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Tom [10]
3 years ago
11

Kyle plays softball. His goal is to have a batting average of .400 or greater by the end of the season. The ratio that relates t

he number of hits to the number of chances, called at-bats, is used to determine batting average in baseball and softball.Batting averagenumber of hits number of at bats So far this season, Kyle has 25 hits. He has been at bat 65 times. He expects to have 10 more at-bats this season
Mathematics
1 answer:
gregori [183]3 years ago
4 0

Answer:

The  number of  hits is  T  \ge 30 \ hits

Step-by-step explanation:

From the question we are told that

   The ratio of the number of hits to the number of chances is equivalent to the batting average

   The  number of chances is j =  65 + 10 =  75

    The  number of hit so far is  n =  25

Generally the batting average can be represented mathematically as

      A =  \frac{n + k}{ j}

Here k is the number of additional hits Kyle need to get in order to obtain the batting average of  0.40 or more

 So  

        \frac{25 + k}{ 75}  \ge 0.40

=>     k \ge  5 \  hits

Hence the number of hits for Kyle  to  get a batting average of 0.400  or greater by the end of the season is  

     T  \ge k +  n

=>   T  \ge 5 +  30

=>   T  \ge 30 \ hits

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A library subscribes to two different weekly news magazines, each of which is supposed to arrive in Wednesday's mail. In actuali
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Answer:

y                 0             1              2          3

P(Y=y)   0.0676   0.3549   0.3875   0.19

Step-by-step explanation:

P(Wed) = 0.26

P(Thurs) = 0.39

P(Fri) = 0.25

P(Sat) = 0.10

Y = No. of days beyond Wednesday it takes for both magazines to arrive i.e. 0,1,2,3

Y=0 means the magazines will arrive on Wednesday

Y=1 means the magazines will arrive till Thursday

Y=2 means the magazines will arrive till Friday

Y=3 means the magazines will arrive till Saturday

The possible combinations for Y are

Y(W,W) Y(W,T) Y(W,F) Y(W,S)

Y(T,W) Y(T,T) Y(T,F) Y(T,S)

Y(F,W) Y(F,T) Y(F,F) Y(F,S)

Y(S,W) Y(S,T) Y(S,F) Y(S,S)

So, we can classify these possible outcomes as Y=0,1,2,3.

Y(0) = Y(W,W) (both magazines take 0 days to arrive beyond Wednesday)

Y(1) = Y(W,T), Y(T,T), Y(T,W) (both magazines take 1 day to arrive beyond Wednesday)

Y(2) = Y(W,F), Y(T,F), Y(F,F) Y(F,W) Y(F,T) (both magazines arrive till Friday)

Y(3) = Y(W,S), Y(T,S), Y(F,S), Y(S,W), Y(S,T), Y(S,F), Y(S,S) (both magazines arrive till Saturday)

To calculate the PMF, we need to calculate the probability for each of the points in Y(0,1,2,3).

Y(0) = Y(W,W)

       = 0.26 x 0.26

Y(0) = 0.0676

Y(1) = Y(W,T) + Y(T,T) + Y(T,W)

      = (0.26 x 0.39) + (0.39 x 0.39) + (0.39 x 0.26)

      = 0.1014 + 0.1521 + 0.1014

Y(1) = 0.3549

Y(2) = Y(W,F) + Y(T,F) + Y(F,F) + Y(F,W) + Y(F,T)

  =(0.26 x 0.25) + (0.39 x 0.25) + (0.25 x 0.25) + (0.25 x 0.26) + (0.25 x 0.39)

  = 0.065 + 0.0975 + 0.0625 + 0.065 + 0.0975

Y(2) = 0.3875

Y(3) = Y(W,S) + Y(T,S) + Y(F,S) + Y(S,W) + Y(S,T) + Y(S,F) + Y(S,S)

      = (0.26 x 0.10) + (0.39 x 0.10) + (0.25 x 0.10) + (0.10 x 0.26) + (0.10 x 0.39) + (0.10 x 0.25) + (0.10 x 0.10)

       = 0.026 + 0.039 + 0.025 + 0.026 + 0.039 + 0.025 + 0.010

Y(3) = 0.19

y                 0             1              2          3

P(Y=y)   0.0676   0.3549   0.3875   0.19

The PMF plot is attached as a photo here.

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