Be Fit Exercise Company marks up their inventory by 50%.

On Monday, Avi walked from his home to the school track and walked 4 laps around the track for a total of 3.5 miles. On Saturday

anyanavicka [17]

Answer:

3.5 = x + 4*y

4 = 2x + 2*y

Step-by-step explanation:

The total distance ran by Avi is the sum of the distance from his house to the track, and the total distance ran in the tracks (that is, the distance of one lap times the number of laps Avi ran).

If we call 'd' the total distance and 'k' the number of laps, we have the equation:

d = x + k*y

So with the data given, we can have the following system:

3.5 = x + 4*y

4 = 2x + 2*y

In the second equation, we have 2x because Avi ran from his home to the track, and in the end ran back home, so Avi ran the distance x two times.

4 0

3 years ago

You flip 3 coins. Is the probability of obtaining 2 heads and 1 tail in any order the same as the probability of obtaining a hea

konstantin123 [22]

Answer:

the probability of obtaining 2 heads and 1 tail in any order is higher than the probability of obtaining a head, than a head then a tail

Step-by-step explanation:

since the probability (p) of obtaining 2 head and 1 tail in any order is

P1 = p of obtaining ( H , H ,T) + p of obtaining ( H , T ,H ) + p of obtaining ( T , H ,H ) = 3*p of obtaining ( H , T ,H)

assuming a fair coin then p heads = p tails = 0.5

thus since each flip is independent from the others

p( H , H ,T)=p( H , T ,H )= p( T , H ,H )= P=0.5*0.5*0.5= 1/8

thus P1 =3*1/8=3/8

while the probability of obtaining a head, than a head then a tail is

P2= p of obtaining ( H , T ,H)= 1/8

then P1=3/8 >P2=1(8

therefore the probability of obtaining 2 heads and 1 tail in any order is higher than the probability of obtaining a head, than a head then a tail

6 0

3 years ago

Plz help me I have no idea how to do this plz help!

Vika [28.1K]

You need to multiply the amount by the tax

8 0

3 years ago

O GRAPHS, FUNCTIONS, AND SEQUENCES

WINSTONCH [101]

Always use the formula (y2-y1)/(x2-x1) to find a line’s slope, given two points:

(-1- -8)/(4-5) = 7/-1 = -7

-7 is the slope

7 0

2 years ago

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

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

so

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

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

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

so

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

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