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Nonamiya [84]
3 years ago
12

What is the value of 0.1561 rounded to the nearest tenth

Mathematics
2 answers:
Ipatiy [6.2K]3 years ago
7 0

Answer:

0.2

Step-by-step explanation:

nika2105 [10]3 years ago
6 0

Answer:

0.2

Step-by-step explanation:

The place after the decimal is the tenths place. If the place next to it is 5 (the Hundredths place) you round up or down depending on the next number (the thousandths place) , in this case is 6.

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Graph the linear equations. Find three points that solve the equation -3x + 2y= 2
fredd [130]

Answer:

(0,1) , (2,4) , (4,7)

Explanation:

convert the equation into slope-intercept form (this would be y=3/2x+1)

now plug in random numbers for x and solve for y by multiplying them by 3/2 and adding 1.

7 0
3 years ago
A regular polygon with an interior angle sum of 9000°? If so, what is it?
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3 0
3 years ago
The cost of renting a chain saw is $3.90 per hour plus $6.50 for a cán
puteri [66]
35.75
explanation-
3.90 ph x 7.5 hours = 29. 25
plus the gas can 6.50
29.25+6.50= 35.75
5 0
3 years ago
A person jogs 1/2 miles in 1/12 hours. The person's speed is how many miles per hour?
ipn [44]

Answer:

6 miles per hour

Step-by-step explanation:

speed = distance/time

speed = 1/2÷1/12 which is the same as

1/2 x 12/1 = 6

6 miles per hour.

or ou could change the 1/2 to become 0.5

and the 1/12 to become 0.083333333

and divide 0.5 ÷ 0.083333333 = 6.000000024

rounded to one decima place become 6.0

6 miles per hour

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
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