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oksian1 [2.3K]
3 years ago
6

Write an expression to represent the sum modeled in the following number line.

Mathematics
1 answer:
Lyrx [107]3 years ago
7 0
That was my brother who said that to you I'm sorry bout that
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faltersainse [42]

OK.  I used my calculator to evaluate sec(85 degrees).

My calculator doesn't have a "sec" button on it. 
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So I used my calculator to find cos(85), and then I hit the
" 1/x " key, and got 11.474, which I knew to be sec(85).

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3 years ago
-x^4y^2+7x^3y^3-3xy^5+2x^2y^4 in descending powers of x
muminat

Answer:

  -x^4y^2 +7x^3y^3 +2x^2y^4 -3xy^5

Step-by-step explanation:

The powers of x in the terms of the given expression are ...

  4, 3, 1, 2

so, we want to swap the last two terms to put them in the desired order:

  -x^4y^2 +7x^3y^3 +2x^2y^4 -3xy^5

6 0
3 years ago
PLS HELP SP=3x+1, and LN=10x−6. Find SP.<br><br> A. 9<br> B. 4<br> C. 21<br> D. 7
Umnica [9.8K]

Step-by-step explanation:

2(sp)=ln

6x+2=10x-6

8=4x

x=2

therefore sp=3×2+1=6+1=7

Option D

5 0
3 years ago
The product of two given numbers is 126 both of them are divisible by 3 but neither of them is 3.The larger of the two numbers i
Mice21 [21]

I believe the answer is 21

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4 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

3 0
3 years ago
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