer:

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