All categories

2 years ago

Kkkmnkf;d;klsnfksdjlkfjk;asjdk;fjkasjd;f. fdjnsa;fdadf

Mathematics

1 answer:

Pavel [41]2 years ago

4 0

Sjdlslsooccjjsjshhcndbsyyqtaoxppcjshwtzjcjcoslahbfidowhsbcc

You might be interested in

What is 5800 divided by 28

Taya2010 [7]

The answer is 207.14

6 0

3 years ago

Read 2 more answers

What is the distance between the point (10,12) and (-4,-36)

dimulka [17.4K]

Well using distance formula

$d \: = \sqrt{(x {}^{2} } - x {}^{1} ) + (y {}^{2} + y {}^{1} )$
The answer I got was Distance = 45.9 what ever unit of measure it might ask for. Feel free to check my math just in case.

4 0

3 years ago

3x-2y=6<br><br> y=3/2x-3<br><br> Find the slope, x-intercept, and y-intercept for both equations.

lys-0071 [83]

The slope-intercept form: y = mx + b

m - slope

b - y-intercept

We have 3x - 2y = 6 <em>subtract 3x from both sides</em>

-2y = -3x + 6 <em>divide both sides by -2</em>

y = 3/2 x - 3

slope: m = 3/2

y-intercept: b = -3

x-intercept for y = 0. Substitute:

3/2 x - 3 = 0 <em>add 3 to both sides</em>

3/2 x = 3 <em>multiply both sides by 2/3</em>

x = 2

<h3>Answer:</h3><h3>slope = 3/2</h3><h3>y-intercept = -3</h3><h3>x-intercept = 2</h3>

----------------------------------------------------------------

y = 3/2 x - 3 it's the same function like first function.

8 0

3 years ago

The probability that an American CEO can transact business in a foreign language is .20. Twelve American CEOs are chosen at rand

NNADVOKAT [17]

Answer with Step-by-step explanation:

We are given that

The probability that an American CEO can transact business in foreign language=0.20

The probability than an American CEO can not transact business in foreign language= $1-0.20=0.80$

Total number of American CEOs chosen=12

a. The probability that none can transact business in a foreign language= $12C_0(0.20)^0(0.80)^{12}$

Using binomial theorem $nC_r(1-p)^{n-r}p^r$

The probability that none can transact business in a foreign language= $\frac{12!}{0!(12-0)!}(0.8)^{12}=(0.8)^{12}$

b.The probability that at least two can transact business in a foreign language= $1-P(x=0)-p(x=1)=1-((0.8)^{12}+12C_1(0.8)^{11}(0.2))=1-((0.8)^{12}+12(0.8)^{11}}(0.2))$

c.The probability that all 12 can transact business in a foreign language= $12C_{12}(0.8)^0(0.2)^{12}$

The probability that all 12 can transact business in a foreign language= $\frac{12!}{12!}(0.2)^{12}=(0.2)^{12}$

5 0

3 years ago

{(-1, -1) (3.8) (2, 4)(2, 3), (7. -7) }

vodka [1.7K]

jStep-by-step explanation:

I cannot see the question

4 0

3 years ago