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VMariaS [17]
3 years ago
11

If the length of the hypotenuse of a right triangle is 39 millimeters and the length of one leg is 15 millimeters what is the le

ngth of the other leg (b)
Mathematics
1 answer:
Genrish500 [490]3 years ago
6 0

Answer:

36 millimeters

Step-by-step explanation:

From Pythagoras theorem, the square of the hypotenuse is equal to the sum of the square of the two other legs

In mathematical terms;

a^2 = b^2 + c^2

Let a represent the hypotenuse = 39 mm and the length of one leg, say c, is 15 mm

Slotting in the values of a and b

39^2 = b^2 + 15^2

1521 = b^2 + 225

collect like terms

1521 - 225 = b^2

1296 = b^2

Take the square root of both sides

36 = b

Therefore b = 36 mm

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Answer:

a) P(x = 0) = 64.69%

b) P(x ≥ 1) = 35.31%

c) E(x) = 0.42

d) var(x) = 0.3906

Step-by-step explanation:

The given problem can be solved using binomial distribution since:

  • There are n repeated trials independent of each other.
  • There are only two possibilities: exceedence happens or  exceedence doesn't happen.
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The binomial distribution is given by

P(x) = ⁿCₓ pˣ (1 - p)ⁿ⁻ˣ

Where n is the number of trials, x is the variable of interest and p is the probability of success.

For the given scenario. the six daily arrivals are the number of trials

Number of trials = n = 6

The probability of success = 7% = 0.07

a) Find the probability that on one day no planes have an exceedence.

Here we have x = 0, n = 6 and p = 0.07

P(x = 0) = ⁶C₀(0.07⁰)(1 - 0.07)⁶⁻⁰

P(x = 0) = (1)(0.07⁰)(0.93)⁶

P(x = 0) = 0.6469

P(x = 0) = 64.69%

b) Find the probability that at least 1 plane exceeds the localizer.

The probability that at least 1 plane exceeds the localizer is given by

P(x ≥ 1) = 1 - P(x < 1)

But we know that P(x < 1) = P(x = 0) so,

P(x ≥ 1) = 1 - P(x = 0)

We have already calculated P(x = 0) in part (a)

P(x ≥ 1) = 1 - 0.6469

P(x ≥ 1) = 0.3531

P(x ≥ 1) = 35.31%

c) What is the expected number of planes to exceed the localizer on any given day?

The expected number of planes to exceed the localizer is given by

E(x) = n×p

Where n is the number of trials and p is the probability of success

E(x) = 6×0.07

E(x) = 0.42

Therefore, the expected number of planes to exceed the localizer on any given day is 0.42

d) What is the variance for the number of planes to exceed the localizer on any given day?

The variance for the number of planes to exceed the localizer is given by

var(x) = n×p×q

Where n is the number of trials and p is the probability of success and q is the probability of failure.

var(x) = 6×0.07×(1 - 0.07)

var(x) = 6×0.07×(0.93)

var(x) = 0.3906

Therefore, the variance for the number of planes to exceed the localizer on any given day is 0.3906.

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==========================================================

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Therefore, \sin(45^{\circ}) = \frac{\sqrt{2}}{2}

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