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12345 [234]
1 year ago
9

A fair sortition trial is carried out, and one of the candidates is assigned the number 32,041. If each digit can be chosen from

0-4, and if each of the possible sequences is assigned to a candidate, how many candidates are there?
Mathematics
2 answers:
erik [133]1 year ago
6 0
The answer to this question is 3,125
Kitty [74]1 year ago
3 0

Answer: 3125


Step-by-step explanation:

Given: A fair sortition trial is carried out, and one of the candidates is assigned the number = 32,041

If each digit can be chosen from 0-4, and if each of the possible sequences is assigned to a candidate, then the number of candidates (repetition of digits allowed)=5\times5\times5\times5\times5

=3125

Hence, the number of candidates there are = 3125.




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heather is training for a long-distance run. her data points listed below represent the days of practice, x, and the number of m
MAXImum [283]
Interpolation equation
Poit 1: (a,b)
Point 2: (c,d)
Intermediate point (x,y)

(d - b)/(c - a) = (y - b) / (x - a)

(a,b) = (2, 4.2)
(c,d) = (4, 5.6)

x = 3

(5.6 - 4.2) / (4 - 2)  =  (y - 4.2) / (3 -2)

y =  [1.4/2]*[1] + 4.2

y = 0.7 + 4.2 = 4.9

Answer 4.9 miles

 




3 0
2 years ago
Read 2 more answers
Calculating conditional probabilities - random permutations. About The letters (a, b, c, d, e, f, g) are put in a random order.
Evgesh-ka [11]

A="b is in the middle"

B="c is to the right of b"

C="The letter def occur together in that order"

a) b can be in 7 places, but only one is the middle. So, P(A)=1/7

b) X=i, "b is in the i-th position"

Y=j, "c is in the j-th position"

P(B)=\displaystyle\sum_{i=1}^{6}(P(X=i)\displaystyle\sum_{j=i+1}^{7}P(Y=j))=\displaystyle\sum_{i=1}^{6}\frac{1}{7}(\displaystyle\sum_{j=i+1}^{7}\frac{1}{6})=\frac{1}{42}\displaystyle\sum_{i=1}^{6}(\displaystyle\sum_{j=i+1}^{7}1)=\frac{6+5+4+3+2+1}{42}=\frac{1}{2}

P(B)=1/2

c) X=i, "d is in the i-th position"

Y=j, "e is in the j-th position"

Z=k, "f is in the i-th position"

P(C)=\displaystyle\sum_{i=1}^{5}( P(X=i)P(Y=i+1)P(Z=i+2))=\displaystyle\sum_{i=1}^{5}(\frac{1}{7}\times\frac{1}{6}\times\frac{1}{5})=\frac{1}{210}\displaystyle\sum_{i=1}^{5}(1)=\frac{1}{42}

P(C)=1/42

P(A∩C)=2*(1/7*1/6*1/5*1/4)=1/420

P(B\cap C)=\displaystyle\sum_{i=1}^{3} P(X=i)P(Y=i+1)P(Z=i+2)\displaystyle\sum_{j=i+3}^{6}P(V=j)P(W=j+1)=\displaystyle\sum_{i=1}^{3}\frac{1}{6}\frac{1}{7}\frac{1}{5}(\displaystyle\sum_{j=1+3}^{6}\frac{1}{4}\frac{1}{3})=1/420

P(B∩A)=3*(1/7*1/6)=1/14

P(A|C)=P(A∩C)/P(C)=(1/420)/(1/42)=1/10

P(B|C)=P(B∩C)/P(C)=(1/420)/(1/42)=1/10

P(A|B)=P(B∩A)/P(B)=(1/14)/(1/2)=1/7

P(A∩B)=1/14

P(A)P(B)=(1/7)*(1/2)=1/14

A and B are independent

P(A∩C)=1/420

P(A)P(C)=(1/7)*(1/42)=1/294

A and C aren't independent

P(B∩C)=1/420

P(B)P(C)=(1/2)*(1/42)=1/84

B and C aren't independent

8 0
1 year ago
A customer/member owes $10.51 and pays $50.00 in cash. The change due is $39.49. Provide the amount owed using the smallest numb
Zigmanuir [339]

The number is 14 bill or coins

Step-by-step explanation:

The method to use the smallest number of bills and coins possible is to utilize the largest denominations possible.

The balance is $39.49

$39.49

-$20.00 -------------a $20 bill is the largest bill not larger than the total ($39.49)

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

$19.49

-$10.00--a $10 bill is the largest bill not larger than the remaining ($19.49)

----------

$9.49

-$5.00------a $5 bill is the largest bill not larger than the remaining ($9.45)

----------

$4.49

-$1.00-------a $1 bill is the largest bill not larger than the remaining ($4.49)

---------

$3.49

-$1.00-------a $1 bill is the largest bill not larger than the remaining ($3.49)

----------

$2.49

-$1.00------a $1 bill is the largest bill not larger than the remaining ($2.49)

----------

$1.49

-$1.00------a $1 bill is the largest bill not larger than the remaining ($1.49)

-----------

$0.49

-$0.25---- a quarter is the largest coin not larger than the remaining ($0.49)

----------

$0.24

-$0.10----a dime is the largest coin not larger than the remaining ($0.24)

---------

$0.14

-$0.10--a dime is the largest coin not larger than the remaining($0.14)

---------

$0.04

- a penny, ($0.01)  is the largest coin not larger than the remaining ($0.04). Here you have 4 pennies

Now you have Bill and Coin as;

1 $20 bill

1 $10 bill

1 $5 bill

4 $1 bill

1 quarter

2 dime

4 penny

14 bill or coins

Learn More

Bill and coins :brainly.com/question/3871376

Keyword : owes, bills, coins

#LearnwithBrainly

5 0
1 year ago
The total operating cost to run the organic strawberry farm varies directly with its number of acres. What is the constant of va
kow [346]

Answer:

16\ acres

Step-by-step explanation:

we know that

A relationship between two variables, x, and y, represent a proportional variation if it can be expressed in the form k=\frac{y}{x} or y=kx

In a proportional relationship the constant of proportionality k is equal to the slope m of the line and the line passes through the origin

Let

x ----> the number of acres

y ---> total operating costs

we have the ordered pair (20,551,520)

For x=20 acres, y=$551,520

<em>Find the value of the constant of proportionality k</em>

k=\frac{y}{x}

substitute the values of x and y

k=\frac{551,520}{20}=\$27,576\ per\ acre

The linear equation is equal to

y=27,576x

Find out how many acres are on a farm with a total operating cost of $441,216

so

For y=$441,216

substitute in the linear equation the value of y and solve for x

441,216=27,576x

x=441,216/27,576

x=16\ acres

5 0
2 years ago
Read 2 more answers
A company currently has 200 units of a product on hand that it orders every 2 weeks when the salesperson visits the premises. De
disa [49]

Answer: 289 units

Step-by-step explanation:

Given the following :

Inventory (I) = 180

Lead time (L) = 7 days

Review time (T) = 2 weeks = 14 days

Demand (D) = 20

Standard deviation (σ) = 5

Zscore for 95% probability = 1.645

Units to be ordered :

D(T + L) + z(σT+L)

(σT+L) = √(T + L)σ²

= √(14 + 7)5²

= √(21)25

= 22.9

D(T + L) + z(σT+L) - I

20(14 + 7) + 1.645(22.9 + 7) - I

= 420 + 49.1855 - 180

= 289.1855

= 289 quantities

5 0
2 years ago
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