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taurus [48]
3 years ago
15

Please answer 25 points!

Mathematics
2 answers:
erastovalidia [21]3 years ago
6 0

Answer:

-3

Step-by-step explanation:

that's just the correct answer, I know.

Alekssandra [29.7K]3 years ago
4 0

Answer:

-3

Step-by-step explanation:

You might be interested in
Can someone explain how to do this and add the formula please
lisabon 2012 [21]

Answer:

5 is 35 and 6 is 50

Step-by-step explanation:

To solve 5, you need to know how many degrees is the line, which is 180 because lines are 180 degrees. You then subtract 110 from 180 which equals to 70. What you have left now is 2k. A number times 2 will equal to 70, so k=35.

For 6, if you ever see two lines intersecting each other, it means that it is 360 degrees. We already have one angle, which is 120. The other angle is d+70 degrees, and any angle that is reflecting the other angle will have the same value. So, d equals 50

(sorry if you still can't understand it I'm not good at English

6 0
2 years ago
Gilbert and Brian drove 604 miles in 11.6 hours. Gilbert drove the first part of the trip and averaged 45 miles per hour. Brian
AVprozaik [17]

The length of hours Gilbert drove is 7.5 hours.

The length of hours Brian drove is 4.1 hours.

<u>Step-by-step explanation</u>:

  • Gilbert averaged = 45 miles per hour
  • Brain averaged = 65 miles per hour
  • Total hours = 11.6 hours
  • Total distance = 604 miles

Let 'x' be the hours Gilbert drove.

Let 'y' be the hours Brian drove.

x+y = 11.6  -----(1)

45x+65y = 604 -----(2)

Multiply (1) by 45 and subtract (2) from (1)

 45x+45y = 522

-(<u>45x+65y = 604</u>)

 <u>      -20y = -82 </u>

y = 82/20

y = 4.1 hours

The length of hours Brian drove = 4.1 hours

Substitute y=4.1 in (1)

x+4.1 = 11.6

x = 7.5 hours

The length of hours Gilbert drove = 7.5 hours

3 0
3 years ago
A student solved this problem and said the answer is about 7 quarts. Eddie has paint left over from a project. He has 2 fraction
Ilia_Sergeevich [38]
The answer would be C.
6 0
3 years ago
B Christian installed an electric pump to pump water from a borehole into a 30 000 litre cement dam. If the water is pumped at a
skelet666 [1.2K]

Given the amount of water needed to fill the cement dam and the water pump rate, the time needed to fill the dam is 400 minutes or 6hours 40minutes.

Option B)6h 40min is the correct answer.

<h3>How long does it take to fill the dam?</h3>

Given that;

  • Amount of water needed to fill the dam A = 30000 litres
  • Pump rate r = 75 litres per minute
  • Time needed to fill the dam T = ?

To determine how long it take to fill the dam, we say;

Time need = Amount of water needed ÷ Pump rate

T = A ÷ r

T = 30000 litres ÷ 75 litres/minute

T = 400 minutes

Note that; 60min = 1hrs

Hence,

T = 6hours 40minutes

Given the amount of water needed to fill the cement dam and the water pump rate, the time needed to fill the dam is 400 minutes or 6hours 40minutes.

Option B)6h 40min is the correct answer.

Learn more word problems here: brainly.com/question/2610134

#SPJ1

8 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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