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Natalka [10]
4 years ago
10

the price to send a large envelope first class mail is 88 cents for the first ounce and 17 cents for each additional ounce.How m

uch did a large envelope weigh that coast $2.07 to send?​
Mathematics
1 answer:
GarryVolchara [31]4 years ago
4 0

Answer:

subtract 0.88 from 2.07 and then from that divide the answer by 0.17

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Sodas in a can are supposed to contain an average 12 oz. This particular brand has a standard deviation of 0.1 oz, with an avera
hoa [83]

Answer:

The probability is  P(X <  12) =  0.99286

Step-by-step explanation:

From the question we are told that

        The population mean is \mu  =  12 \ oz

         The  standard deviation is  \sigma =  0.1 \ oz

          The sample mean is  \= x =  12.1 \ oz

          The sample size is  n = 6 packs

   

The standard error of the mean is mathematically represented as

              \sigma_{\= x } =  \frac{\sigma}{\sqrt{n} }

substituting values

            \sigma_{\= x } =  \frac{0.1}{\sqrt{6} }

            \sigma_{\= x } =  0.0408

Given that the can’s contents follow a Normal distribution then then  the probability that the mean contents of a six-pack are less than 12 oz is mathematically represented as

         P(X <  12) =  P ( \frac{X - \mu }{ \sigma_{\= x }}  < \frac{\= x - \mu }{ \sigma_{\= x }}  )

Generally  \frac{X  - \mu }{ \sigma_{ \= x }}   =  Z (The  \ standardized \  value \ of  \ X )

So

         P(X <  12) =  P ( Z < \frac{\= x - \mu }{ \sigma_{\= x }}  )

substituting values

       P(X <  12) =  P ( Z < \frac{12.2 -12 }{0.0408}  )

      P(X <  12) =  P ( Z < 2.45   )

From the normal distribution table the value of P ( Z < 2.45   ) is  

           P (Z < 2.45)0.99286

=>   P(X <  12) =  0.99286

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4 years ago
HELP PLEASE I HAVE 5 MINUTES PLEASE HELP! (IGNORE THE HIGHLIGHTED ANSWER I PRESSED IT ON ACCIDENT)
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4 0
3 years ago
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50 pointer guys!!! Penny buys 3 bouquets of flowers for $3 each and 4 bouquets for $2 each. Which expression gives the total cos
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Answer:

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3 years ago
What is the approximate volume of a cone with a height of 11 ft and radius of 3 ft?
nadya68 [22]
Substitute the values of the radius <span>(<span><span>r=3</span>),</span></span><span> the </span>height <span>(<span><span>h=11</span>),</span></span><span> and an approximation of </span>Pi <span>(<span>3.14)</span></span><span> into the </span>formula<span> to find the </span>volume<span> of the </span>cone<span>.
</span><span>V≈+<span>1/3</span>⋅3.14⋅<span>3^2</span>⋅11

</span><span>Simplify.
</span>Write 3.14<span> as a </span>fraction<span> with </span>denominator <span>1.</span><span> 
</span><span>V≈+<span>1/3</span>⋅<span>3.14/1</span></span> <span>

</span>Combine <span>1/3</span><span> and </span><span>3.14/1</span><span> to get </span><span><span>3.14/3</span>.</span><span> 
</span><span>V≈+<span>3.14/3</span></span><span> 
</span>
Replace back in to larger expression<span>. 
</span><span>V≈+<span>3.14/3</span>⋅<span>3^2</span>⋅11</span><span> 
</span>
Divide 3.14<span> by </span>3<span> to get </span><span>1.04666667.
</span><span>V≈+1.04666667⋅<span>3^2</span>⋅11

</span>Raise 3<span> to the </span>power<span> of </span>2<span> to get </span><span>9.
</span><span>V≈+1.04666667⋅9⋅11

</span>Multiply 1.04666667<span> by </span>9<span> to get </span><span>9.42.
</span><span>V≈+9.42⋅11

</span>Multiply 9.42<span> by </span>11<span> to get </span><span>103.62.
</span><span>V≈+103.62

</span>Remove trailing zeros<span> from </span><span>103.62.
</span><span>V≈+103.62</span><span>f<span>t^3

Volume:
</span></span>V≈+103.62ft^3
8 0
3 years ago
Read 2 more answers
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