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2 years ago

13

A telephone pole that is 14 ft. tall has fallen against a house. If the top of the telephone pole touches the house 10 ft. above

the ground, what is the angle that the telephone pole makes with the ground?

1 answer:

Karo-lina-s [1.5K]2 years ago

7 0

Answer:

45°

Step-by-step explanation:

made a special triangle 45-45-90,

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Find the mean of these numbers : 35,42,41,32,30

abruzzese [7]

Answer:

156

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The difference between 6 digits smallest number and 4 digits greatest number is <br>

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12

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2 years ago

A men's department store sells 3 different suit jackets, 6 different shirts, 8 different ties, and 4 different pairs of pants. h

Ksenya-84 [330]

Jackets = 3
Shirt = 6
Ties = 8
Pants = 4

Number of combination = 3 x 6 x 8 x 4 = 576

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2 years ago

The population of a small town in central Florida has shown a linear decline in the years 1992-2000. In 1992 the population was

wlad13 [49]

Answer:

P = -390t +48,000
P = 43,710 in 2003

Step-by-step explanation:

A) We are given two pairs of (t, P): (0, 48000), (8, 44880). We can use these with the 2-point form of the equation of a line:

y = (y2 -y1)/(x2 -x1)(x -x1) +y1

P = (44880 -48000)/(8 -0)(t -0) +48000

P = -3120/8t +48000

P = -390t +48000

__

B) In 2003, t = 11, so we have ...

P = -390(11) +48000 = 43710

The population in 2003 is predicted to be 43,710.

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3 years ago

Suppose that the length of a phone call in minutes is an exponential random variable with parameter λ = 1/ 8. If someone arrives

Elis [28]

Answer:

a) P=0.535

b) P=0.204

c) P=0.286

Step-by-step explanation:

The exponential distribution is expressed as

$F(x>t)=e^{-\lambda t}$

In this example, λ=1/8=0.125 min⁻¹.

a) The probability of having to wait more than 5 minutes

$F(x>5)=e^{-0.125*5}=e^{-0.625}=0.535$

b) The probability of having to wait between 10 and 20 minutes

$F(1020)=e^{-0.125*10}-e^{-0.125*20}\\\\F(10$

c) The exponential distribution is memory-less, so it is independent of past events.

If you have waited 5 minutes, the probability of waiting more than 15 minutes in total is the same as the probability of waiting 15-5=10 minutes.

$F(x>15|t^*=5)=F(x>15)/F(x>5)=\frac{e^{-0.125*15}}{e^{-0.125*5}}=e^{-0.125*(15-5)}\\\\ F(x>15|t^*=5)=e^{-0.125*(10)}=F(x>10)=0.286$

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2 years ago