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Murrr4er [49]
3 years ago
13

Let A be the set of all strings of decimal digits of length 5. For example 00312 and 19483 are two strings in A. You pick a stri

ng from A at random. What is the probability that the string has no 4 in it
Mathematics
1 answer:
lord [1]3 years ago
4 0

Answer:

0.59049

Step-by-step explanation:

Set of all digit strings of length 5

Digit range = (0 - 9) = 10

Length of digit = 5

Total possible outcomes = 10^5 = 100000

String without a 4 ;

Digit range = (0,1,2,3,5,6,7,8,9) = 9

Length of digit = 5

Required outcome = 9^5 = 59049

Required outcome / Total possible outcomes

59049 / 100000

= 0.59049

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Serggg [28]

Answer:

57 squares

Step-by-step explanation:

38/2=19

19*3=57

3 0
3 years ago
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a language arts test is worth 100 points there is a total of 26 questions . there are spelling word questions that ae worth 2 po
natka813 [3]

10 question of spelling words and 16 questions of vocabulary are present

<em><u>Solution:</u></em>

Let "x" be the number of spelling word questions

Let "y" be the number of vocabulary word question

<em><u>There is a total of 26 questions. Therefore, we get</u></em>

number of spelling word questions + number of vocabulary word question = 26

x + y = 26 --------- eqn 1

<em><u>There are spelling word questions that worth 2 points each and vocabulary word question worth 5 points each</u></em>

The language arts test is worth 100 points

Therefore, we frame a equation as:

number of spelling word questions x 2 + number of vocabulary word question x 5 = 100

x \times 2 + y \times 5 = 100

2x + 5y = 100 --------- eqn 2

<em><u>Let us solve eqn 1 and eqn 2</u></em>

From eqn 1,

x = 26 - y ------- eqn 3

<em><u>Substitute eqn 3 in eqn 2</u></em>

2(26 - y) + 5y = 100

52 - 2y + 5y = 100

3y = 100 - 52

3y = 48

<h3>y = 16</h3>

<em><u>Substitute y = 16 in eqn 3</u></em>

x = 26 - 16

<h3>x = 10</h3>

Thus 10 question of spelling words and 16 questions of vocabulary are present

4 0
3 years ago
Assume V and W are​ finite-dimensional vector spaces and T is a linear transformation from V to​ W, T: Upper V right arrow Upper
scZoUnD [109]

Answer:

Thus for the vectors v_1, v_2, v_p there are scalars c_1, c_2, c_p not all zeros, such that c_1v_1 +c_2v_2+... +c_pv_p = 0. It means that the vectors v_1, v_2, v_p are linearly dependent in contradiction with the fact that the vectors form a basis for H. So the assumption that T(v_1), T(v_2),..., T(v_p) are linearly dependent is false, proving the required.  

Step-by-step explanation:

Let B = {v_1 ,v_2,..., v_p} be a basis of H, that is dim H = p and for any v ∈ H there are scalars c_1 , c_2, c_p, such that v = c_1*v_1 + c_2*v_2 +....+ C_p*V_p It follows that  

T(v) = T(c_1*v_1 + c_2v_2 + ••• + c_pV_p) = c_1T(v_1) +c_2T(v_2) + c_pT(v_p)

so T(H) is spanned by p vectors T(v_1),T(v_2), T(v_p). It is enough to prove that these vectors are linearly independent. It will imply that the vectors form a basis of T(H), and thus dim T(H) = p = dim H.  

Assume in contrary that T(v_1 ), T(v_2), T(v_p) are linearly dependent, that is there are scalars c_1, c_2, c_p not all zeros, such that  

c_1T(v_1) + c_2T(v_2) +.... + c_pT(v_p) = 0

T(c_1v_1) + T(c_2v_2) +.... + T(c_pv_p) = 0

T(c_1v_1+ c_2v_2 ... c_pv_p) = 0  

But also T(0) = 0 and since T is one-to-one, it follows that c_1v_1 + c_2v_2 +.... + c_pv_p = O.

Thus for the vectors v_1, v_2, v_p there are scalars c_1, c_2, c_p not all zeros, such that c_1v_1 +c_2v_2+... +c_pv_p = 0. It means that the vectors v_1, v_2, v_p are linearly dependent in contradiction with the fact that the vectors form a basis for H. So the assumption that T(v_1), T(v_2),..., T(v_p) are linearly dependent is false, proving the required.  

8 0
3 years ago
Find the slope of the line that passes through the points<br><br> A(0, 2) and B(4, –1).
kati45 [8]

Slope=M=y2-y1/X2-X1=

-1 -2 /4-0=

-3/4  =

-0.75

5 0
3 years ago
How many solutions does the system pf equations below have?​
Oliga [24]

Answer:

Your answer is One solution

Hope that this is helpful. Tap the crown button, Like & Follow me

6 0
3 years ago
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