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frez [133]
3 years ago
8

There are 135 people in a sport centre. 73 people use the gym. 59 people use the swimming pool. 31 people use the track. 19 peop

le use the gym and the pool. 9 people use the pool and the track. 16 people use the gym and the track. 4 people use all three facilities. Given that a randomly selected person uses the gym and the track, what is the probability they do not use the swimming pool?
Mathematics
1 answer:
Anna71 [15]3 years ago
6 0

Answer:

1/9

Step-by-step explanation:

135 in Sport centre: Total

59:swimming pool

31:track

19 both swimming and gym

16 gym and track

4 all three facilities

4 people use all three facilities, then

16 - 4 = 12 people use the gym and the track and do not use the pool;

9 - 4 = 5 people use the pool and the track and do not use the gym;

19 - 4 = 15 people use the gym and the pool and do not use the track.

At least two facilities use 4 + 12 + 5 + 15 = 36 people, 4 of them use all three facilities. Thus, the probability that a randomly selected person which uses at least two facilities, uses all the facilities is

4/36=1/9

Hope this helps!!!

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If the average yield of cucumber acre is 800 kg, with a variance 1600 kg, and that the amount of the cucumber follows the normal
lorasvet [3.4K]

Answer:

a

   The  percentage is

            P(x_1 <  X <  x_2 ) =   51.1 \%

b

   The probability is  P(Z >  2.5 ) =  0.0062097

Step-by-step explanation:

From the question we are told that

        The  population mean is  \mu =  800

        The  variance is  var(x) =  1600 \ kg

        The  range consider is  x_1 =  778 \ kg  \  x_2 =  834 \ kg

         The  value consider in second question is  x =  900 \ kg

Generally the standard deviation is mathematically represented as

        \sigma =  \sqrt{var (x)}

substituting value

        \sigma =  \sqrt{1600}

       \sigma = 40

The percentage of a cucumber give the crop amount between 778 and 834 kg  is mathematically represented as

       P(x_1 <  X <  x_2 ) =  P( \frac{x_1 -  \mu }{\sigma} <  \frac{X - \mu }{ \sigma } < \frac{x_2 - \mu }{\sigma }   )

    Generally  \frac{X - \mu }{ \sigma } = Z (standardized \  value  \  of  \  X)

So

      P(x_1 <  X <  x_2 ) =  P( \frac{778 -  800 }{40} < Z< \frac{834 - 800 }{40 }   )

      P(x_1 <  X <  x_2 ) =  P(z_2 < 0.85) -  P(z_1 <  -0.55)

From the z-table  the value for  P(z_1 <  0.85) =  0.80234

                                            and P(z_1 <  -0.55) =   0.29116  

So

             P(x_1 <  X <  x_2 ) =   0.80234 - 0.29116

             P(x_1 <  X <  x_2 ) =   0.51118

The  percentage is

            P(x_1 <  X <  x_2 ) =   51.1 \%

The probability of cucumber give the crop exceed 900 kg is mathematically represented as

             P(X > x ) =  P(\frac{X - \mu }{\sigma }  > \frac{x - \mu }{\sigma } )

substituting values

             P(X > x ) =  P( \frac{X - \mu }{\sigma }  >\frac{900 - 800 }{40 }   )

             P(X > x ) =  P(Z >2.5   )

From the z-table  the value for  P(Z >  2.5 ) =  0.0062097

 

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3 years ago
Which of the following BEST describes how to use the addition property of equality to isolate the variable y
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Answer:  C) add 12 to both sides

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We add 12 to the left side to undo the -12, which will get y by itself.

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The perimeter of a rectangle is 72 cm. The length is 3 more than twice the width What
solmaris [256]
<h3><u>Solution</u></h3>

<u>Given </u><u>:</u><u>-</u>

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  • The length is 3 more than twice the width.

<u>F</u><u>i</u><u>n</u><u>d</u><u> </u><u>:</u><u>-</u>

  • Dimensions of rectangle
<h3 /><h3><u>Explantion</u></h3>

<u>Using </u><u>Formula</u>

\boxed{\underline{\tt{\red{\:(perimeter_{rectang}\:=\:2\times (Length+Width)}}}}

<u>Let,</u>

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<u>According</u><u> to</u><u> question</u><u>,</u>

==> perimeter of Rectangle = 72

==> 2(x+y) = 72

==> x + y = 72/2

==> x + y = 36_________________(1)

<u>Again,</u>

==> x = 2y + 3

==> x - 2y = 3__________________(2)

<u>Subtract</u><u> </u><u>equ(</u><u>1</u><u>)</u><u> </u><u>&</u><u> </u><u>equ(</u><u>2</u><u>)</u>

==> y + 2y = 36 - 3

==> 3y = 33

==> y = 33/3

==> y = 11

<u>keep </u><u>in </u><u>equ(</u><u>1</u><u>)</u>

==> x - 2×11 = 3

==> x = 3 + 22

==> x = 25

<h3><u>Hence</u></h3>

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<h3><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u></h3>

<h3 />

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