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olganol [36]
3 years ago
6

a space shuttle is 50 miles above earth an astronaut on the shuttle sees the sun rise over earth. to the nearest mile, what is t

he distance from the astronaut to the horizon? (hint: earths radius is about 4,000 miles)
Mathematics
1 answer:
Ludmilka [50]3 years ago
6 0
From the astronaut to the horizon is a tangent line to the curvature of the earth.  And from that point on the horizon to the center of the earth is at a right angle to the tangent and is equal to the radius of the earth...so we can say

cosα=r/(r+h) where r is the radius of the earth and h is the height above the surface of the earth...

α=arccos(r/(r+h))

now tanα=d/r where d is the distance from the astronaut to the point on the horizon so:

tan(arccos(r/(r+h))=d/r

d=rtan(arccos(r/(r+h)))  and using r≈4000 and h=50 we get:

d=4000tan(arccos(4000/(4050)))

d≈634.42mi

d≈634 mi (to nearest mile)
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The sum of 33 consecutive even numbers is 78. What is the second number in this sequence?
hjlf

Answer: The second number is 26


Step-by-step explanation:

To solve this problem you must apply the proccedure shown below:

1. Let's call:

x: The first even number.

x+2: the second consecutive even number.

x+4: the third consecutive even number.

2. So, you know that the sum ot the 3 consecutive numbers is 78, therefore, you must add them, as you can see below:

x+(x+2)+(x+4)=78\\3x+6=78\\3x=78-6\\3x=72\\x=\frac{72}{3}\\x=24

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3 0
3 years ago
An urn contains 8 red chips, 10 green chips, and 2 white chips. A chip is drawn and replaced, and then a second chip is drawn.
Harlamova29_29 [7]

Answer:

(A) 0.04

(B) 0.25

(C) 0.40

Step-by-step explanation:

Let R = drawing a red chips, G = drawing green chips and W = drawing white chips.

Given:

R = 8, G = 10 and W = 2.

Total number of chips = 8 + 10 + 2 = 20

P(R) = \frac{8}{20}=\frac{2}{5}\\P(G)=  \frac{10}{20}=\frac{1}{2}\\P(W)=  \frac{2}{20}=\frac{1}{10}

As the chips are replaced after drawing the probability of selecting the second chip is independent of the probability of selecting the first chip.

(A)

Compute the probability of selecting a white chip on the first and a red on the second as follows:

P(1^{st}\ white\ chip, 2^{nd}\ red\ chip)=P(W)\times P(R)\\=\frac{1}{10}\times \frac{2}{5}\\ =\frac{1}{25} \\=0.04

Thus, the probability of selecting a white chip on the first and a red on the second is 0.04.

(B)

Compute the probability of selecting 2 green chips:

P(2\ Green\ chips)=P(G)\times P(G)\\=\frac{1}{2} \times\frac{1}{2}\\ =\frac{1}{4}\\ =0.25

Thus, the probability of selecting 2 green chips is 0.25.

(C)

Compute the conditional probability of selecting a red chip given the first chip drawn was white as follows:

P(2^{nd}\ red\ chip|1^{st}\ white\ chip)=\frac{P(2^{nd}\ red\ chip\ \cap 1^{st}\ white\ chip)}{P (1^{st}\ white\ chip)} \\=\frac{P(2^{nd}\ red\ chip)P(1^{st}\ white\ chip)}{P (1^{st}\ white\ chip)} \\= P(R)\\=\frac{2}{5}\\=0.40

Thus, the probability of selecting a red chip given the first chip drawn was white is 0.40.

6 0
3 years ago
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