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2 years ago

This probability distribution shows

Mathematics

1 answer:

Charra [1.4K]2 years ago

7 0

Answer:

The probability that a student earns a grade of A is 1/7.

Let E be an event and S be the sample space. The probability of E, denoted by P(E) could be computed as:

P(E) = n(E) / n(S)

As the total number of students = n(S) = 35

Students getting the grade A = n(E) = 5

So, the probability that a student earns a grade of A:

P(E) = n(E) / n(S)

= 5/35

= 1/7

Hence, the probability that a student earns a grade of A is 1/7.

Keywords: probability, sample space, event

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A triangle is formed from the points L(-3, 6), N(3, 2) and P(1, -8). Find the equation of the following lines:

Dima020 [189]

Answer:

Part A) $y=\frac{3}{4}x-\frac{1}{4}$

Part B) $y=\frac{2}{7}x-\frac{5}{7}$

Part C) $y=\frac{2}{7}x+\frac{8}{7}$

see the attached figure to better understand the problem

Step-by-step explanation:

we have

points L(-3, 6), N(3, 2) and P(1, -8)

Part A) Find the equation of the median from N

we Know that

The median passes through point N to midpoint segment LP

step 1

Find the midpoint segment LP

The formula to calculate the midpoint between two points is equal to

$M(\frac{x1+x2}{2},\frac{y1+y2}{2})$

we have

L(-3, 6) and P(1, -8)

substitute the values

$M(\frac{-3+1}{2},\frac{6-8}{2})$

$M(-1,-1)$

step 2

Find the slope of the segment NM

The formula to calculate the slope between two points is equal to

$m=\frac{y2-y1}{x2-x1}$

we have

N(3, 2) and M(-1,-1)

substitute the values

$m=\frac{-1-2}{-1-3}$

$m=\frac{-3}{-4}$

$m=\frac{3}{4}$

step 3

Find the equation of the line in point slope form

$y-y1=m(x-x1)$

we have

$m=\frac{3}{4}$

$point\ N(3, 2)$

substitute

$y-2=\frac{3}{4}(x-3)$

step 4

Convert to slope intercept form

Isolate the variable y

$y-2=\frac{3}{4}x-\frac{9}{4}$

$y=\frac{3}{4}x-\frac{9}{4}+2$

$y=\frac{3}{4}x-\frac{1}{4}$

Part B) Find the equation of the right bisector of LP

we Know that

The right bisector is perpendicular to LP and passes through midpoint segment LP

step 1

Find the midpoint segment LP

The formula to calculate the midpoint between two points is equal to

$M(\frac{x1+x2}{2},\frac{y1+y2}{2})$

we have

L(-3, 6) and P(1, -8)

substitute the values

$M(\frac{-3+1}{2},\frac{6-8}{2})$

$M(-1,-1)$

step 2

Find the slope of the segment LP

The formula to calculate the slope between two points is equal to

$m=\frac{y2-y1}{x2-x1}$

we have

L(-3, 6) and P(1, -8)

substitute the values

$m=\frac{-8-6}{1+3}$

$m=\frac{-14}{4}$

$m=-\frac{14}{4}$

$m=-\frac{7}{2}$

step 3

Find the slope of the perpendicular line to segment LP

Remember that

If two lines are perpendicular, then their slopes are opposite reciprocal (the product of their slopes is equal to -1)

$m_1*m_2=-1$

we have

$m_1=-\frac{7}{2}$

$m_2=\frac{2}{7}$

step 4

Find the equation of the line in point slope form

$y-y1=m(x-x1)$

we have

$m=\frac{2}{7}$

$point\ M(-1,-1)$ ----> midpoint LP

substitute

$y+1=\frac{2}{7}(x+1)$

step 5

Convert to slope intercept form

Isolate the variable y

$y+1=\frac{2}{7}x+\frac{2}{7}$

$y=\frac{2}{7}x+\frac{2}{7}-1$

$y=\frac{2}{7}x-\frac{5}{7}$

Part C) Find the equation of the altitude from N

we Know that

The altitude is perpendicular to LP and passes through point N

step 1

Find the slope of the segment LP

The formula to calculate the slope between two points is equal to

$m=\frac{y2-y1}{x2-x1}$

we have

L(-3, 6) and P(1, -8)

substitute the values

$m=\frac{-8-6}{1+3}$

$m=\frac{-14}{4}$

$m=-\frac{14}{4}$

$m=-\frac{7}{2}$

step 2

Find the slope of the perpendicular line to segment LP

Remember that

If two lines are perpendicular, then their slopes are opposite reciprocal (the product of their slopes is equal to -1)

$m_1*m_2=-1$

we have

$m_1=-\frac{7}{2}$

$m_2=\frac{2}{7}$

step 3

Find the equation of the line in point slope form

$y-y1=m(x-x1)$

we have

$m=\frac{2}{7}$

$point\ N(3,2)$

substitute

$y-2=\frac{2}{7}(x-3)$

step 4

Convert to slope intercept form

Isolate the variable y

$y-2=\frac{2}{7}x-\frac{6}{7}$

$y=\frac{2}{7}x-\frac{6}{7}+2$

$y=\frac{2}{7}x+\frac{8}{7}$

7 0

3 years ago

The top and bottom margins of a poster 66 cm each, and the side margins are 44 cm each. If the area of the printed material on t

jasenka [17]

Answer:

width: 24 cm
height: 36 cm

Step-by-step explanation:

When margins are involved, the smallest area will be the one that has its dimensions in the same proportion as the margins. If x is the "multiplier", the dimensions of the printed area are ...

(4x)(6x) = 384 cm^2

x^2 = 16 cm^2 . . . . . divide by 24

x = 4 cm

The printed area is 4x by 6x, so is 16 cm by 24 cm. With the margins added, the smallest poster will be ...

24 cm by 36 cm

_____

Comment on margins

It should be obvious that if both side margins are 4 cm, then the width of the poster is 8 cm more than the printed width. Similarly, the 6 cm top and bottom margins make the height of the poster 12 cm more than the height of the printed area.

_____

Alternate solution

Let w represent the width of the printed area. Then the printed height is 384/w, and the total poster area is ...

A = (w+8)(384/w +12) = 384 +12w +3072/w +96

Differentiating with respect to w gives ...

A' = 12 -3072/w^2

Setting this to zero and solving for w gives ...

w = √(3072/12) = 16 . . . . matches above solution.

Generic solution

If we let s and t represent the side and top margins, and we use "a" for the printed area, then the above equation becomes the symbolic equation ...

A = (w +s)(a/w +t)

A' = t - sa/w^2

For A' = 0, ...

w = √(sa/t)

and the height is ...

a/w = a/√(sa/t) = √(ta/s)

Then the ratio of width to height is ...

w/(a/w) = w^2/a = (sa/t)/a

width/height = s/t . . . . . . the premise we started with, above

6 0

3 years ago