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3 years ago

A 70 meter path surrounds a rectangular garden. the width of the garden is two thirds it's length. Find its area

1 answer:

6 0

You can set up an equation.
Let x = measure of the length.

Width: 2/3x
Length: x

From the question, we know that the perimeter is 70 and based on our prior knowledge, we know that for a rectangle, the perimeter is equal to 2 times the length added to two times the width.

2(2/3x + x) = 70
4/3x + 2x = 70
10/3x = 70
x = 21

So the length is 21 meters.
And since the width is 2/3 the length, 2/3 x 21 is 14.
So the width is 14 meters.

And if you add 21 + 21 + 14 + 14, it will equal 70 meters.

Now let's find the area.
21 x 14 = 294 square meters.

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What Number is X in this equation?<br><br> 6 ÷ 2( 1 + 2 ) = X

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A movie theater has been charging $10.00 per person and selling about 500 tickets on a typical weeknight. After surveying their

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Answer:

$p(x) = -0.01x + 15$

Step-by-step explanation:

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Determine the demand function

First, calculate the slope:

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$m = \frac{-0.5}{50}$

$m = -0.01$

So, the equation is:

$p = m(x - x_1) + y_1$

$p = -0.01(x - 500) +10$

$p = -0.01x + 5 +10$

$p = -0.01x + 15$

Hence, the function is:

$p(x) = -0.01x + 15$

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3 years ago

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The arrivals of clients at a service firm in Santa Clara is a random variable from Poisson distribution with rate 2 arrivals per

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Answer:

1.76% probability that in one hour more than 5 clients arrive

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.

The arrivals of clients at a service firm in Santa Clara is a random variable from Poisson distribution with rate 2 arrivals per hour.

This means that $\mu = 2$

What is the probability that in one hour more than 5 clients arrive

Either 5 or less clients arrive, or more than 5 do. The sum of the probabilities of these events is decimal 1. So

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

We want P(X > 5). So

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

In which

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

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

$P(X = 0) = \frac{e^{-2}*2^{0}}{(0)!} = 0.1353$

$P(X = 1) = \frac{e^{-2}*2^{1}}{(1)!} = 0.2707$

$P(X = 2) = \frac{e^{-2}*2^{2}}{(2)!} = 0.2707$

$P(X = 3) = \frac{e^{-2}*2^{3}}{(3)!} = 0.1804$

$P(X = 4) = \frac{e^{-2}*2^{4}}{(4)!} = 0.0902$

$P(X = 5) = \frac{e^{-2}*2^{5}}{(5)!} = 0.0361$

$P(X \leq 5) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) = 0.1353 + 0.2702 + 0.2702 + 0.1804 + 0.0902 + 0.0361 = 0.9824$

$P(X > 5) = 1 - P(X \leq 5) = 1 - 0.9824 = 0.0176$

1.76% probability that in one hour more than 5 clients arrive

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3 years ago

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