Solve z=9/13 c for c

A nickle 5% of a dollar. Write the value of a nickel as a decimal and as a fraction.

kkurt [141]

Decimal= 0.05
fraction=5/100 or 1/20

3 0

3 years ago

] Consider a randomly shuffled deck of ten cards labeled {1, 1, 2, 2, 3, 3, 4, 4, 5, 5}, of which game play requires three to be

ollegr [7]

Answer:

B should be worth more points.

Step-by-step explanation:

Given - Consider a randomly shuffled deck of ten cards labeled

{1, 1, 2, 2, 3, 3, 4, 4, 5, 5}, of which game play requires three

to be drawn.

To find - Which one should be worth more points:

A getting a pair of matching cards within the three,

or B getting a set of three cards that can be arranged such

that the sum of two of them is the value of the third.

Proof -

Given that, three cards has to be drawn.

So, total number of ways the card be drawn = ¹⁰C₃

= $\frac{10!}{3! (10-3)!}$

= $\frac{10!}{3! 7!}$

= $\frac{10.9.8.7!}{3! 7!}$

= $\frac{10.9.8}{3.2.1}$

= 120

⇒¹⁰C₃ = 120

Now,

Given that, A getting a pair of matching cards within the three

Sample space = { (1,1,2), (1,1,3).(1,1,4),(1,1,5);

(2,2,1),(2,2,3),(2,2,4),(2,2,5);

(3,3,1),(3,3,2),(3,3,4),(3,3,5);

(4,4,1),(4,4,2),(4,4,3),(4,4,5);

(5,5,1),(5,5,2),(5,5,3),(5,5,4) }

⇒n(A) = 20

Also,

B getting a set of three cards that can be arranged such that the sum of two of them is the value of the third

Sample space = { (1,1,2),(1,2,3),(1,3,4),(1,4,5),(2,2,4),(2,3,5) }

⇒n(B) = 6

Now,

P(A) = $\frac{20}{120} = \frac{1}{6}$ = 0.167

P(B) = $\frac{6}{120} = \frac{1}{20}$ = 0.05

We can see that, P(B) < P(A)

⇒B should be worth more points.

4 0

3 years ago

These are two more of the question

amm1812

3456 TGᗩTᔕ ᑎOT TᕼE ᗩᑎᔕᗯEᖇ I ᒍᑌᔕT ᑎEEᗪ ᑭOIᑎTᔕ

7 0

3 years ago

12. A, B and C are consecutive natural numbers. If 2/7 of the reciprocal of A is equal to ⅓ of the reciprocal of C, find B.

abruzzese [7]

<h3>Answer: 13</h3>

===================================================

Explanation:

The set of natural numbers is {1,2,3,4,...} aka the set of positive whole numbers.

Consecutive natural numbers follow one right after another, like with that example above or something like 7,8,9,...

Since B follows right after A, this means B = A+1. Similarly, C = B+1 = (A+1)+1 = A+2 because C follows right after B.

So we have

A = some unknown natural number
B = A+1
C = A+2

In other words, we have the sequence A,A+1,A+2 to replace the sequence A,B,C in that order.

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

The reciprocal of A is 1/A. Taking 2/7 of this gets us 2/(7A). This result is equal to 1/3 of the reciprocal of C, so,

2/(7A) = 1/3 of 1/C

2/(7A) = 1/(3C)

2/(7A) = 1/(3(A+2))

2/(7A) = 1/(3A+6)

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

Let's cross multiply and solve for A

2/(7A) = 1/(3A+6)

2(3A+6) = 7A*1

6A+12 = 7A

12 = 7A-6A

12 = A

A = 12

Therefore, B = A+1 = 12+1 = 13 and C = A+2 = 12+2 = 14

The sequence A,B,C updates to 12,13,14.

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

Check:

A = 12

reciprocal of A = 1/A = 1/12

D = 2/7 of reciprocal of A = (2/7)*(1/12) = 2/84 = 1/42

C = 14

reciprocal of C = 1/C = 1/14

E = 1/3 of the reciprocal of C = (1/3)*(1/14) = 1/42

Items D and E are equal, so it confirms we have the correct answer.

8 0

3 years ago

Which of the sets of ordered pairs represents a function? (4 points) A = {(−4, 5), (1, −1), (2, −2), (2, 3)} B = {(2, 2), (3, −2

elena55 [62]

A function will not have any repeating x values....it can have repeating y values, just not the x ones

(-4,5),(-1,1),(2,-2),(2,3)
this is not a function because it has 2 sets of points that has x as 2...so it has repeating x values

(2,2),(3,-2),(9,3),(9,-3)
this is not a function because it has 2 sets of points that has an x value of 9...so it also has repeating x values

so both of these are not functions.....neither of them

6 0

3 years ago