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user100 [1]
2 years ago
12

Copy and complete each sequence below. Using words, not numbers, describe how the patterns work. (for example, write, “Double th

e previous number.”)
a. 1,3,6,10, __ , __
b. 1, 1/2, 1/4, 1/8, __ , __
c.1, 3, 9, 27, __ , __
d. 8, 7, 5, 2, 15, 21, __ , __
e. 49, 47, 52, 50, 55, __ ,__
Mathematics
1 answer:
Allisa [31]2 years ago
8 0
A. Add one more than was added to the previous number.
b. Divide by 2 each time.
c. Multiply by 3 each time.
d. Beginning is subtract one more than previously subtracted but the last two(15,,21) don’t fit the pattern.
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AB and BC share point B to form AC. If AC=7 and BC=5, then what is the length of AB?
AnnZ [28]
|AC|=7;\ |BC|=5\\\\|AC|=|AB|+|BC|\\\\|AB|+5=7\ \ \ \ \ |subtract\ 5\ from\ both\ sides\\\\\boxed{|AB|=2}

4 0
3 years ago
Read 2 more answers
A) Write 84 as the product of prime factors.<br> Write the prime factors in ascending order.
timofeeve [1]

84 = 2 x 2 x 3 x 7 which can also be written 2² x 3 x 7.

Ascending order means to keep the factors in order from least to greatest.

3 0
2 years ago
The sum area please.
Ann [662]

Answer:

4xy+24y^2−7x−38y−7

Step-by-step explanation:

(-7+4y) (6y+x+1)

-7*6y= -42y

-7*x= -7x

-7*1= -7

4y*6y= 24y²

4y*x= 4yx

4x*1= 4y

Add them up

3 0
2 years ago
PLEASE HELP WILL GIVE BRAINLIEST Which of the following scale factors will result in an enlargement? A. k &lt; –1 B. –1 &lt; k &
Sergio [31]

Answer: OPTION A.

Step-by-step explanation:

By definition, a dilation can be an enlargement or a reduction of the shape.

An enlargement is when dilation creates a larger image and a reduction is when dilation creates a smaller image.

When, the scale factor is greater than 1, the image is an enlargement and when the scale factor is between 0 and 1, the image is a reduction.

It is important to know that with a negative scale factor the enlargement will will be inverted and it will also be on the other side of the center of dilation.

Knowing this, we can say that the scale factor that will result in an enlargement is:

k

5 0
2 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
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