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3 years ago

A puddle is 0.06 feet deep after 1 hour and 0.03 feet deep after 5 hours. At what rate in

1 answer:

6 0

Assuming it's a constant rate, there is a four hour difference from t=1 and t=5, so subtract the difference in depth (0.03 - 0.06) and divide that by four.

-0.03/4 = -0.0075
So your change in depth in feet per hour is -0.0075 $\frac{ft}{hr}$ .

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Let x denote the lifetime of a mcchine component with an exponential distribution. The mean time for the component failure is 25

aliina [53]

Answer:

0.1353 = 13.53% probability that the lifetime exceeds the mean time by more than 1 standard deviations

Step-by-step explanation:

Exponential distribution:

The exponential probability distribution, with mean m, is described by the following equation:

$f(x) = \mu e^{-\mu x}$

In which $\mu = \frac{1}{m}$ is the decay parameter.

The probability that x is lower or equal to a is given by:

$P(X \leq x) = \int\limits^a_0 {f(x)} \, dx$

Which has the following solution:

$P(X \leq x) = 1 - e^{-\mu x}$

The probability of finding a value higher than x is:

$P(X > x) = 1 - P(X \leq x) = 1 - (1 - e^{-\mu x}) = e^{-\mu x}$

The mean time for the component failure is 2500 hours.

This means that $m = \frac{2500}, \mu = \frac{1}{2500} = 0.0004$

What is the probability that the lifetime exceeds the mean time by more than 1 standard deviations?

The standard deviation of the exponential distribution is the same as the mean, so this is P(X > 5000).

$P(X > x) = e^{-0.0004*5000} = 0.1353$

0.1353 = 13.53% probability that the lifetime exceeds the mean time by more than 1 standard deviations

4 0

3 years ago

H(a) = a^2 + 3a; Find h(-10)

Colt1911 [192]

H(-10)=70
because when you input -10 in for each a, it will equal 70

6 0

3 years ago

Each side of a cube 2.5 cm long. what is the volume?

Firdavs [7]

V ≈ 15.63cm^3

volume of a cube = side × side × side = units^3

Volume = 2.5*2.5*2.5 = 15.63^3

<em>Hoped this helped!</em>

7 0

3 years ago

Read 2 more answers

Need number 12 plz help

dlinn [17]

I think its letter A

6 0

3 years ago

Read 2 more answers

Directions: Complete all 3 questions. Make sure you include a graph, work and conclusion.

Simora [160]

Prove that the quadrilateral whose vertices are I(-2,3), J(2,6), K(7,6), and L(3, 3) is a rhombus.

I think in these problems the first step is to express each side as a vector. A vector is the difference between points. When two sides have the same vector (or negatives) it means they're parallel and congruent. So in a rhombus IJKL the vectors IJ and LK should be the same, as should JK and IL. That much assures a parallelogram; we check IJ and JK are congruent to complete the crowing of the rhombus.

Let's calculate these vectors:

IJ = J - I = (2,6) - (-2,3) = (2 - -2, 6 - 3) = (4, 3)

LK = K - L = (7, 6) - (3, 3) = (4, 3)

IJ = LK, so far so good

(Note: If you haven't got to vectors yet you can just show the two sides are the same length, 5, and have the same slope, 3/4, both of which can be read off the vectors.)

JK = K - J = (7,6) - (2,6) = (5,0)

IL = L - I = (3, 3) - (-2, 3) = (5, 0)

Those are the same too.

Now we have to show IJ ≅ JK

The length of IJ is the cliche √4²+3² = 5, the same as JK, so IJ ≅ JK

We showed all four sides are congruent and we have two pair of parallel sides, so we have a rhombus.

8 0

4 years ago