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Likurg_2 [28]
3 years ago
7

I do not understand what number I am supposed to use for “y=4x+__. I thought it would be 6 because that’s how many hexagons are

showed in the picture but it isn’t.

Mathematics
1 answer:
skad [1K]3 years ago
8 0
Based on the table, it should be 2.
Y=4x+2 works for all x and y values in the table.
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Find the probability of selecting none of the correct six integers in a lottery, where the order in which these integers are sel
dusya [7]

Answer:

Probability of selecting none of the correct six integers:

a) 0.350

b) 0.427

c) 0.489

d) 0.540

Step-by-step explanation:

a) 40

Given:

Number of integers in a lottery  6

Order in which these integers are selected does not matter

To find:

Probability of selecting none of the correct six integers

Solution:

When the order of selection does not matter then we use Combinations.

Given integers = 40

Number of ways to choose 6 from 40.

Let A be the sample space of choosing digits 6 from 40.

Then using Combinations:

(n,k) = n! / r! (n-r)!

n = 40

r = 6

40C6

=(40,6) = 40! / 6! ( 40 - 6)!

           = 40! / 6!34!

           = 40*39*38*37*36*35*34! / 6!34!

           = 2763633600 / 720

          = 3838380

Let E be the event of selecting none of the correct six integers.

So using combinations we can find the total number of ways of selecting none of 6 integers from 40

n = 40 - 6 = 34

r = 6

34C6

=(34,6) = 34! / 6! ( 34 - 6)!

           = 34! / 6! 28!

           = 34 * 33 * 32 * 31 * 30 * 29 * 28! / 6! 28!

           =968330880 / 720

           = 1344904

Probability of selecting none of the correct six integers:

P(E) = E / A

       = 1344904 / 3838380

        = 0.350

Probability of selecting none of the correct six integers is 0.350

b) 48

Following the method used in part a)

(n,k) = n! / r! (n-r)!

n = 48

r = 6

48C6

=(48,6) = 48! / 6! ( 48 - 6)!

            = 48! / 6! ( 42 )!

            = 48*47*46*45*44*43*42! / 6!42!

            = 8835488640 / 720

            = 12271512

Let E be the event of selecting none of the correct six integers.

So using combinations we can find the total number of ways of selecting none of 6 integers from 48

n = 48 - 6 = 42

r = 6

42C6

= (42,6) = 42! / 6! ( 42 - 6)!

              = 42! / 6! 36!

              = 3776965920

              = 5245786

P(E) = E / A

       = 5245786/12271512

       = 0.427

c) 56

(n,k) = n! / r! (n-r)!

n = 56

r = 6

56C6

=(56,6) = 56! / 6! ( 56- 6)!

            = 56! / 6! ( 50 )!

            = 56*55*54*53*52*51*50! / 6! 50!

            = 23377273920/6

            = 32468436

Let E be the event of selecting none of the correct six integers.

So using combinations we can find the total number of ways of selecting none of 6 integers from 56

n = 56 - 6 = 50

(50,6) = 50! / 6! ( 50- 6)!

           = 50*49*48*47*46*45*44! / 44! 6!

           = 11441304000 / 6

           = 15890700

P(E) = E / A

       = 15890700 / 32468436

      = 0.489

d) 64

(n,k) = n! / r! (n-r)!

n = 64

r = 6

64C6

=(64,6) = 64! / 6! ( 64 - 6)!

            = 64! / 6! ( 58 )!

            = 64*63*62*61*60*59*58! / 6! 58!

            = 53981544960 / 720

            = 74974368

Let E be the event of selecting none of the correct six integers.

So using combinations we can find the total number of ways of selecting none of 6 integers from 64

n = 64 - 6 = 58

(58,6) = 58! / 6! ( 58- 6)!

           = 58*57*56*55*54*53*52! / 52! 6!

           = 29142257760/ 6

           = 40475358

P(E) = E / A

       = 40475358/ 74974368

      = 0.540

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Answer:

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Step-by-step explanation:

f(x) = 1-x^2

g(x) = 1 - x

f(x) - g(x) = 1-x^2  - (1 - x )

Distribute the minus sign

              = 1-x^2 -1 +x

Combine like terms

             = -x^2 +x

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Answer:

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