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ser-zykov [4K]
3 years ago
15

A professor has recorded exam grades for 10 students in his​ class, but one of the grades is no longer readable. If the mean sco

re on the exam was 82 and the mean of the 9 readable scores is 84​, what is the value of the unreadable​ score?
Mathematics
1 answer:
Mkey [24]3 years ago
6 0

Answer:

64

Step-by-step explanation:

mean=\frac{sum\ of\ total\ number\ of\ score}{total\ number\ of\ students}

we have given that mean of 9 students is 84

so total score of 9 students = mean×9

                                              =84×9=756

and we have given mean score of exam is 82 and there is total 10 students so the total score of 10 students =10×82

                                                      =820

so the unreadable score = score of 10 students -score of 9 students =820-756=64

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Now replace y with 6 - x in both equations.

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Now we solve the lower equation for c.

c = 2(6 - x) + 4x

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Since we have two equations solved for c, we substitute to get

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This is an equation in only x, so we can solve for x.

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Now we solve for y.

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about 70,000 people live in a circular region with a 30-mile radius. find the population density in people per square mile.
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3 years ago
What is the area of this triangle ?
oee [108]

Answer:

Area of triangle is 9.88 units^2

Step-by-step explanation:

We need to find the area of triangle

Given E(5,1), F(0,4), D(0,8)

We will use formula:

Area\,\,of\,\,triangle =\sqrt{s(s-a)(s-b)s-c)} \\where\,\, s = \frac{a+b+c}{2}

We need to find the lengths of side DE, EF and FD

Length of side DE = a = \sqrt{(x_{2}-x_{1})^2+(y_{2}-y_{1})^2}

Length of side DE = a = =\sqrt{(5-0)^2+(1-8)^2}\\=\sqrt{(5)^2+(-7)^2}\\=\sqrt{25+49}\\=\sqrt{74}\\=8.60

Length of side EF = b = \sqrt{(x_{2}-x_{1})^2+(y_{2}-y_{1})^2}

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Length of side FD = c = =\sqrt{(0-0)^2+(8-4)^2}\\=\sqrt{(0)^2+(4)^2}\\=\sqrt{0+16}\\=\sqrt{16}\\=4

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s= 8.6+5.8+4/2

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So, area of triangle is 9.88 units^2

4 0
3 years ago
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