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3 years ago

What is the coefficient of the fourth term in the expression? 5 − 10x + 15y − 20xy

Mathematics

2 answers:

allsm [11]3 years ago

8 0

Answer:

Step-by-step explanation:

the coefficient of a term is the numeric value including the sign in front of the variable/ s

the fourth term in the expression is - 20xy

The coefficient of this term is - 20 → C

mario62 [17]3 years ago

6 0

The answer is C. -20

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Given f(x) = x + 5 and g(x) = 11x - 5, what is f(x) + g(x)?

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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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2 years ago

5.8×10^−10 ? <br><br> Solve please! (Scientific Notation)

worty [1.4K]

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2 years ago

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