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Alekssandra [29.7K]
2 years ago
5

If a certian perosn has a mass of 250kg on earth, would that person's mass be on the moon, if the moon's gravity is 1.62m/s2? *

Mathematics
1 answer:
IceJOKER [234]2 years ago
6 0

Answer:

Weight On the Moon: 41.34 kilograms

Step-by-step explanation:

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Evaluate -a+(-b)−a+(−b)minus, a, plus, left parenthesis, minus, b, right parenthesis where a = 6.05a=6.05a, equals, 6, point, 05
swat32

Answer: -9.66

Step-by-step explanation:

Here the given expression,   -a + (-b)

Where  a = 6.05 and b =3.61

By putting the given values of a and b in the given expression,

We get,

- 6.05 + ( -3.61)

= -6.05 - 3.61

= - (6.05 + 3.61)

= - 9.66

3 0
3 years ago
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A Government company claims that an average light bulb lasts 270 days. A researcher randomly selects 18 bulbs for testing. The s
Fofino [41]

Answer:

31.92% probability that 18 randomly selected bulbs would have an average life of no more than 260 days

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal probability distribution

When the distribution is normal, we use 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 a proportion p in a sample of size n, the sampling distribution of the sample proportion will be approximately normal with mean \mu = p and standard deviation s = \sqrt{\frac{p(1-p)}{n}}

In this question, we have that:

\mu = 270, \sigma = 90, n = 18, s = \frac{90}{\sqrt{18}} = 21.2

What is the probability that 18 randomly selected bulbs would have an average life of no more than 260 days?

This is the pvalue of Z when X = 260. So

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

By the Central Limit Theorem

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

Z = \frac{260 - 270}{21.2}

Z = -0.47

Z = -0.47 has a pvalue of 0.3192.

31.92% probability that 18 randomly selected bulbs would have an average life of no more than 260 days

5 0
3 years ago
There are 125 students in sixth grade. 2/3 of the sixth graders ride the bus. How many sixth graders ride the bus?​
Luden [163]
125 ÷ 3 = 41.6666666667

41.66666666667 X 2
= 83.3333......
Technically not correct, because you cannot split sixth graders into pieces, so you would probably round it to 84 students.
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den301095 [7]

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