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poizon [28]
3 years ago
9

Find the values of y = c(x) = for x = 0, 0.008, 0.027, 0.064, 0.125, 0.216, 0.343, 0.512, , 0.729, and 1.

Mathematics
1 answer:
kondaur [170]3 years ago
7 0

c(x) is a function, then we will solve the problem for two different functions.

1. If c(x)=\frac{x}{0.5}+x

Then

y=c(x)

y=\frac{x}{0.5}+x

Using each of the values for x we have.

When x=0 the result is:   y=\frac{0}{0.5}+0=0

When x=0.008 the result is: y=\frac{0.008}{0.5}+0.008=0.016+0.008=0.024

When x=0.027 the result is: y=\frac{0.027}{0.5}+0.027=0.054+0.027=0.081

When x=0.064 the result is: y=\frac{0.064}{0.5}+0.064=0.128+0.064=0.192

When x=0.125 the result is: y=\frac{0.125}{0.5}+0.125=0.25+0.125=0.375

When x=0.216 the result is: y=\frac{0.216}{0.5}+0.216=0.432+0.216=0.648

When x=0.343 the result is: y=\frac{0.343}{0.5}+0.343=0.686+0.343=1.029

When x=0.512 the result is: y=\frac{0.512}{0.5}+0.512=1.024+0.512=1.536

When x=0.729 the result is: y=\frac{0.729}{0.5}+0.729=1.458+0.729=2.187

When x=1 the result is: y=\frac{1}{0.5}+1=2+1=3

2. If c(x)=sin(x)+2x

Then

y=c(x)

y=sin(x)+2x

Using each of the values for x we have.

When x=0 the result is:   y=sin(0)+2(0)=0

When x=0.008 the result is: y=sin(0.008)+2(0.008)=0.000139+0.016=0.016139

When x=0.027 the result is: y=sin(0.027)+2(0.027)=0.000471+0.054=0.054471

When x=0.064 the result is: y=sin(0.064)+2(0.064)=0.001117+0.128=0.129117

When x=0.125 the result is: y=sin(0.125)+2(0.125)=0.002182+0.25=0.252182

When x=0.216 the result is: y=sin(0.216)+2(0.216)=0.003769+0.432=0.435769

When x=0.343 the result is: y=sin(0.343)+2(0.343)=0.005986+0.686=0.691986

When x=0.512 the result is: y=sin(0.512)+2(0.512)=0.008936+1.024=1.032936

When x=0.729 the result is: y=sin(0.729)+2(0.729)=0.012723+1.458=1.470723

When x=1 the result is: y=sin(1)+2(1)=0.017452+2=2.017452

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Step-by-step explanation:

For each dog, there are only two possible outcomes. Either they have completed emotional support training, or they have not. So we use the binomial probability distribution to solve this problem.

However, we are working with samples that are considerably big. So i am going to aproximate this binomial distribution to the normal.

Binomial probability distribution

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Can be approximated to a normal distribution, using the expected value and the standard deviation.

The expected value of the binomial distribution is:

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The standard deviation of the binomial distribution is:

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In a set with mean \mu and standard deviation \sigma, the zscore of a measure X is given by:

Z = \frac{X - \mu}{\sigma}

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When we are approximating a binomial distribution to a normal one, we have that \mu = E(X), \sigma = \sqrt{V(X)}.

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\mu = E(X) = np = 100*0.23 = 23

\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{100*0.23*0.77} = 4.21

What is the probability that under 30% of the dogs are emotional support trained?

30% of 100 is 0.3*100 = 30

So this is the pvalue of Z when X = 30.

Z = \frac{X - \mu}{\sigma}

Z = \frac{30 - 23}{4.1}

Z = 1.71

Z = 1.71 has a pvalue of 0.9564.

So there is a 95.64% probability that under 30% of the dogs are emotional support trained

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