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inn [45]
2 years ago
15

PLEASE HELP ME? I AM BEING TIMED ON THIS QUESTION

Mathematics
2 answers:
Alex777 [14]2 years ago
6 0
The slope would be B) 0.05
irga5000 [103]2 years ago
4 0
The answer would be b)0.05
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Mark Brainliest please

I think it looks like the image
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VikaD [51]

Answer:

<em>-1,3,5,9,13</em>

Step-by-step explanation:

2(-3)+5=-6+5=-1

2(-1)+5=-2+5=3

2(0)+5=0+5=5

2(2)+5=4+5=9

2(4)+5=8+5=13

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2 years ago
3x∧2±6x±x±20 sole by factorization
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1 year ago
The Office of Student Services at a large western state university maintains information on the study habits of its full-time st
Vera_Pavlovna [14]

Answer:

0.8254 = 82.54% probability that the mean of this sample is between 19.25 hours and 21.0 hours

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 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.

Mean of 20 hours, standard deviation of 6:

This means that \mu = 20, \sigma = 6

Sample of 150:

This means that n = 150, s = \frac{6}{\sqrt{150}}

What is the probability that the mean of this sample is between 19.25 hours and 21.0 hours?

This is the p-value of Z when X = 21 subtracted by the p-value of Z when X = 19.5. So

X = 21

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

By the Central Limit Theorem

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

Z = \frac{21 - 20}{\frac{6}{\sqrt{150}}}

Z = 2.04

Z = 2.04 has a p-value of 0.9793

X = 19.5

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

Z = \frac{19.5 - 20}{\frac{6}{\sqrt{150}}}

Z = -1.02

Z = -1.02 has a p-value of 0.1539

0.9793 - 0.1539 = 0.8254

0.8254 = 82.54% probability that the mean of this sample is between 19.25 hours and 21.0 hours

3 0
2 years ago
Solve −(3x − 2) = −4(x + 1) − 4.
Stella [2.4K]
-(3x-2)=-4(x+1)-4
-3x+2=-4x-4-4
-3x+4x=-4-4-2
x=-10
3 0
2 years ago
Read 2 more answers
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