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Luda [366]
2 years ago
10

) Express the number 482.073 Standard Form?

Mathematics
2 answers:
vfiekz [6]2 years ago
7 0

Hey there!

482.073 is expressed in

Standard Form

The standard form of 482.073 is 482.073. (four hundred eighty-two point zero seven three)

I hope you find my answer helpful.

Good luck!

Have a great day!

\bf{StargazingWithJoy}

Inessa [10]2 years ago
3 0

Answer:

It is already in standard form

Step-by-step explanation:

Four hundred eighty-two and seventy-three thousandths. = 482.073

It is already in standard form

Hope this helps!

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Answer:

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Levart [38]
<h3>Answer:   12 square units</h3>

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2 years ago
Suppose a continuous probability distribution has an average of μ=35 and a standard deviation of σ=16. Draw 100 times at random
yulyashka [42]

Answer:

To use a Normal distribution to approximate the chance the sum total will be between 3000 and 4000 (inclusive), we use the area from a lower bound of 3000 to an upper bound of 4000 under a Normal curve with its center (average) at 3500 and a spread (standard deviation) of 160 . The estimated probability is 99.82%.

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, we have that the mean is \mu*n and the standard deviation is s = \sigma \sqrt{n}

In this problem, we have that:

\mu = 100*35 = 3500, \sigma = \sqrt{100}*16 = 160

This probability is the pvalue of Z when X = 4000 subtracted by the pvalue of Z when X = 3000.

X = 4000

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

By the Central Limit Theorem

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

Z = \frac{4000 - 3500}{160}

Z = 3.13

Z = 3.13 has a pvalue of 0.9991

X = 3000

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

Z = \frac{3000 - 3500}{160}

Z = -3.13

Z = -3.13 has a pvalue of 0.0009

0.9991 - 0.0009 = 0.9982

So the correct answer is:

To use a Normal distribution to approximate the chance the sum total will be between 3000 and 4000 (inclusive), we use the area from a lower bound of 3000 to an upper bound of 4000 under a Normal curve with its center (average) at 3500 and a spread (standard deviation) of 160 . The estimated probability is 99.82%.

5 0
3 years ago
Plz help, I'm taking an Advanced class (Algebra) Will give brainliest ✔✔✔✔✔✔✔✔✔
bija089 [108]

Part A:

The average rate of change refers to a function's slope. Thus, we are going to need to use the slope formula, which is:

m = \dfrac{y_2 - y_1}{x_2 - x_1}

  • (x_1, y_1) and (x_2, y_2) are points on the function

You can see that we are given the x-values for our interval, but we are not given the y-values, which means that we will need to find them ourselves. Remember that the y-values of functions refers to the outputs of the function, so to find the y-values simply use your given x-value in the function and observe the result:

h(0) = 3(5)^0 = 3 \cdot 1 = 3

h(1) = 3(5)^1 = 3 \cdot 5 = 15

h(2) = 3(5)^2 = 3 \cdot 25 = 75

h(3) = 3(5)^3 = 3 \cdot 125 = 375


Now, let's find the slopes for each of the sections of the function:

<u>Section A</u>

m = \dfrac{15 - 3}{1 - 0} = \boxed{12}

<u>Section B</u>

m = \dfrac{375 - 75}{3 - 2} = \boxed{300}


Part B:

In this case, we can find how many times greater the rate of change in Section B is by dividing the slopes together.

\dfrac{m_B}{m_A} = \dfrac{300}{12} = 25


It is 25 times greater. This is because 3(5)^x is an exponential growth function, which grows faster and faster as the x-values get higher and higher. This is unlike a linear function which grows or declines at a constant rate.

7 0
3 years ago
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