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miss Akunina [59]
3 years ago
15

Suppose f(7) = 5​, f'(7) = 8​, g(7) = 3​, and g'(7) = 4. Find h(7) and h'(7)​, where h(x ) = 4f (x) + 3g(x).

Mathematics
1 answer:
dusya [7]3 years ago
7 0

Answer:

h(7) = 29

h'(7) = 44

Step-by-step explanation:

If h(x) =4f(x)+3g(x), to find h(7) we can substitute the values of f(7) and g(7) and we get:

h(7)=4f(7)+3g(7)\\h(7)=4(5)+3(3)\\h(7)=20+9\\h(7)=29

To find the derivative, we know that the derivative of a sum of functions equals the sum of the derivatives of those functions.

This would mean that h'(x)=4f'(x)+3g'(x), we can substitute the values for f'(7) and g'(7)

h'(7)=4f'(7)+3g'(7)\\h'(7)=4(8)+3(4)\\h'(7)=32+12\\h'(7)=44

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A distribution of values is normal with a mean of 220 and a standard deviation of 13. From this distribution, you are drawing sa
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The interval containing the middle-most 48% of sample means is between 218.59 to 221.41.

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.

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

In this problem, we have that:

\mu = 220, \sigma = 13, n = 35, s = \frac{13}{\sqrt{35}} = 2.1974

Find the interval containing the middle-most 48% of sample means:

50 - 48/2 = 26th percentile to 50 + 48/2 = 74th percentile. So

74th percentile

value of X when Z has a pvalue of 0.74. So X when Z = 0.643.

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

By the Central Limit Theorem

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

0.643 = \frac{X - 220}{2.1974}

X - 220 = 0.643*2.1974

X = 221.41

26th percentile

Value of X when Z has a pvalue of 0.26. So X when Z = -0.643

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

-0.643 = \frac{X - 220}{2.1974}

X - 220 = -0.643*2.1974

X = 218.59

The interval containing the middle-most 48% of sample means is between 218.59 to 221.41.

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3 years ago
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