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kifflom [539]
3 years ago
5

How I get the LCM for 2,15

Mathematics
2 answers:
tamaranim1 [39]3 years ago
7 0
30 is the least common multiple for 2 and 15
 
nordsb [41]3 years ago
7 0
LCM of  2  and 15

2    /   2       15                      2 into 2 is 1.

3  /     1      15                      3 into 15 is 5

5  /      1      5                        5 into 5 is 1

     /      1      1


LCM = 2 * 3 * 5 = 30

LCM = 30
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Step-by-step explanation:

Slope intercept form is y = mx + b where m is the slope, b is the y-intercept, and y and x are their respective coordinates.

The questions gives us a coordinate pair and a slope so all we need to find is the y-intercept to find the equation.

Begin by plugging in what we know.

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There are two college entrance exams that are often taken by students, Exam A and Exam B. The composite score on Exam A is appro
elena55 [62]

Answer:

B.The score on Exam A is better, because the percentile for the Exam A score is higher.

Step-by-step explanation:

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.

In this problem, we have that:

Two exams. The exam that you did score better is the one in which you had a higher zscore.

The composite score on Exam A is approximately normally distributed with mean 20.1 and standard deviation 5.1.

This means that \mu = 20.1, \sigma = 5.1.

You scored 24 on Exam A. So

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

Z = \frac{24 - 20.1}{5.1}

Z = 0.76

The composite score on Exam B is approximately normally distributed with mean 1031 and standard deviation 215.

This means that \mu = 1031, \sigma = 215.

You scored 1167 on Exam B, s:

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

Z = \frac{1167 - 1031}{215}

Z = 0.632

You had a better Z-score on exam A, so you did better on that exam.

The correct answer is:

B.The score on Exam A is better, because the percentile for the Exam A score is higher.

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4 years ago
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