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Mnenie [13.5K]
4 years ago
11

Find and simplify the difference quotient for the given function

Mathematics
1 answer:
iren2701 [21]4 years ago
3 0
\dfrac{f(x+h)+f(x)}{h}

Replace f(x) and f(x+h) by using the equation given for the problem.

=\dfrac{(2(x+h)^2-(x+h)-1)-(2x^2-x-1)}{h}

Open all parenthesis

= \dfrac{2x^2+4xh+2h^2-x-h-1-2x^2+x+1}{h}

Factor the h

= \dfrac{h(4x+2h-1)+2x^2-2x^2+1-1+x-x}{h}

Other terms not in the parenthesis cancel each other out

\dfrac{h(4x+2h-1)}{h}

Now, simplify

=4x+2h-1
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Data from the Centers for Disease Control and Prevention indicate that weights of American adults in 2005 had a mean of 167 poun
sesenic [268]

Answer:

92.65% probability that the total weight in a random sample of 47 American adults exceeds 7500 pounds in 2005.

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 the theorem can also be used, with mean n*\mu and standard deviation s = \sqrt{n}*\sigma

In this problem, we have that:

n = 47, \mu = 47*167 = 7849, s = \sqrt{47}*35 = 240

Use this information to estimate the probability that the total weight in a random sample of 47 American adults exceeds 7500 pounds in 2005.

This is 1 subtracted by the pvalue of Z when X = 7500. So

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

By the Central Limit Theorem

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

Z = \frac{7500 - 7849}{240}

Z = -1.45

Z = -1.45 has a pvalue of 0.0735

1 - 0.0735 = 0.9265

92.65% probability that the total weight in a random sample of 47 American adults exceeds 7500 pounds in 2005.

5 0
3 years ago
Kevin has $1427 in a savings account. His brother, Harley has saved an amount with a 4 that is 1/10 the value of the 4 in Kevin'
Paul [167]
I'm guessing 400$????
3 0
3 years ago
Five family members will bowl this month. The prices at Dans alley are $5 per game and $3 for shoe rental. The total budget is $
Svetllana [295]
115 less then or equal to 3(5) + 25 + 5x

the family can bowl a total of 15 games, or 3 games each
4 0
3 years ago
19x+rx=-37+w <br><br><br> solve for x
ipn [44]

Factor the left side:-

x (19 + r) = -37 + w

x = ( -37 + w) / (19 + r) Answer


3 0
3 years ago
Read 2 more answers
Someone please help. It is easy
Juli2301 [7.4K]
Alright. So we have 2 eggs at 2.49 each. That is $4.98. We have .25 pounds cheese priced at $13.40 per pound. 13.40 x .25 = $3.35. Then we have juice at $5.49. If we add those up we get: 4.98+3.35+5.49=$13.82. Then we get $0.75 in change, so if we add $13.82 and 0.75 we get: $14.57

The answer is $14.57.
Brainliest???
5 0
3 years ago
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