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DaniilM [7]
3 years ago
15

For f(x)=x^5-x^3 x what is the equation of the tangent line at x=3?

Mathematics
1 answer:
Zigmanuir [339]3 years ago
7 0
F(x) = x^5 - x^3 + x
f'(x) = 5x^4 - 3x^2 + 1
f(3) = (3)^5 - (3)^3 + 3 = 243 - 27 + 3 = 219
f'(3) = 5(3)^4 - 3(3)^2 + 1 = 5(81) - 3(9) + 1 = 405 - 27 + 1 = 379

Let the required equation of the tangent line be
y = mx + c; where y = 219, m = 379, x = 3
219 = 379(3) + c = 1,137
c = 219 - 1,137 = -918

Therefore, required equation is
y = 379x - 918
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An internet start-up predicts its revenue by a straight-line method with a slope of $7000 per month. In its 6 th month it had re
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A charity receives 2025 contributions. Contributions are assumed to be mutually independent and identically distributed with mea
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Answer:

The 90th percentile for the distribution of the total contributions is $6,342,525.

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal probability distribution

Problems of normally distributed samples are solved using the z-score formula.

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

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

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean \mu and standard deviation \sigma, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean \mu and standard deviation s = \frac{\sigma}{\sqrt{n}}.

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

For sums of size n, the mean is \mu*n and the standard deviation is s = \sqrt{n}*\sigma

In this question:

n = 2025, \mu = 3125*2025 = 6328125, \sigma = \sqrt{2025}*250 = 11250

The 90th percentile for the distribution of the total contributions

This is X when Z has a pvalue of 0.9. So it is X when Z = 1.28. Then

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

By the Central Limit Theorem

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

1.28 = \frac{X - 6328125}{11250}

X - 6328125 = 1.28*11250

X = 6342525

The 90th percentile for the distribution of the total contributions is $6,342,525.

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