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PolarNik [594]
3 years ago
7

A standard number die is tossed. Find P(4 or even).

Mathematics
1 answer:
zmey [24]3 years ago
4 0

Answer:

50%

Step-by-step explanation:

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Could anyone help me with this problem i'm stuck <br><br> h(17)= ?
nikdorinn [45]

Answer:

\sqrt{17t}

Step-by-step explanation:

Consider the right side of the table

It informs us of the values for h(t) for different values of t

The top one informs us what h(t) is when t = 17, that is

h(t) = \sqrt{17t}, when t = 17, thus

h(17) = \sqrt{17t}

4 0
3 years ago
Andrew has one book that is two and 378 inches thick and a second book that is 3.56 inches thick if he stacks the books about ho
Georgia [21]
2.378 + 3.56 = 5.938

two and 378 inches =  2.378 
8 0
3 years ago
Find the volume of the composite solid. Round your answer to the nearest hundredth.
Vera_Pavlovna [14]

Answer: see my work

Step-by-step explanation:

volume is lwh

volume is 3*3*10

volume is 9*10

volume is cubic cm

volume is 90

volume is 90 cubic cm

8 0
2 years ago
Read 2 more answers
Help!! I hate geometryyyy
miskamm [114]

Answer:

AFB, BDC, and FBD

Step-by-step explanation:


3 0
4 years ago
For each function below, determine whether or not the function is injective and whether or not the function is surjective. (You
Harlamova29_29 [7]

Answer: First, injective means that for every y, there is only one x such f(x) = y, and surjective means that if f is a function that goes from the set {x} to the set {y}, for every element y in the codomain there is at least one element x in the domain such f(x) = y

(a) f:R----> + R given by f(x) = x2 (i guess you written x^{2})

The function is not injective, because f(-2) = f(2), and is surjective, because the codomain is +R, and x^{2} is always a positive real number.

(b) f:N----> + N given by f(n) = n2  (i guess you written n^{2})

As the naturals have no negative numbers, in this case the function is injective, but isn't surjective, because there is no number that when squared is equal to 5, for example.

(c) f: Zx Z → Z given by f(n, k) = n +k:

here the domain is of the form (n,k) and the function is n +k, so the numbers (z,w) and (w,z) return the same value when evaluated in f, then f is not injective. And is easy to see that f is surjective, because in the sum you can reach al the integers on the codomain.

3 0
3 years ago
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