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Assoli18 [71]
3 years ago
12

suppose 200 students take an exam, graded on a scale of 0-100%. The median exam grade was 78%, the mean exam grade was 68%, and

the range of scores was 40%. would you expect this distribution to be symmetric or skewed? Does the distribution have a high or low variation?
Mathematics
1 answer:
romanna [79]3 years ago
4 0
Because the median of the distribution is greater than the mean of the distribution.
Therefore, the distribution is skewed to the left.

Because the range is row (40%), the distribution has a low variation.
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Answer:

Step-by-step explanation:

Given that for a biology project, you measure the weight in grams and the tail length in millimeters of a group of mice.

The correlation is r = 0.9

If you had measured tail length in centimeters instead of millimeters, what would be the correlation?

For this we must understand the meaning of correlation and formula

Correlation is a measure of linear association between the two variables and always lie between -1 and 1

This is got by dividing covariance of (x,y) by product of std deviation of x and y

So because we changed the units of length correlation coefficient would not change.

It would remain the same as 0.90

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The answer is 6 if you need an explanation I have one

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The equation y = (-16t – 2)(t-1) represents the height in feet of a beach ball thrown by a child as a function of time, `t`, in
olasank [31]

2 ft

Step-by-step explanation:

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Answer:

122.6875 (122 ft 8.25 in)

Step-by-step explanation:

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To find the measure of ∠A in ∆ABC, use the___(Pythagorean Theorem, Law of Sines, Law of Cosines). To find the length of side HI
nadya68 [22]

<u>Part 1) </u>To find the measure of ∠A in ∆ABC, use

we know that

In the triangle ABC

Applying the law of sines

\frac{a}{sin\ A}=\frac{b}{sin\ B}=\frac{c}{sin\ C}

in this problem we have

\frac{a}{sin\ A}=\frac{b}{sin\ theta}\\ \\a*sin\ theta=b*sin\ A\\ \\ sin\ A=\frac{a*sin\ theta}{b} \\ \\ A=arc\ sin (\frac{a*sin\ theta}{b})

therefore

<u>the answer  Part 1) is</u>

Law of Sines

<u>Part 2) </u>To find the length of side HI in ∆HIG, use

we know that

In the triangle HIG

Applying the law of cosines

g^{2}=h^{2}+i^{2}-2*h*i*cos\ G

In this problem we have

g=HI

G=angle Beta

substitute

HI^{2}=h^{2}+i^{2}-2*h*i*cos\ Beta

HI=\sqrt{h^{2}+i^{2}-2*h*i*cos\ Beta}

therefore

<u>the answer Part 2) is</u>

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