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AURORKA [14]
3 years ago
12

Is it possible capture the correlation between continuous and categorical variable?

Mathematics
1 answer:
Bas_tet [7]3 years ago
7 0
It is possible to<span> capture the correlation (or lack thereof) between continuous and categorical variable using Analysis of Covariance (</span>ANCOVA) technique to capture association among continuous and categorical variables.
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Find a constant c for which p(z ≥ c.= 0.1587. round the answer to two decimal places.
Vikki [24]
What you see in the z score table is P(z< a constant), for P( z≥ a number), subtract the probability from 1:
1-0.1587=0.8413
on the z score table, you will see that p(z<1.00)=0.8413
so c=1.00
8 0
2 years ago
The points in the table lie on a line. Find the slope of the line.
wolverine [178]
Answer

Y=5x+2

I’m not sure but I think that’s the answer
8 0
3 years ago
I need help :3 !!!!!!
Vikki [24]

Answer:

b = 43

Since 43 is congruent to b

4 0
2 years ago
Read 2 more answers
A manufacturer knows that their items have a normally distributed length, with a mean of 15.4 inches, and standard deviation of
Masteriza [31]

Answer:

0.9452 = 94.52% probability that their mean length is less than 16.8 inches.

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.

Mean of 15.4 inches, and standard deviation of 3.5 inches.

This means that \mu = 15.4, \sigma = 3.5

16 items are chosen at random

This means that n = 16, s = \frac{3.5}{\sqrt{16}} = 0.875

What is the probability that their mean length is less than 16.8 inches?

This is the p-value of Z when X = 16.8. So

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

By the Central Limit Theorem

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

Z = \frac{16.8 - 15.4}{0.875}

Z = 1.6

Z = 1.6 has a p-value of 0.9452.

0.9452 = 94.52% probability that their mean length is less than 16.8 inches.

5 0
2 years ago
Can anyone answer this question please<br> 6[y-2]/7-12=2[y-7]/3
Kaylis [27]

Answer:

y =47.5.

Step-by-step explanation:

First eliminate the fractions by multiplying through by the LCM of 7 and 3 which is 21:

21* 6[y-2]/7-21*12 = 21*2[y-7]/3

18(y - 2) - 252 = 14(y - 7)

18y -36 - 252 = 14y - 98

18y - 14y = -98 + 36 + 252

4y = 190

y = 190/4

y = 47.5.

8 0
2 years ago
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