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Dmitry_Shevchenko [17]
3 years ago
14

En la granja hay 509 puercos anota el número antecesor y el número sucesor de la cantidad.​

Mathematics
1 answer:
nasty-shy [4]3 years ago
5 0

Answer: que

Step-by-step explanation:

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The legs of a right triangle are 18 centimeters and 80 centimeters long. What is the length of the hypotenuse
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a^2 + b^2 = c^2\\\\18^2 + 80^2 = c^2\\\\6724 = c^2\\\\c = \sqrt{6724}\\\\ c = 82

The hypotenuse is 82 centimeters long.
4 0
3 years ago
Read 2 more answers
Let X represent the amount of gasoline (gallons) purchased by a randomly selected customer at a gas station. Suppose that the me
Alexus [3.1K]

Answer:

a) 18.94% probability that the sample mean amount purchased is at least 12 gallons

b) 81.06% probability that the total amount of gasoline purchased is at most 600 gallons.

c) The approximate value of the 95th percentile for the total amount purchased by 50 randomly selected customers is 621.5 gallons.

Step-by-step explanation:

To solve this question, we use 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 can apply the theorem, with mean \mu and standard deviation s = \sqrt{n}*\sigma

In this problem, we have that:

\mu = 11.5, \sigma = 4

a. In a sample of 50 randomly selected customers, what is the approximate probability that the sample mean amount purchased is at least 12 gallons?

Here we have n = 50, s = \frac{4}{\sqrt{50}} = 0.5657

This probability is 1 subtracted by the pvalue of Z when X = 12.

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

By the Central Limit theorem

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

Z = \frac{12 - 11.5}{0.5657}

Z = 0.88

Z = 0.88 has a pvalue of 0.8106.

1 - 0.8106 = 0.1894

18.94% probability that the sample mean amount purchased is at least 12 gallons

b. In a sample of 50 randomly selected customers, what is the approximate probability that the total amount of gasoline purchased is at most 600 gallons.

For sums, so mu = 50*11.5 = 575, s = \sqrt{50}*4 = 28.28

This probability is the pvalue of Z when X = 600. So

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

Z = \frac{600 - 575}{28.28}

Z = 0.88

Z = 0.88 has a pvalue of 0.8106.

81.06% probability that the total amount of gasoline purchased is at most 600 gallons.

c. What is the approximate value of the 95th percentile for the total amount purchased by 50 randomly selected customers.

This is X when Z has a pvalue of 0.95. So it is X when Z = 1.645.

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

1.645 = \frac{X- 575}{28.28}

X - 575 = 28.28*1.645

X = 621.5

The approximate value of the 95th percentile for the total amount purchased by 50 randomly selected customers is 621.5 gallons.

5 0
3 years ago
Explain how modeling partial products can be used to find the products of greater numbers ?
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You can break large numbers into a sum of a multiple(s) of 10 and the last digit of the number. For example, you can break 26 as 20+6, or 157 as 100+50+7.

Then, using the distributive property, you can turn the original multiplication into a sum of easier multiplications. For example, suppose we want to multiply 26 and 37. This is quite challenging to do in your mind, but you can break the numbers as we said above:

26\times 37=(20+6)(30+7) = 20\times 30+20\times 7+30\times 6+6\times 7

All these multiplications are rather easy, because they either involve multiples of 10 of single-digit numbers:

20\times 30+20\times 7+30\times 6+6\times 7 = 600+140+180+42=962

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3 years ago
An exponential function is expressed in the form y=axb^x. The relation represents a growth when ________ and a decay when_______
ArbitrLikvidat [17]

Answer:

Growth when: b>1.

Decay when: 0<b<1.

Step-by-step explanation:

Any function in the form f(x) =a\times b^x , where a > 0, b > 0 and b not equal to 1 is called an exponential function with base b.

if 0 < b < 1.  It is an example of an exponential decay.

The general shape of an exponential with b > 1 is an example of exponential growth.

Hence,

An exponential function is expressed in the form y=a\times b^x The relation represents a growth when  b >1 and a decay when 0<b<1.

4 0
2 years ago
Math Stuff please help 6th grade quiz!!
Whitepunk [10]
Between 1 and 3 cus -0.17 and before that it’s like 22 and 25 (-)
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2 years ago
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