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xeze [42]
4 years ago
12

A pennant is shaped like a right triangle with a hypotenuse of 10feet. The length of one side of the pennant is two feet longer

than the length of the other side. Find the length of the two sides of the pennant.
Mathematics
1 answer:
marissa [1.9K]4 years ago
6 0

Answer:

6 ft and 8 ft

Step-by-step explanation:

let x be the length of one leg then (x + 2) is the other leg.

Using Pythagoras' identity in the right triangle, that is

x² + (x + 2)² = 10² ← expand left side and simplify

x² + x² + 4x + 4 = 100 ( subtract 100 from both sides )

2x² + 4x - 96 = 0 ( divide all terms by 2 )

x² + 2x - 48 = 0 ← in standard form

(x + 8)(x - 6) = 0 ← in factored form

Equate each factor to zero and solve for x

x + 8 = 0 ⇒ x = - 8

x - 6 = 0 ⇒ x = 6

But x > 0 ⇒ x = 6

Thus the 2 sides are 6 ft and x + 2 = 6 + 2 = 8 ft

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9.375

Step-by-step explanation:

150 divided by 9.375 gives you 16

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A ship at position (1, 0) on a nautical chart (with north in the positive y direction) sights a rock at position (6, 5). What is
OLga [1]

Answer:

The vector joining the ship to the rock is t= 7 i + 5 j

The direction is 0.9505 radians east of north.

Step-by-step explanation:

The position vector of the ship:

r= 1 i + 0 j

The position vector of the ship:

s= 6 i + 5 j

The vector joining the ship to the rock is:

t = r + s

t = (1 i + 0 j) + (6 i + 5 j)

t = 7 i + 5 j

The bearing of the rock to the ship is:

Θ= \frac{ \pi}{2} - arctan (\frac{5}{7})= 0.9505 radians

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What does it mean when a polynomial equation is in standard​ form?
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In Northern Yellowstone Lake, earthquakes occur at a mean rate of 1.3 quakes per year. Let X be the number of quakes in a given
hichkok12 [17]

Answer:

a) Earthquakes are random and independent events.

b) There is an 85.71% probability of fewer than three quakes.

c) There is a 0.51% probability of more than five quakes.

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:

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 time interval.

In this problem, we have that:

In Northern Yellowstone Lake, earthquakes occur at a mean rate of 1.3 quakes per year, so \mu = 1.3

(a) Justify the use of the Poisson model.

Earthquakes are random and independent events.

You can't predict when a earthquake is going to happen, or how many are going to have in a year. It is an estimative.

(b) What is the probability of fewer than three quakes?

This is P(X < 3)

P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2)

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

P(X = 0) = \frac{e^{-1.3}*1.3^{0}}{(0)!} = 0.2725

P(X = 1) = \frac{e^{-1.3}*1.3^{1}}{(1)!} = 0.3543

P(X = 2) = \frac{e^{-1.3}*1.3^{2}}{(2)!} = 0.2303

P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2) = 0.2725 + 0.3543 + 0.2303 = 0.8571

There is an 85.71% probability of fewer than three quakes.

(c) What is the probability of more than five quakes?

This is P(X > 5)

We know that either there are 5 or less earthquakes, or there are more than 5 earthquakes. The sum of the probabilities of these events is decimal 1.

So

P(X \leq 5) + P(X > 5) = 1

P(X > 5) = 1 - P(X \leq 5)

In which

P(X \leq 5) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5)

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

P(X = 0) = \frac{e^{-1.3}*1.3^{0}}{(0)!} = 0.2725

P(X = 1) = \frac{e^{-1.3}*1.3^{1}}{(1)!} = 0.3543

P(X = 2) = \frac{e^{-1.3}*1.3^{2}}{(2)!} = 0.2303

P(X = 3) = \frac{e^{-1.3}*1.3^{3}}{(3)!} = 0.0980

P(X = 4) = \frac{e^{-1.3}*1.3^{4}}{(4)!} = 0.0324

P(X = 5) = \frac{e^{-1.3}*1.3^{5}}{(5)!} = 0.0084

P(X \leq 5) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) = 0.2725 + 0.3543 + 0.2303 + 0.0980 + 0.0324 + 0.0084 = 0.9949

Finally

P(X > 5) = 1 - P(X \leq 5) = 1 - 0.9949 = 0.0051

There is a 0.51% probability of more than five quakes.

5 0
3 years ago
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