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Bezzdna [24]
3 years ago
15

Each row in a window display of compact disk cartons contains two more boxes than the row above. The first row has one box. Comp

lete the table to find the number of rows in a display containing 144 boxes.
Mathematics
2 answers:
elixir [45]3 years ago
8 0
Hey there!

You know that the pattern increases by 2 each time so just add two each time until you get to 144

1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21 +23 = 144

R1 R2 R3 R4 R5 R6 R7 R8 R9 R10 R11 R12

So there will be 12 rows in the display

Hope this helps!

~~Cutelion918~~
kherson [118]3 years ago
3 0
The pattern increases by 2 each time.

1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21 +23 = 144

R1 R2 R3 R4 R5 R6 R7 R8 R9 R10 R11 R12

There are 12 rows in the display.

I hope this helped! c:
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Urgent!!!!
LekaFEV [45]

Answer:

One pair of socks is $370.

Step-by-step explanation:

Sean bought <em>three pairs</em> of the same socks and a shoe polish for $1450. If the

polish cost $340, how much is the cost of <em>one pair</em> of socks.

6x - 340 = 1450

6x = 1450 -340

6x = 1110

2x = 1110 / 3

2x = 370

8 0
3 years ago
Read 2 more answers
6(6x−3)= helpppppppppppppp distibutive property again
bagirrra123 [75]

Answer:

6(6x-3)

=36x-18

Step-by-step explanation:

6(6x-3(

multiply 6 x 6 and 6 x 3

36x - 18

5 0
3 years ago
Let X denote the amount of time a book on two-hour reserve is actually checked out, and suppose the cdf is the following. F(x) =
Troyanec [42]

Answer:

a) P (x <= 3 ) = 0.36

b) P ( 2.5 <= x <= 3  ) = 0.11

c) P (x > 3.5 ) = 1 - 0.49 = 0.51

d) x = 3.5355

e) f(x) = x / 12.5

f) E(X) = 3.3333

g) Var (X) = 13.8891  , s.d (X) = 3.7268

h) E[h(X)] = 2500

Step-by-step explanation:

Given:

The cdf is as follows:

                           F(x) = 0                  x < 0

                           F(x) = (x^2 / 25)     0 < x < 5

                           F(x) = 1                   x > 5

Find:

(a) Calculate P(X ≤ 3).

(b) Calculate P(2.5 ≤ X ≤ 3).

(c) Calculate P(X > 3.5).

(d) What is the median checkout duration ? [solve 0.5 = F()].

(e) Obtain the density function f(x). f(x) = F '(x) =

(f) Calculate E(X).

(g) Calculate V(X) and σx. V(X) = σx =

(h) If the borrower is charged an amount h(X) = X2 when checkout duration is X, compute the expected charge E[h(X)].

Solution:

a) Evaluate the cdf given with the limits 0 < x < 3.

So, P (x <= 3 ) = (x^2 / 25) | 0 to 3

     P (x <= 3 ) = (3^2 / 25)  - 0

     P (x <= 3 ) = 0.36

b) Evaluate the cdf given with the limits 2.5 < x < 3.

So, P ( 2.5 <= x <= 3 ) = (x^2 / 25) | 2.5 to 3

     P ( 2.5 <= x <= 3  ) = (3^2 / 25)  - (2.5^2 / 25)

     P ( 2.5 <= x <= 3  ) = 0.36 - 0.25 = 0.11

c) Evaluate the cdf given with the limits x > 3.5

So, P (x > 3.5 ) = 1 - P (x <= 3.5 )

     P (x > 3.5 ) = 1 - (3.5^2 / 25)  - 0

     P (x > 3.5 ) = 1 - 0.49 = 0.51

d) The median checkout for the duration that is 50% of the probability:

So, P( x < a ) = 0.5

      (x^2 / 25) = 0.5

       x^2 = 12.5

      x = 3.5355

e) The probability density function can be evaluated by taking the derivative of the cdf as follows:

       pdf f(x) = d(F(x)) / dx = x / 12.5

f) The expected value of X can be evaluated by the following formula from limits - ∞ to +∞:

         E(X) = integral ( x . f(x)).dx          limits: - ∞ to +∞

         E(X) = integral ( x^2 / 12.5)    

         E(X) = x^3 / 37.5                    limits: 0 to 5

         E(X) = 5^3 / 37.5 = 3.3333

g) The variance of X can be evaluated by the following formula from limits - ∞ to +∞:

         Var(X) = integral ( x^2 . f(x)).dx - (E(X))^2          limits: - ∞ to +∞

         Var(X) = integral ( x^3 / 12.5).dx - (E(X))^2    

         Var(X) = x^4 / 50 | - (3.3333)^2                         limits: 0 to 5

         Var(X) = 5^4 / 50 - (3.3333)^2 = 13.8891

         s.d(X) = sqrt (Var(X)) = sqrt (13.8891) = 3.7268

h) Find the expected charge E[h(X)] , where h(X) is given by:

          h(x) = (f(x))^2 = x^2 / 156.25

  The expected value of h(X) can be evaluated by the following formula from limits - ∞ to +∞:

         E(h(X))) = integral ( x . h(x) ).dx          limits: - ∞ to +∞

         E(h(X))) = integral ( x^3 / 156.25)    

         E(h(X))) = x^4 / 156.25                       limits: 0 to 25

         E(h(X))) = 25^4 / 156.25 = 2500

8 0
3 years ago
Find an equation for the perpendicular bisector of the line segment whose endpoints are
Luba_88 [7]

Answer:

y = 3x + 7

Step-by-step explanation:

3 0
3 years ago
How to work out this problem
vova2212 [387]
Divide the pentagon into a triangle and rectangle.  

Rectangle:  12in*6in=72in^2
Triangle:  (12in*5in)/2=30in^2
 72in^2+30in^2=102in^2
8 0
3 years ago
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