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Annette [7]
3 years ago
13

g A 5 foot tall man walks at 10 ft/s toward a street light that is 20 ft above the ground. What is the rate of change of the len

gth of his shadow when he is 25 ft from the street light

Mathematics
1 answer:
vredina [299]3 years ago
7 0

Answer:

-\frac{10}{3}ft/s

Step-by-step explanation:

We are given that

Height of man=5 foot

\frac{dy}{dt}=-10ft/s

Height of street light=20ft

We have to find the rate of change of the length of his shadow when he is 25 ft form the street light.

ABE and CDE are similar triangle because all right triangles are similar.

\frac{20}{5}=\frac{x+y}{x}

4=\frac{x+y}{x}

4x=x+y

4x-x=y

3x=y

3\frac{dx}{dt}=\frac{dy}{dt}

\frac{dx}{dt}=\frac{1}{3}(-10)=-\frac{10}{3}ft/s=-\frac{10}{3}ft/s

Hence, the rate of change of the length of his shadow when he is 25 ft from the street light=-\frac{10}{3}ft/s

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What is the median of the following salary list $32,019, $21,971, $27,512, $43,702, $38,860, $25,997
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A random variable x follows a normal distribution with mean d and standard deviation o=2. It is known that x is less than 5 abou
Vaselesa [24]

Answer:

The mean of this distribution is approximately 3.96.

Step-by-step explanation:

Here's how to solve this problem using a normal distribution table.

Let z be the

\displaystyle z = \frac{x - \mu}{\sigma}.

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\displaystyle z = \frac{5 - \mu}{2}.

To solve for \mu, the mean of this distribution, the only thing that needs to be found is the value of z. Since

The problem stated that P(X \le 5) = 69.85\% = 0.6985. Hence, P(Z \le z) = 0.6985.

The problem is that the normal distribution tables list only the value of P(0 \le Z \le z) for z \ge 0. To estimate  z from P(Z \le z) = 0.6985, it would be necessary to find the appropriate

Since P(Z \le z) = 0.6985 and is greater than P(Z \le 0) = 0.50, z > 0. As a result, P(Z \le z) can be written as the sum of P(Z < 0) and P(0 \le Z \le z). Besides, P(Z < 0) = P(Z \le 0) = 0.50. As a result:

\begin{aligned}&P(Z \le z)\\ &= P(Z < 0) + P(0 \le Z \le z) \\ &= 0.50 + P(0 \le Z \le z)\end{aligned}.

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\begin{aligned}&P(0 \le Z \le z) \\ &= P(Z \le z) - 0.50 \\&= 0.6985 - 0.50 \\&=0.1985 \end{aligned}.

Lookup 0.1985 on a normal distribution table. The corresponding z-score is 0.52. (In other words, P(0 \le Z \le 0.52) = 0.1985.)

Given that

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Solve the equation \displaystyle z = \frac{x - \mu}{\sigma} for the mean, \mu:

\displaystyle 0.52 = \frac{5 - \mu}{2}.

\mu = 5 - 2 \times 0.52 = 3.96.

3 0
3 years ago
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