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Feliz [49]
3 years ago
5

Wires manufactured for use in a computer system are specified to have resistances between 0.14 and 0.16 ohms. The actual measure

d resistances of the wires produced by company A have a normal probability distribution with mean 0.15 ohm and standard deviation 0.005 ohm. (Round your answers to four decimal places.) (a) What is the probability that a randomly selected wire from company A's production will meet the specifications
Mathematics
1 answer:
kodGreya [7K]3 years ago
5 0

Answer:

Hence the probability that a randomly selected wire from company A's production will meet the specifications is 0.95455.

Step-by-step explanation:

a)P(0.14 < x < 0.16 ) = P[(0.14 - 0.15)/ 0.005) < (x - \mu) /\sigma  < (0.16 - 0.15) / 0.005) ]

=P(0.14 < x < 0.16 ) = P[(0.14 - 0.15)/ 0.005)< ((x - 0.15) /0.005) < (0.16 - 0.15) / 0.005) ]

=P(z  

Using z table,  

= 0.9773 - 0.02275  

= 0.95455.

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What this doesnt make sense
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3 years ago
A particular fruit's weights are normally distributed, with a mean of 406 grams and a standard deviation of 27 grams.The heavies
Tom [10]

Given

mean of 406 grams and a standard deviation of 27 grams.

Find

The heaviest 14% of fruits weigh more than how many grams?

Explanation

given

mean = 406 gms

standard deviation = 27 gms

using standard normal table ,

\begin{gathered} P(Z>z)=14\% \\ 1-P(Zso , [tex]\begin{gathered} x=z\times\sigma+\mu \\ x=1.08\times27+406 \\ x=435.16 \end{gathered}

Final Answer

Therefore , The heaviest 14% of fruits weigh more than 435.16 gms

8 0
1 year ago
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The value of x in terms of p is 15/p.

The value of x when p is -5 is -3.

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6 0
3 years ago
5^(-x)+7=2x+4 This was on plato
Setler79 [48]

Answer:

Below

I hope its not too complicated

x=\frac{\text{W}_0\left(\frac{\ln \left(5\right)}{2e^{\frac{3\ln \left(5\right)}{2}}}\right)}{\ln \left(5\right)}+\frac{3}{2}

Step-by-step explanation:

5^{\left(-x\right)}+7=2x+4\\\\\mathrm{Prepare}\:5^{\left(-x\right)}+7=2x+4\:\mathrm{for\:Lambert\:form}:\quad 1=\left(2x-3\right)e^{\ln \left(5\right)x}\\\\\mathrm{Rewrite\:the\:equation\:with\:}\\\left(x-\frac{3}{2}\right)\ln \left(5\right)=u\mathrm{\:and\:}x=\frac{u}{\ln \left(5\right)}+\frac{3}{2}\\\\1=\left(2\left(\frac{u}{\ln \left(5\right)}+\frac{3}{2}\right)-3\right)e^{\ln \left(5\right)\left(\frac{u}{\ln \left(5\right)}+\frac{3}{2}\right)}

Simplify\\\\\mathrm{Rewrite}\:1=\frac{2e^{u+\frac{3}{2}\ln \left(5\right)}u}{\ln \left(5\right)}\:\\\\\mathrm{in\:Lambert\:form}:\quad \frac{e^{\frac{2u+3\ln \left(5\right)}{2}}u}{e^{\frac{3\ln \left(5\right)}{2}}}=\frac{\ln \left(5\right)}{2e^{\frac{3\ln \left(5\right)}{2}}}

\mathrm{Solve\:}\:\frac{e^{\frac{2u+3\ln \left(5\right)}{2}}u}{e^{\frac{3\ln \left(5\right)}{2}}}=\frac{\ln \left(5\right)}{2e^{\frac{3\ln \left(5\right)}{2}}}:\quad u=\text{W}_0\left(\frac{\ln \left(5\right)}{2e^{\frac{3\ln \left(5\right)}{2}}}\right)\\\\\mathrm{Substitute\:back}\:u=\left(x-\frac{3}{2}\right)\ln \left(5\right),\:\mathrm{solve\:for}\:x

\mathrm{Solve\:}\:\left(x-\frac{3}{2}\right)\ln \left(5\right)=\text{W}_0\left(\frac{\ln \left(5\right)}{2e^{\frac{3\ln \left(5\right)}{2}}}\right):\\\quad x=\frac{\text{W}_0\left(\frac{\ln \left(5\right)}{2e^{\frac{3\ln \left(5\right)}{2}}}\right)}{\ln \left(5\right)}+\frac{3}{2}

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3 years ago
Simplify 8/10 to find the sum/product solve the problem
slamgirl [31]
8/10 simplified is 4/5
5 0
3 years ago
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