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tangare [24]
3 years ago
14

What is the better estimate for the mass of a textbook, 1 gram or 1 kilogram

Mathematics
2 answers:
frez [133]3 years ago
8 0
I kilogram, 1 gram would be a speck of dust 
Tems11 [23]3 years ago
3 0
1 kilogram<span>,because one </span>gram<span> would be about a piece of paper.</span>
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Helllllp plz!!!! I really need it!!!!
ahrayia [7]

Answer:

The second one

Step-by-step explanation:

Its not the third one because it is an equilateral

its not the first one because it is not obtuse

The triangle is isosceles

obtuse means the angles are big

the angles are small so they are acute

I hope this helps you

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3 years ago
Read 2 more answers
In a geometric sequence, a2=2, a3=16, and a4=128.
Oduvanchick [21]

Answer:

an = 1/4(8)^n-1

Step-by-step explanation:

Given the following in a geometric sequence, a2=2, a3=16, and a4=128.

nth term of a sequence = ar^n-1

a is the first term

r is the common ratio

r = a3/a2 = a4/a3

r = 16/2 = 128/16

r = 8

a2 = ar

2 = 8a

a = 2/8

a = 1/4

The nth term by substituting the parameters will be;

an = 1/4(8)^n-1

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3 years ago
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Lina20 [59]
The answer would be 150
6 0
3 years ago
Please answer all parts of the question and all work shown.
faust18 [17]

Answer:

a. 0.4931

b. 0.2695

Step-by-step explanation:

Given

Let BG represents Boston Globe

NYT represents New York Times

P(BG) = 0.55

P(BG') = 1 - 0.55 = 0.45

P(NYT) = 0.6

P(NYT') = 1 -0.6 = 0.4

Number of headlines = 5

Number of depressed articles = 3 (at most)

a.

Let P(Read) = Probability that he reads the news the first day

P(Read) = P(He reads BG) and P(He reads NYT)

For the professor to read BG, then there must be at most 3 depressing news

i.e P(0) + P(1) + P(2) + P(3)

But P(0) + P(1) + .... + P(5) = 1 (this is the sample space)

So,

P(0) + P(1) + P(2) + P(3) = 1 - P(4) - P(5)

P(4) or P(BG = 4) is given as the binomial below

(BG + BG')^n where n = 5, r = 4

So, P(BG = 4) = C(5,4) * 0.55⁴ * 0.45¹

P(BG = 5). = (BG + BG')^n where n = 5, r = 5

So, P(BG = 5) = C(5,5) * 0.55^5 * 0.45°

P(0) + P(1) + P(2) + P(3)= 1 - P(BG = 4) - P(BG = 5)

P(0) + P(1) + P(2) + P(3) = 1 - C(5,4) * 0.55⁴ * 0.45¹ - C(5,5) * 0.55^5 * 0.45°

P(0) + P(1) + P(2) + P(3) = 0.7438

For the professor to read NYT, then there must be at most 3 depressing news

i.e P(0) + P(1) + P(2) + P(3)

But P(0) + P(1) + .... + P(5) = 1 (this is the sample space)

So,

P(0) + P(1) + P(2) + P(3) = 1 - P(4) - P(5)

P(4) or P(NYT = 4) is given as the binomial below

(NYT+ NYT')^n where n = 5, r = 4

So, P(NYT = 4) = C(5,4) * 0.6⁴ * 0.4¹

P(NYT = 5). = (NYT + NYT')^n where n = 5, r = 5

So, P(NYT = 5) = C(5,5) * 0.6^5 * 0.4°

P(0) + P(1) + P(2) + P(3)= 1 - P(NYT = 4) - P(NYT = 5)

P(0) + P(1) + P(2) + P(3) = 1 - C(5,4) * 0.6⁴ * 0.4¹ - C(5,5) * 0.6^5 * 0.4°

P(0) + P(1) + P(2) + P(3) = 0.6630

P(Read) = P(He reads BG) and P(He reads NYT)

P(Read) = 0.7438 * 0.6630

P(Read) = 0.4931

b.

Given

n = Number of week = 7

P(Read) = 0.4931

R(Read') = 1 - 0.4931 =

He needs to read at least half the time means he reads for 4 days a week

So,

P(Well-informed) = (Read + Read')^n where n = 7, r = 4

P(Well-informed) = C(7,4) * (0.4931)⁴ * (1-0.4931)³ = 0.2695

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There were 72 more maple trees before the bug problem then after because there was 108 maple trees before the bug problem and 36 maple trees after the bug problem
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