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ruslelena [56]
3 years ago
12

Mason wrote the numbers 0 through 14 on scraps of paper and placed the scraps in a box. He then randomly chose one of the scraps

from the box. What is the probability that Manson chose a number greater than or equal to 6? Plz explain and show work.
A. 3/7
B. 4/7
C. 3/5
D. 2/5
Mathematics
2 answers:
lisabon 2012 [21]3 years ago
8 0

There are 0-14 numbers inclusive, so that means there are 15 scraps of paper.

Probability of getting greater than or equal to six (6, 7, 8, 9, 10, 11, 12, 13, 14) is

\frac{9}{15}

(as there are 15 scraps)

\frac{9}{15}  =  \frac{3}{5}

so the answer is C.

kupik [55]3 years ago
4 0

Answer:

C.3/5

Step-by-step explanation:

The numbers that would fit are 6,7,8,9,10,11,12,13,14. There are 15 numbers int he box. 9/15=3/5

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Which property of equality would be used to solve 3x = 81?
slega [8]
Division x=81/3 that’s right
4 0
3 years ago
Read 2 more answers
The path of a football can be modelled by the equation, h=-10d2+120d , where h represents the height, in metres, of the football
Marta_Voda [28]

Answer:

12 m

Step-by-step explanation:

The path of a football has been modeled by the equation:

h= -10d^2+120d

where h represents the height and d represents the horizontal distance.

When the ball lands, it means that its height is back at 0 metres. This means that we have to find horizontal distance, d, when height, h, is 0.

=> 0= -10d^2+120d

=> 10d^2 - 120d = 0

d(10d - 120) = 0

∴ d = 0 m

and

10d - 120 = 0

=> d = 120 / 10 = 12 m

There are two solutions for d when h = 0 m.

The first solution (d = 0 m) is a case where the ball has not been thrown at all. This means the ball has not moved away from the football player and it is still on the ground.

The second solution is the answer to our problem (d = 12 m). The ball lands at a horizontal distance of 12 m

3 0
3 years ago
Find the geometric means in the following sequence. –14, ? , ? , ? , ? , –235,298
timama [110]
The answer should be something like 12, 33, 56, 66, 77
6 0
3 years ago
Gage's math teacher entered the seventh-grade student in a math competition. There was an enrollment fee of $30 and also $11 cha
Amanda [17]

Answer:

110 test

Step-by-step explanation:

Let

x ----> the number  of packet of 10 test

y ----> the total cost

we know that

The number of packet of 10 test purchase multiplied by $11 plus the enrollment fee of $30 must be equal to $151

so

The linear equation that represent this scenario is

y=11x+30

we have

y=\$151

substitute

151=11x+30

Solve for x

subtract 30 both sides

11x=151-30

11x=121

Divide by 11 both sides

x=11

The number of packets purchase was 11

To find out the number of test, multiply the number of packets by 10

11(10)=110\ test

4 0
3 years ago
Part 4: Identify the vertex, focus and directrix of each. Then sketch the graph.
Nezavi [6.7K]

Answer: 1) Vertex: (6, -2)    Focus: (6, -7/4)     Directrix: y = -9/4

              2) Vertex: (-2, -1)   Focus: (-7/4, -1)     Directrix: x = -9/4

<u>Step-by-step explanation:</u>

Rewrite the equation in vertex format y = a(x - h)² + k   or   x = a(y - k)² + h by completing the square. Divide the b-value by 2 and square it - add that value to both sides of the equation.

  • (h, k) is the vertex
  • p is the distance from the vertex to the focus
  • -p is the distance from the vertex to the directrix

    \bullet\quad a=\dfrac{1}{4p}

1) y = x² - 12x + 34

y-34=x^2-12x\\\\y-34+\bigg(\dfrac{-12}{2}\bigg)^2=x^2-12x+\bigg(\dfrac{-12}{2}\bigg)^2\\\\\\y-34+36=(x-6)^2\\\\y+2=(x-6)^2\\\\y=(x-6)^2-2\qquad \rightarrow \qquad a=1\quad (h,k)=(6,-2)\\

a=\dfrac{1}{4p}\quad \rightarrow \quad 1=\dfrac{1}{4p}\quad \rightarrow \quad p=\dfrac{1}{4}\\\\\\\text{Focus = Vertex + p}\\.\qquad \quad =\dfrac{-8}{4}+\dfrac{1}{4}\\\\.\qquad \quad =-\dfrac{7}{4}\quad \rightarrow \quad\text{Focus}=\bigg(6,-\dfrac{7}{4}\bigg)\\\\\\\text{Directrix: y= Vertex - p}\\.\qquad \qquad y=\dfrac{-8}{4}-\dfrac{1}{4}\\\\.\qquad \qquad y=-\dfrac{9}{4}

*******************************************************************************************

2) x = y² + 2y - 1

x+1=y^2+2y\\\\x+1+\bigg(\dfrac{2}{2}\bigg)^2=y^2+2y+\bigg(\dfrac{2}{2}\bigg)^2\\\\\\x+1+1=(y+1)^2\\\\x+2=(y+1)^2\\\\x=(y+1)^2-2\qquad \rightarrow \qquad a=1\quad (h,k)=(-2,-1)\\

a=\dfrac{1}{4p}\quad \rightarrow \quad 1=\dfrac{1}{4p}\quad \rightarrow \quad p=\dfrac{1}{4}\\\\\\\text{Focus = Vertex + p}\\.\qquad \quad =\dfrac{-8}{4}+\dfrac{1}{4}\\\\.\qquad \quad =-\dfrac{7}{4}\quad \rightarrow \quad\text{Focus}=\bigg(-\dfrac{7}{4},-1\bigg)\\\\\\\text{Directrix: x= Vertex - p}\\.\qquad \qquad y=\dfrac{-8}{4}-\dfrac{1}{4}\\\\.\qquad \qquad x=-\dfrac{9}{4}

4 0
3 years ago
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