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IgorLugansk [536]
3 years ago
8

There is a spinner with 14 equal areas, numbered 1 through 14. If the spinner is spun one time, what is the probability that the

result is a multiple of 5 or a multiple of 2?
Mathematics
1 answer:
nadezda [96]3 years ago
7 0
Multiples of 5 are A{5,10}
multiples of 2 are B{2,4,6,8,10,12,14}

Multiples of 2 OR 5 are {2,4,5,6,8,10,12,14} for 8 possible outcomes.

Therefore the probability of spinning a multiple of 2 or 5
P(2x or 5x)=8/14 = 4/7
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I think the best answer is A.
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horrorfan [7]

Complete Question

The complete question is shown on the first uploaded image

Answer:

a

 g(h(x)) = \sqrt[4]{x^2 + x +5} + 3

b

 h(g(x))  = \sqrt{x}  + 7\sqrt[4]{x} + 17

c

 h(h(x)) =  [x^2 + x + 5 ]^2 + x^2 + x + 10

Step-by-step explanation:

From the question we are told that

    h(x) =  x^2  + x  + 5

and  

    g(x) = \sqrt[4]{x} + 3

Considering first question

Now we are told  g(h(x))

i.e

        g(h(x)) =  [x^2 + x + 5 ]^{\frac{1}{4} } + 3

=>     g(h(x)) = \sqrt[4]{x^2 + x +5} + 3

Considering second  question

   Now we are told  h(g(x))

i.e

    h(g(x)) =  [x^{\frac{1}{4} } + 3]^2 +  x^{\frac{1}{4} } + 3 + 5

=> h(g(x)) =  x^{\frac{1}{2} } + 6x^{\frac{1}{4} } + 9+ x^{\frac{1}{4}}  + 8

=>  h(g(x))  = x^{\frac{1}{2}} + 7x^{\frac{1}{4}} + 17

=>  h(g(x))  = \sqrt{x}  + 7\sqrt[4]{x} + 17

Considering third question

         h(h(x))= [x^2 + x + 5]^2 + [x^2 + x + 5 ] +  5

=>       h(h(x)) =  [x^2 + x + 5 ]^2 + x^2 + x + 10

8 0
4 years ago
Plsss help me Find The Area Of This Shape. Show Your Work.​
ra1l [238]

Answer:

28

Step-by-step explanation:

Area (trapezoid) = 1/2 x (6 + 8) x 4

= 2 x 14

= 28

4 0
2 years ago
1. *
slavikrds [6]

Answer:

I am not so clear about this question

3 0
3 years ago
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