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natima [27]
3 years ago
6

Multiple choice PLEASE HELP

Mathematics
1 answer:
bezimeni [28]3 years ago
3 0

Answer:

I think it is B

Step-by-step explanation:

Sorry  if it is wrong

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D
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The mean June midday temperature in Desertville is 36°C and the standard deviation is 3°C.Assuming this data is normally distrib
AleksandrR [38]

Answer:

The value is  E(X) = 4 \ days

Step-by-step explanation:

From the question we are told that

   The mean is  \mu  =  36^oC

    The standard deviation \sigma =  3^oC

Generally the probability that in June , the midday  temperature is between  

39°C and 42°C is mathematically represented as

      P(39 <  X <  42) = P(\frac{39 - 36}{3}  < \frac{X - \mu }{\sigma} < (\frac{42 - 36}{3} )

\frac{X -\mu}{\sigma }  =  Z (The  \ standardized \  value\  of  \ X )

      P(39 <  X <  42) = P(1  < Z

=>   P(39 <  X <  42) = P(Z   < 2) - P( Z

From the z table  the area under the normal curve to the left corresponding to  1 and  2  is

     P(Z   < 2)  = 0.97725

and

    P(Z   < 1)  = 0.84134

    P(39 <  X <  42) = 0.97725  - 0.84134

=>  P(39 <  X <  42) = 0.13591

Generally number of days in June would you expect the midday temperature to be between 39°C and 42°C

      E(X) =  n  *  P(39 <  X 42 )

Here n is the number of days in June which is  n = 30

       E(X) =  30  *   0.13591

=>    E(X) = 4 \ days

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3 years ago
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Answer:

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Step-by-step explanation:

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ra1l [238]
The answer to your question is C
8 0
3 years ago
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Hi, sorry to bother you, but can you answer this question about probability?
liberstina [14]

Answer:

sorry I don't get the question

Step-by-step explanation:

4 0
3 years ago
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