Solve the equation<br>(5p-8)/2=(7p+4)/6

a Put the following equation of a line into slope-intercept form, simplifying all fractions. x + 3y = -24

Alex73 [517]

Here’s the answer to your question.

4 0

2 years ago

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Convert .2 to a fraction

shepuryov [24]

.2 converted to a fraction would be 1/5

3 0

2 years ago

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Please answer this correctly without making mistakes

faltersainse [42]

Answer:

so to get a third you divide it by 3

first convert it to fraction

so it is 26/3

so do 26/3 divided by 3

so we do keep switch flip

26/3*1/3

so answer is 26/9 or 2 8/9

Step-by-step explanation:

6 0

2 years ago

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If the distribution is really (5.43,0.54)

defon

Answer:

0.7486 = 74.86% observations would be less than 5.79

Step-by-step explanation:

I suppose there was a small typing mistake, so i am going to use the distribution as N (5.43,0.54)

Problems of normally distributed samples can be 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.

The general format of the normal distribution is:

N(mean, standard deviation)

Which means that:

$\mu = 5.43, \sigma = 0.54$

What proportion of observations would be less than 5.79?

This is the pvalue of Z when X = 5.79. So

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

$Z = \frac{5.79 - 5.43}{0.54}$

$Z = 0.67$

$Z = 0.67$ has a pvalue of 0.7486

0.7486 = 74.86% observations would be less than 5.79

7 0

2 years ago

A set of test scores is normally distributed with a mean of 130 and a standard deviation of 30 What score is necessary to reach

nlexa [21]

Answer:

A score of 150.25 is necessary to reach the 75th percentile.

Step-by-step explanation:

Problems of normal distributions can be 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.

A set of test scores is normally distributed with a mean of 130 and a standard deviation of 30.

This means that $\mu = 130, \sigma = 30$