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AfilCa [17]
3 years ago
12

The final exam of a particular class makes up 40% of the final grade, and Moe is failing the class with an average (arithmetic m

ean) of 45% just before taking the final exam. What grade does Moe need on his final exam in order to receive the passing grade average of 60% for the class?
Mathematics
1 answer:
amm18123 years ago
6 0

Answer:

%82.5

Step-by-step explanation:

  1. The final exam of a particular class makes up 40% of the final grade
  2. Moe is failing the class with an average (arithmetic mean) of 45% just before taking the final exam.

From point 1 we know that Moe´s grade just before taking the final exam represents 60% of the final grade. Then, using the information in the point 2 we can compute Moe´s final grade as follows:

FG=0.40*FE+0.60*0.45,

where FG is Moe´s Final Grade and FE is Moe´s final exam grade. Then,

\frac{ FG-0.60*0.45}{0.40}=FE.

So, in order to receive the passing grade average of 60% for the class Moe needs to obtain in his exam:

FE=\frac{ 0.60-0.60*0.45}{0.40}=0.825

That is, he need al least %82.5 to obtain a passing grade.

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Studentka2010 [4]

Answer:

<h3>(-4, 1)</h3>

Step-by-step explanation:

-x-3=y, therefore x = -y-3

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Find the value of y

Substitute the x in -3x - 8y = 4 with -y-3

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5 0
3 years ago
Read 2 more answers
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

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Step-by-step explanation:

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