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dalvyx [7]
3 years ago
15

What is the VARIABLE in the expression 8p - 5?*

Mathematics
2 answers:
Aleksandr-060686 [28]3 years ago
8 0

Answer:

p is the variable

Step-by-step explanation:

a variable is a letter that represents an unspecified number.

saveliy_v [14]3 years ago
6 0

Answer:

p is the variable

Step-by-step explanation:

p is the variable because "p" is the only unknown

hope this helps :) have a nice day !1

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What's 0.00045 in scientific notation
tino4ka555 [31]
0.00045 = 4.5 × 10^-4
8 0
3 years ago
Read 2 more answers
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
Ms. Lackey gave each student in her class a calculator.Each calculator weighed 16 ounces. If Ms.Lackey gave each of her 20 stude
jeka94
Everyone got a calculator.....each weighed 16 oz.......gave them to 20 students.....20 * 16 = 320 total oz

16 oz = 1 lb

320/16 = 20 lbs <==

or, if u just think about it...u gave 16 oz to each student.....16 oz = 1 lb...so u gave 1 lb to each student....20 students = 20 lbs
4 0
3 years ago
Joe mauer throws a baseball from first base to second at a rate of 90 miles per hour. How fast did Joe throw the ball in feet pe
deff fn [24]
Lol that's me ...........
6 0
3 years ago
A 600g packet of biscuits costs $2.40. Calculate How much would a 200g packet cost?
sp2606 [1]

Answer:

$0.8

Step-by-step explanation:

There is a relation of direct proportion between quantity of biscuit and its cost.

Let 200 g biscuit will cost $x.

Therefore,

\frac{600}{2.40}  =  \frac{200}{x} \\  \\ x =  \frac{2.40 \times 200}{600}   \\  \\ x = 0.4 \times 2 \\  \\ x = 0.8

3 0
3 years ago
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