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DedPeter [7]
3 years ago
13

Apollo 14 landed on part of a vast ejecta blanket that surrounds the Sea of Rains. The Sea of Rains was formed by the volcanic f

looding of a large impact basin that is referred to as the Imbrium Basin. The Imbrium Basin is over 1000 km in diameter. (i) (1 mark] The rocks brought back by Apollo 14 astronauts were used to determine the time at which a very specific event occurred in the history of the Moon. What was that event? (ii) (1 mark) How many interior rings of mountains on the floor of the Imbrium Basin likely got completely covered over by the volcanic flooding of the basin? Explain the reason behind your answer. (iii) (1 mark] (True/False) Apollo 14 was the first mission to bring along a four-wheeled battery-powered roving vehicle that could be driven across the lunar surface by astronauts. The Apollo 15 landing site was hundreds of kilometers away from the Apollo 14 landing site. Figure 1 shows a picture of the Apollo 15 landing site. Apollo 15 landed on the dark flat area at the upper right of the picture. (iv) (1 mark] (True/False) The Apollo 15 lunar module touched down on a lowlands surface located close to the mountainous rim of the above-mentioned Sea of Rains. (v) [1 mark] The mountains seen at lower right of Figure 1 are part of a high mountain range named the lunar Apennines. Is this mountainous region a highlands terrain or a lowlands terrain? Did the Apollo 15 astronauts make any trips from their Lunar Module to the base of this mountain range? (iv) (1 mark] (True/False) The Apollo 15 lunar module touched down on a lowlands surface located close to the mountainous rim of the above-mentioned Sea of Rains. (v) (1 mark] The mountains seen at lower right of Figure 1 are part of a high mountain range named the lunar Apennines. Is this mountainous region a highlands terrain or a lowlands terrain? Did the Apollo 15 astronauts make any trips from their Lunar Module to the base of this mountain range? (vi) [1 mark] (True/False) The Apollo 15 landing site is also close to Hadley rille which can be seen in Figure 1. Hadley rille is an example of an arcuate rille. (vii) (2 marks How was Hadley rille formed? • Apollo 15 Figure 1: The landing site of Apollo 15 as photographed from lunar orbit.
Mathematics
1 answer:
ycow [4]3 years ago
7 0

Answer:

bcbcb fdf vd bgdb b fgbf dvscsdvdsv md vd vgd gd fgb dgb sffdefdfdvse

Step-by-step explanation:

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What is the bases, height, perimeter, and the area from the trapezoid​?<br><br>please help me
Sergio039 [100]

Answer:

base : 30 cm

height : 20 cm

perimeter : 92 cm

Area          :  400 cm

Step-by-step explanation:

base : 30 cm

height : 20 cm

perimeter : 27 + 30 + 25 + 10

                 : 92 cm

Area : (1/2 ( a + b) ) x h

        : (1/2 ( 10 + 30) ) x 20

        : (20) x 20

        : 400 cm

3 0
3 years ago
10 times 100 fhg[';g;8utfdkibjlif
adoni [48]

Answer:

1,000 nmvfdmld

Step-by-step explanation:

5 0
3 years ago
Read 2 more answers
Find the sum of the first 6 terms of 3 - 6 + 12 + …
Natalka [10]

Answer:

S_6 = -63

Step-by-step explanation:

The sequence above is a geometric sequence.

The common ratio (r) = \frac{-6}{3} = \frac{12}{-6} = -2

The common ratio < 1, therefore, the formula for the sum of nth terms of the sequence would be: S_n = \frac{a_1(1 - r^n)}{1 - r}

a1 = 3

r = -2

n = 6

Plug in the values into the formula

S_6 = \frac{3(1 - (-2^6)}{1 - (-2)}

S_6 = \frac{3(1 - (64)}{1 + 2}

S_6 = \frac{3(-63)}{3}

S_6 = -63

6 0
3 years ago
A. ?? <br> B. ?? <br> C. ??<br> Please help me!!!
patriot [66]
Bonjour!

For A, the two angles are congruent by the Corresponding Angles are Congruent Theorem/Postulate.

For B, the two angles are congruent by the Vertical Angels are Congruent Theorem/Postulate.

As for C, the two angles are congruent by the Same Side Exterior Theorem/Postulate.(Though I am uncertain if that is the correct name for this one)

Hope that helps :)
4 0
3 years ago
In the last quarter of​ 2007, a group of 64 mutual funds had a mean return of 2.2 2.2​% with a standard deviation of 6.3 6.3​%.
Svet_ta [14]

Answer:

a) 0.625; b) 16.879; c) 8.442

Step-by-step explanation:

Since this is a normal distribution, we want the value of the mean, μ:  2.2; and the value of the standard deviation, σ:  6.3.

For the 40th percentile, we look in a z chart.  We want to find the value as close to 0.40 as we can get; this is 0.4013, and it corresponds to a z score of z = -0.25.

Our formula for a z score is z=\frac{X-\mu}{\sigma}.  Using our values, we have

-0.25 = (X-2.2)/6.3

Multiply both sides by 6.3:

6.3(-0.25) = X-2.2

-1.575 = X-2.2

Add 2.2 to each side:

-1.575+2.2 = X-2.2+2.2

0.625 = X

For the 99th percentile, the value in the z chart closest to 0.99 is 0.9901, which corresponds to a z score of z = 2.33:

2.33 = (X-2.2)/6.3

Multiply both sides by 6.3:

6.3(2.33) = X-2.2

14.679 = X-2.2

Add 2.2 to each side:

14.679+2.2 = X-2.2+2.2

16.879 = X

For the IQR, we find the values for the 75th percentile (Q3) and the 25th percentile (Q1).  The value in a z chart closest to 0.75 is 0.7486, which corresponds to a z score of z = 0.67:

0.67 = (X-2.2)/6.3

Multiply both sides by 6.3:

6.3(0.67) = X-2.2

4.221 = X-2.2

Add 2.2 to each side:

4.221+2.2 = X-2.2+2.2

6.421 = X

The value in a z chart closest to 0.25 is 0.2514, which corresponds to a z score of z = -0.67:

-0.67 = (X-2.2)/6.3

Multiply both sides by 6.3:

6.3(-0.67) = X-2.2

-4.221 = X-2.2

Add 2.2 to each side:

-4.221+2.2 = X-2.2+2.2

-2.021 = X

This makes the interquartile range

6.421--2.021 = 8.442

8 0
3 years ago
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