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USPshnik [31]
4 years ago
11

Part A: How many positive integers are there between 42 and 97 inclusive? I do not understand how the formula works! Part B: The

re is a line whose length is 200 feet. If points are placed every 2 feet starting from one end, how many points are on the line?
Mathematics
2 answers:
avanturin [10]4 years ago
8 0

Answer:

Step-by-step explanation:

To find the answer to Part A we can do (97 - 42) + 1 = 56. The reason we do + 1 is because 97 - 42 is just counting the numbers in between 42 and 97 but it leaves out 42 and since it's inclusive we need to include 42.

Part B: We can do (200 / 2) - 1 = 99. The reason we do - 1 is because 200 / 2 includes the endpoint of the line but since we don't want to include the endpoint of the line we do - 1.

chubhunter [2.5K]4 years ago
6 0

Answer:

Part A = 56 Integers

Part B = 99 Points

Step-by-step explanation:

Part A ~ Since it's inclusive, at the end we must put n + 1 -->

97 - 42 = 55 => 55 = n => 56 Integers

Part B ~ n - 1  Since it is <u>exclusive</u>. so, =>

200/2 = 100, 100 = n (n-1) => 99 Points

Hope this helps!

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Answer:

A. Adjacent

Step-by-step explanation:

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4 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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