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Alekssandra [29.7K]
3 years ago
9

Plz help me ............

Mathematics
2 answers:
Leviafan [203]3 years ago
7 0
2 times 3.14 times 4
8 times 3.14
C=25.12
Effectus [21]3 years ago
3 0
C=2•3.14•4
C=8•3.14
C=25.12 yd.
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The right answer for the question that is being asked and shown above is that: "D. all real numbers less than or equal to 0." This is the statement that <span>best describes the range of the function f(x) = 2/3(6)x after it has been reflected over the x-axis</span>
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Solve this equation with the variable on both sides<br>6x - 2 = 7x+1​
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Answer:

x=-3

Step-by-step explanation:

6x - 2 =7x +1

6x - 7x =1+2

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A horse's resting heart rate is
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A popular soft drink is sold in 2-liter (2000 milliliter) bottles. Because of variation in the filling process, bottles have a m
andrezito [222]

Answer:

a) 0.27% probability that the mean is less than 1995 milliliters.

b) 2002.3 milliliters or more will occur only 10% of the time for the sample of 100 bottles.

Step-by-step explanation:

To solve this problem, it is important 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 random variable X, with mean \mu and standard deviation \sigma, a large sample size can be approximated to a normal distribution with mean \mu and standard deviation s = \frac{\sigma}{\sqrt{n}}

In this problem, we have that:

\mu = 2000, \sigma = 18

A) If the manufacturer samples 100 bottles, what is the probability that the mean is less than 1995 milliliters?

So n = 100, s = \frac{18}{\sqrt{100}} = 1.8

This probability is the pvalue of Z when X = 1995. So

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

By the Central Limit Theorem

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

Z = \frac{1995 - 2000}{1.8}

Z = -2.78

Z = -2.78 has a pvalue of 0.0027

So 0.27% probability that the mean is less than 1995 milliliters.

B) What mean overfill or more will occur only 10% of the time for the sample of 100 bottles?

This is the value of X when Z has a pvalue of 1-0.1 = 0.9.

So it is X when Z = 1.28

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

1.28 = \frac{X - 2000}{1.8}

X - 2000 = 1.8*1.28

X = 2002.3

2002.3 milliliters or more will occur only 10% of the time for the sample of 100 bottles.

3 0
3 years ago
What is the better deal work a restaurant a make 18 dollars per hour no tipping allowed or restaurant b 10.50 per hour and you a
Lyrx [107]

Answer:

18$

Step-by-step explanation:

6 0
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