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kodGreya [7K]
2 years ago
7

Anna is covering the circular pool with a heavy-duty coer for the winter. The pool has a diameter of 24 feet. The cover extends

12 inches beyond the edge of the pool, and a rope runs along the edge of the cover to secure the cover in place. What is the area of the cover? What is the length of the rope?
Mathematics
2 answers:
MrRa [10]2 years ago
7 0
12 multiplied by 24.
Finger [1]2 years ago
3 0
12 multiplied by 24.
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Nutka1998 [239]

Answer:

y = 1/2x+4

Step-by-step explanation:

First we find the slope

m = (y2-y1)/(x2-x1)

    = (3-6)/(-2-4)

    = -3/-6

   =1/2

The equation of a line in slope intercept form is

y = mx+b  where m is the slope and b is the y intercept

y =1/2x+b

Substitute one of the point into the equation

6 = 1/2(4) +b

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y = 1/2x+4

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Tom's tank hold 12.5
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Answer:

$24.5 = 1 tank

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Of the entering class at a​ college, ​% attended public high​ school, ​% attended private high​ school, and ​% were home schoole
Veronika [31]

Answer:

(a) The probability that the student made the​ Dean's list is 0.1655.

(b) The probability that the student came from a private high school, given that the student made the Dean's list is 0.2411.

(c) The probability that the student was not home schooled, given that the student did not make the Dean's list is 0.9185.

Step-by-step explanation:

The complete question is:

Of the entering class at a college, 71% attended public high school, 21% attended private high school, and 8% were home schooled. Of those who attended public high school, 16% made the Dean's list, 19% of those who attended private high school made the Dean's list, and 15% of those who were home schooled made the Dean's list.

a) Find the probability that the student made the Dean's list.

b) Find the probability that the student came from a private high school, given that the student made the Dean's list.

c) Find the probability that the student was not home schooled, given that the student did not make the Dean's list.

Solution:

Denote the events as follows:

<em>A</em> = a student attended public high school

<em>B</em> = a student attended private high school

<em>C</em> = a student was home schooled

<em>D</em> = a student made the Dean's list

The provided information is as follows:

P (A) = 0.71

P (B) = 0.21

P (C) = 0.08

P (D|A) = 0.16

P (D|B) = 0.19

P (D|C) = 0.15

(a)

The law of total probability states that:

P(X)=\sum\limits_{i} P(X|Y_{i})\cdot P(Y_{i})

Compute the probability that the student made the​ Dean's list as follows:

P(D)=P(D|A)P(A)+P(D|B)P(B)+P(D|C)P(C)

         =(0.16\times 0.71)+(0.19\times 0.21)+(0.15\times 0.08)\\=0.1136+0.0399+0.012\\=0.1655

Thus, the probability that the student made the​ Dean's list is 0.1655.

(b)

Compute the probability that the student came from a private high school, given that the student made the Dean's list as follows:

P(B|D)=\frac{P(D|B)P(B)}{P(D)}

             =\frac{0.21\times 0.19}{0.1655}\\\\=0.2410876\\\\\approx 0.2411

Thus, the probability that the student came from a private high school, given that the student made the Dean's list is 0.2411.

(c)

Compute the probability that the student was not home schooled, given that the student did not make the Dean's list as follows:

P(C^{c}|D^{c})=1-P(C|D^{c})

               =1-\frac{P(D^{c}|C)P(C)}{P(D^{c})}\\\\=1-\frac{(1-P(D|C))\times P(C)}{1-P(D)}\\\\=1-\frac{(1-0.15)\times 0.08}{(1-0.1655)}\\\\=1-0.0815\\\\=0.9185

Thus, the probability that the student was not home schooled, given that the student did not make the Dean's list is 0.9185.

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Ahat [919]

Adding a value to the X value shifts the graph that many units to the left.

X+2 adds 2 to x, so the graph would shift 2 unites to the left.

The answer is:

The graph of g(x) is the graph of ​f(x)​ translated 2 units left.

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