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2 years ago

14

A 20-ft flagpole casts a shadow of 12 feet long. How tall is a nearby building that casts a shadow of 20 feet long? Round to one

decimal place.

1 answer:

olga_2 [115]2 years ago

3 0

A^2 + b^2 = c^2
20^2 + 12^2 = c^2
400+144 = c^2
544 = c^2
Sqrt(544) = Sqrt(c^2)
23.3 = c

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Need help with my homework 2 problems

Solnce55 [7]

4(x+2) , 2(2x+4) , 8+4x , 1/2(8x+16) are all the same as 4x+ 8.

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2 years ago

an underground mine elevator reaches a total depth of 99 meters. it descends 17% of the total distance every minute. how many me

malfutka [58]

17% of 99 is 16.83. Since that's only the distance it goes in one minute we have to multiply that by 3. 16.83 × 3 = 50.49. Since the elevator is moving down then it will be negative. The answer will be -50.49 meters.

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3 years ago

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Which is true about the solution to the system of inequalities shown? y less than or equal to 1/3x-1 and y less than or equal to

tekilochka [14]

Y≤x/3-1 AND y≤x/3-3

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y≤x/3-3

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2 years ago

(x – 2) / 4 – (3x + 5) / 7 = – 3, x = ?<br><br> WILL GIVE BRAINLIEST

balu736 [363]

Answer:

under

Step-by-step explanation:

change the underminator, we have

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2 years ago

In the Ardmore Hotel, 20 percent of the guests (the historical percentage) pay by American Express credit card. (a) What is the

xxMikexx [17]

Answer:

(a) The expected number of guests until the next one pays by American Express credit card is 4.

(b) The probability that the first guest to use an American Express is within the first 10 to checkout is 0.0215.

Step-by-step explanation:

The random variable <em>X</em> can be defined as the number of guests until the next one pays by American Express credit card

The probability that a guest paying by American Express credit card is, <em>p</em> = 0.20.

The random variable <em>X</em> follows a Geometric distribution since it is defined as the number of trials before the first success.

The probability mass function of <em>X</em> is:

$P(X=x)=(1-p)^{x}p;\ x=0,1,2,3...,\ 0$

(a)

The expected value of a Geometric distribution is:

$E(X)=\frac{1-p}{p}$

Compute the expected number of guests until the next one pays by American Express credit card as follows:

$E(X)=\frac{1-p}{p}$

$=\frac{1-0.20}{0.20}$

$=4$

Thus, the expected number of guests until the next one pays by American Express credit card is 4.

(b)

Compute the probability that the first guest to use an American Express is within the first 10 to checkout as follows:

$P(X=10)=(1-0.20)^{10}\times0.20$

$=0.1073741824\times 0.20\\=0.02147483648\\\approx0.0215$

Thus, the probability that the first guest to use an American Express is within the first 10 to checkout is 0.0215.

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3 years ago