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laiz [17]
3 years ago
10

How many hours are there in 1 month if there are 31 days in a month. Will mark brainliest.

Mathematics
2 answers:
Alona [7]3 years ago
4 0

Answer:

744

Step-by-step explanation:

<h3>in one day there is 24 hours if you multiply 24*31 it will be 744hours in total .</h3><h3 /><h3>24*31=744</h3><h3 /><h3>so you multiply the amount of time with the amount of days :)</h3><h3 /><h3>hope you get it !</h3><h3 /><h3>please mark me brilliant if im right</h3><h3 /><h3>thank you :3</h3>
Karo-lina-s [1.5K]3 years ago
3 0

Answer: 744 hours in a month

Step-by-step explanation:

31x24=744

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You are preparing a community garden with your youth group to have enough room for the spots for all the people who already sign
Annette [7]

Answer:

<u>92 feet.</u>

Step-by-step explanation:

Given:

The length of the garden needs to be 22 feet longer than the width of the garden.

The perimeter of the fence is a total of 324 feet

Question asked:

What will the length be of garden ?

Solution:

As length of the garden needs to be 22 feet longer than the width,

<u>Let width</u> = x

Then,<u> length </u>will be = x+22

As here perimeter of the fence is given, we can find length and width of garden by using:

Perimeter of rectangle = 2(length+width)

324=2(x+22+x)\\324=2(2x+22)\\324=4x+44\\

Subtracting both sides by 44

324-44=4x+44-44\\280=4x\\

Dividing both sides by 4

70=x

Width = x = 70 feet

Then, length will be = x+22 = 70 + 22 = 92 feet

Therefore, the length of the garden will be 92 feet.

3 0
3 years ago
Answer correctly for the brainliest answer
Dahasolnce [82]

Similar triangles so AC:BC=AE:DE

15+6=21 = AE

15:12.5=21:?

to get from 15 to 12.5 you divide by 15 and times by 12.5 so do this the to the other side

21/15=1.4

1.4 x 12.5=17.5

So DE=17.5

4 0
3 years ago
A person is on the outer edge of a carousel with a radius of 20 feet that is rotating counterclockwise around a point that is ce
Scilla [17]
Answer: Assuming the riders starts at the position (20, 0) on the x-axis, the exact position of the rider will be (20cos75, 20sin75) or about (5.18, 19.32).

The angle for 5pi/12 radians is 75 degrees. Therefore, to find the position we can use the sine and cosine of 75 to find the x and y value of the coordinate.

For the y-value, we can write and solve:
sin75 = x/20

For the x-value, we can write and solve:
cos75 = x/20
3 0
3 years ago
100 points the right answer please the set designer for a play painted some background scenery on a large piece of plywood. He u
tiny-mole [99]

Answer:the answer is 14

Step-by-step explanation:

6 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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