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miv72 [106K]
3 years ago
14

I need to find the derivative. I’m not very good at this so I am answer asap would be sosososo helpful

Mathematics
1 answer:
Bingel [31]3 years ago
5 0

Answer:

See below.

Step-by-step explanation:

First, notice that this is a composition of functions. For instance, let's let f(x)=\sqrt{x} and g(x)=7x^2+4x+1. Then, the given equation is essentially f(g(x))=\sqrt{7x^2+4x+1}. Thus, we can use the chain rule.

Recall the chain rule: \frac{d}{dx}[f(g(x))]=f'(g(x))\cdot g'(x). So, let's find the derivative of each function:

f(x)=\sqrt{x}=x^\frac{1}{2}

We can use the Power Rule here:

f'(x)=(x^\frac{1}{2})'=\frac{1}{2}(x^{-\frac{1}{2}  })=\frac{1}{2\sqrt{x}}

Now:

g(x)=7x^2+4x+1

Again, use the Power Rule and Sum Rule

g'(x)=(7x^2+4x+1)'=(7x^2)'+(4x)'+(1)'=14x+4

Now, we can put them together:

y'=f'(g(x))\cdot g'(x)=\frac{1}{2\sqrt{g(x)}} \cdot (14x+4)

y'=\frac{14x+4}{2\sqrt{7x^2+4x+1} } =\frac{7x+2}{\sqrt{7x^2+4x+1} }

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Answer:

Mean = (2.2 + 2.4 + 2.5 + 2.5 + 2.6 + 2.7)/6 = 2.48

Standard deviation = √(summation(x - mean)²/n

n = 6

Summation(x - mean)² = (2.2 - 2.48)^2 + (2.4 - 2.48)^2 + (2.5 - 2.48)^2 + (2.5 - 2.48)^2 + (2.6 - 2.48)^2 + (2.7 - 2.48)^2 = 0.1484

Standard deviation = √(0.1484/6

s = 0.16

Standard error = s/√n = 0.16/√6 = 0.065

Part B

Confidence interval is written as sample mean ± margin of error

Margin of error = z × s/√n

Since sample size is small and population standard deviation is unknown, z for 98% confidence level would be the t score from the student t distribution table. Degree of freedom = n - 1 = 6 - 1 = 5

Therefore, z = 3.365

Margin of error = 3.365 × 0.16/√6 = 0.22

Confidence interval is 2.48 ± 0.22

Part C

We would set up the hypothesis test. This is a test of a single population mean since we are dealing with mean

For the null hypothesis,

H0: µ = 2.3

For the alternative hypothesis,

H1: µ > 2.3

This is a right tailed test

Since the number of samples is small and no population standard deviation is given, the distribution is a student's t.

Since n = 6

Degrees of freedom, df = n - 1 = 6 - 1 = 5

t = (x - µ)/(s/√n)

Where

x = sample mean = 2.48

µ = population mean = 2.3

s = samples standard deviation = 0.16

t = (2.48 - 2.3)/(0.16/√6) = 2.76

We would determine the p value using the t test calculator. It becomes

p = 0.02

Assuming significance level, alpha = 0.05.

Since alpha, 0.05 > than the p value, 0.02, then we would reject the null hypothesis. Therefore, At a 5% level of significance, the sample data showed significant evidence that the mean absolute refractory period for all mice when subjected to the same treatment increased.

Step-by-step explanation:

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Set is defined as the arrangement of elements. They can be represented using the venn diagram.

Given the following sets

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