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Naddika [18.5K]
3 years ago
12

Please help I don't understand T^T

Mathematics
2 answers:
yarga [219]3 years ago
5 0

Answer:

I can't see the picture

Step-by-step explanation:

Are you sure that you uploaded it?

Ivahew [28]3 years ago
5 0

Answer:

B)

-\dfrac{190}{63}x

Step-by-step explanation:

You are adding two like terms which are fractions with denominators 9 and 7.

You need a common denominator. The LCD of 9 and 7 is 63.

-\dfrac{31}{9}x + \dfrac{3}{7}x = -\dfrac{31}{9} \times \dfrac{7}{7}x + \dfrac{3}{7} \times \dfrac{9}{9}x = -\dfrac{217}{63}x + \dfrac{27}{63}x = -\dfrac{190}{63}x

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Is this for area and perimeter? If so, first make the fraction a decimal, (that would be 3.5). Then, for permeter, add 3.5 + 3.5 + 9 + 9, which is 25, for area, multiply 3.5 by 9, which is 31.5. Hope this helped!
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What is the decimal expansion of 7/2
valentina_108 [34]

Answer:

0.583

Step-by-step explanation:

you just divide ig

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The missing number is 4
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4 years ago
Life Expectancies In a study of the life expectancy of people in a certain geographic region, the mean age at death was years an
Sphinxa [80]

Answer:

The probability that the mean life expectancy of the sample is less than X years is the p-value of Z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}, in which \mu is the mean life expectancy, \sigma is the standard deviation and n is the size of the sample.

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.

We have:

Mean \mu, standard deviation \sigma.

Sample of size n:

This means that the z-score is now, by the Central Limit Theorem:

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

Find the probability that the mean life expectancy will be less than years.

The probability that the mean life expectancy of the sample is less than X years is the p-value of Z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}, in which \mu is the mean life expectancy, \sigma is the standard deviation and n is the size of the sample.

8 0
3 years ago
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