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KonstantinChe [14]
2 years ago
8

If F (x) =8+11 X-3X², find a. What is F (5)? b. What is F (x + b)? c. What is F (-3)?

Mathematics
1 answer:
stepladder [879]2 years ago
5 0

Answer:

a) -12

b) 8+11x+11b-3x^2-3b^2-6xb

c)-52

Step-by-step explanation:

in a) we have to substitute the value of x with 5

in b) we substitute the value of x with (x+b) and then simplify

in c) just have to substitute x with -3

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24.5 x 0.25 = 6.125 acres that will be used for the new park.


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Write an equation that has ONE SOLUTION to the equation: y=5x+3
Aneli [31]

Answer:(0,3)

Step-by-step explanation:

8 0
3 years ago
A filling machine fills bottles of nail polish with amounts that are normallydistributed with mean 18 mL and standard deviation
natulia [17]

Answer:

a. 0.2033 = 20.33% probability that one of the randomly selected bottles is filled with less than 17.5mL of nail polish

b. 0.0019 = 0.19% probability that the average fill of the twelve bottles is more than 17.5mL

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 normal distributions can be solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the z-score 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 p-value, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem establishes 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.

Mean 18 mL and standard deviation 0.6 mL.

This means that \mu = 18, \sigma = 0.6

a. What is the probability that one of the randomly selected bottles is filled with less than 17.5mL of nail polish?

This is the p-value of Z when X = 17.5. So

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

Z = \frac{17.5 - 18}{0.6}

Z = -0.83

Z = -0.83 has a p-value of 0.2033

0.2033 = 20.33% probability that one of the randomly selected bottles is filled with less than 17.5mL of nail polish.

b. What is the probability that the average fill of the twelve bottles is more than 17.5mL?

Twelve bottles, so now n = 12, s = \frac{0.6}{\sqrt{12}}

The probability is:

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

By the Central Limit Theorem

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

Z = \frac{17.5 - 18}{\frac{0.6}{\sqrt{12}}}

Z = -2.89

Z = -2.89 has a p-value of 0.0019

0.0019 = 0.19% probability that the average fill of the twelve bottles is more than 17.5mL

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quester [9]
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(Assuming I guessed your question correctly.)
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