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sertanlavr [38]
3 years ago
10

Richie and bethany play basketball and practice shooting free throws after school. During one practice session, Richie attempted

15 free throws and made 12 of them.
a. bethany made eight free throws for every 3 that she missed. Did bethany do better than richie? how do you know.
Mathematics
1 answer:
mamaluj [8]3 years ago
5 0

Answer:

<em>Bethany did not do better than Richie because her average score was less than his.</em>

Step-by-step explanation:

<u>Average Value</u>

The average or mean value of m successes out of n tries is given by:

\displaystyle \bar x=\frac{m}{n}

During a practice session playing basketball, Richie made m=12 free throws from n=15 attempts. His average score was:

\displaystyle \bar x=\frac{12}{15}=0.8

Bethany made 8 free throws for every 3 that she missed, this means she made 8 out of 11 attempts. Her average score was:

\displaystyle \bar x=\frac{8}{11}=0.72

Bethany did not do better than Richie because her average score was less than his.

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Which of the following are factor pairs for 12?
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Answer:

<h2> FACTOR pairs are a set of two integers that when multiplied together gives a particular product.</h2>

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Indicate true or false as to whether the following equation is quadratic 2x-4x+1=0
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3 0
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Read 2 more answers
Calculate pt3 such that a line from pt1 to pt3 is perpendicular to the line from pt1 to pt2, and the distance between pt1 and pt
Leni [432]
Let the point_1 = p₁ = (1,4)
and      point_2 = p₂ = (-2,1)
and      Point_3 = p₃ = (x,y)

The line from point_1 to point_2 is L₁ and has slope = m₁
The line from point_1 to point_3 is L₂ and has slope = m₂
m₁ = Δy/Δx = (1-4)/(-2-1) = 1
m₂ = Δy/Δx = (y-4)/(x-1)
L₁⊥L₂ ⇒⇒⇒⇒ m₁ * m₂ = -1
∴ (y-4)/(x-1) = -1 ⇒⇒⇒ (y-4)= -(x-1)
(y-4) = (1-x) ⇒⇒⇒⇒⇒⇒⇒⇒⇒⇒⇒⇒⇒⇒ equation (1)

The distance from point_1 to point_2 is d₁
The distance from point_1 to point_3 is d₂
d = \sqrt{Δx^2+Δy^2}
d₁ = \sqrt{(-2-1)^2+(1-4)^2}
d₂ = \sqrt{(x-1)^2+(y-4)^2}
d₁ = d₂
∴ \sqrt{(-2-1)^2+(1-4)^2} = \sqrt{(x-1)^2+(y-4)^2} ⇒⇒ eliminating the root
∴(-2-1)²+(1-4)² = (x-1)²+(y-4)²
 (x-1)²+(y-4)² = 18
from equatoin (1)  y-4 = 1-x
∴(x-1)²+(1-x)² = 18            ⇒⇒⇒⇒⇒ note: (1-x)² = (x-1)²
2 (x-1)² = 18
(x-1)² = 9
x-1 = \pm \sqrt{9} = \pm 3
∴ x = 4 or x = -2
∴ y = 1 or y = 7

Point_3 = (4,1)  or  (-2,7)












8 0
3 years ago
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