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Bad White [126]
4 years ago
7

juan plays on a school baseball team in the last 9 games juan was at bat 28 times and got 6 hits what is the experimental probab

ility that juan will get a hit during his next time at bat express your answer as a fraction in the simplest form???
Mathematics
2 answers:
almond37 [142]4 years ago
4 0

Answer: \dfrac{3}{14}

Step-by-step explanation:

Given : The number of times Juan was at bat= 28

The number of times he got hits = 6

Now, the experimental probability hat Juan will get a hit during his next time at bat :-

\dfrac{\text{Favorable outcomes}}{\text{Total outcomes}}\\\\=\dfrac{6}{28}=\dfrac{3}{14}

Hence, the experimental probability that juan will get a hit during his next time at bat : \dfrac{3}{14}

vredina [299]4 years ago
3 0
<span> 3/14 this will be your answer!!!...</span>
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Consider the first order differential equation
AURORKA [14]

Answer:

For y(-7) =6.4

        The  largest interval  is  between  

                                       -\infty \to -5

For   y(-2.5) = -0.5.

          The  largest interval  is  between  

                                        -5 \to 5

For  y(0) = 0

          The  largest interval  is  between  

                                        -5 \to 5

For y(4.5) = -2.1.

            The  largest interval  is  between  

                                        -5 \to 5

For y(14)= 1.7.

                          The  largest interval  is  between  

                                        9 \to \infty        

   

Step-by-step explanation:

From m the question we are told that

     The first order differential equation is   \frac{y' -  t}{ t^2 -25} =  \frac{e^t}{t-9}

Now the first step is to obtain the domain of the differential equation

Now to do that let consider the denominators

  Now generally

                           t^2 - 25 \ne  0                                           side calculation

                   =>     t\ne \pm5                                                      t^2 - 25 =  0

                                                                                          t = \pm 5

Also              t-9\ne 0                                                     t -9 = 0

        =>          t\ne 9                                                            t= 9    

This means that this  first order differential equation is discontinuous at

       t =  -5 ,  \   \  t =  5 \ \  t =  9

  This  \  is \  illustrated \  below    \  \\      ------------------\\\. \ \ \  | \ \ \   \ \ \   \ \ \ \  \ \ \ \ \ | \ \ \ \ \ \ \ \ \ \ \ \ \ |\\. \  -5 \ \ \ \ \  \ \ \ \ \ \ \  5 \ \ \ \ \ \ \ \ \ \ \ \ \   9

So

For y(-7) =6.4

        The  largest interval  is  between  

                                       -\infty \to -5

For   y(-2.5) = -0.5.

          The  largest interval  is  between  

                                        -5 \to 5

For  y(0) = 0

          The  largest interval  is  between  

                                        -5 \to 5

For y(4.5) = -2.1.

            The  largest interval  is  between  

                                        -5 \to 5

For y(14)= 1.7.

                          The  largest interval  is  between  

                                        9 \to \infty        

   

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Answer:

The expression is 9(n + 8) - 6, the value is 111 ⇒ answer A

Step-by-step explanation:

* Lets explain how to solve the problem

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" six less than nine times the sum of a number and eight"

- When we solve we will start by the last part

- The last part is the sum of a number and 8

- Assume that the number is n

∵ The sum is the answer of adding

∴ The sum of a number and eight is ⇒ n + 8

- Nine times the sum

∵ The sum is n + 8

∵ Nine times means multiply the sum by 9

∴ Nine times the sum is ⇒ 9(n + 8)

- Six less than nine times the sum

∵ Nine times the sum is 9(n + 8)

∵ Six less than means subtract 6 from nine times the sum

∴ Six less than nine times the sum is ⇒ 9(n + 9) - 6

*<em> </em><em>The expression is 9(n + 8) - 6</em>

- Lets evaluate the expression when n = 5

∵ 9(n + 8) - 6

∵ n = 5

∴ The value = 9(5 + 8) - 6

∴ The value = 9(13) - 6

∴ The value = 117 - 6

∴ The value = 111

* <em>When n = 5 the value of the expression is 111</em>

8 0
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(−∞,∞) Sorry I didn’t quite understand
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