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blondinia [14]
3 years ago
11

One day, River writes down the numbers 1, 2,...., 999 What is the sum of all the digits that River wrote down? And what is the n

umber of all the digits River wrote down
Mathematics
1 answer:
padilas [110]3 years ago
3 0

Answer: He wrote down 999 digits and the sum is

499,500

Step-by-step explanation:

1+2+3 going all they up to 999 and then added together.

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This question relates to concepts covered in Lectures 1 & 2. You can use any of the excel files posted to work through the q
liubo4ka [24]

Answer:

mean of this demand distribution = 100

Step-by-step explanation:

To find the mean of this demand distribution;

Mean = Expected vale = E[x]

for discrete provability function,

we say E[x] = ∑(x.p(x))

x     p(x)     x.p(x)

10     0.1     1

30    0.4    12

60    0.4    24

90    0.7    63

∴ ∑(x.p(x)) = ( 1 + 12 + 24 + 63 )

∑(x.p(x)) = 100

7 0
3 years ago
Name:
saveliy_v [14]
Bread A because it was more likely to have it ..
4 0
3 years ago
P(A) = .6, P(B) = .3, P(Β | A) = .5
Reika [66]

Answer:

A

Step-by-step explanation:

P(B|A)=\frac{A ~and~B}{P(A)} \\0.5=\frac{P(A and B)}{0.6} \\P(A~and~B)=0.5 \times 0.6=0.30

8 0
2 years ago
Find the inverse of the function:<br>f(x) = 3x + 6<br>​
natta225 [31]

Answer:

      x = -2

Step-by-step explanation:

Pull out like factors :

  -3x - 6  =   -3 • (x + 2)

Solve :    -3   =  0

This equation has no solution.

A a non-zero constant never equals zero.

Solve  :    x+2 = 0

Subtract  2  from both sides of the equation :

                     x = -2

4 0
3 years ago
Let E = {(x, y) e R2|xy &gt; 0}. Determine whether E is a subspace of R2 . X3
gayaneshka [121]

Answer:

E is not a subspace of \mathbb{R}^2

Step-by-step explanation:

E is not a subspace of  \mathbb{R}^2

In order to see this, we must find two points (a,b), (c,d) in  E such that (a,b) + (c,d) is not in E.

Consider

(a,b) = (1,1)

(c,d) = (-1,-1)

It is easy to see that both (a,b) and (c,d) are in E since 1*1>0 and (1-)*(-1)>0.  

But (a,b) + (c,d) = (1-1, 1-1) = (0,0)

and (0,0) is not in E.

By the way, it can be proved that in any vector space all sub spaces must have the vector zero.

5 0
3 years ago
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