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qaws [65]
3 years ago
11

MP Persevere with Problems Suppose 20

Mathematics
1 answer:
grigory [225]3 years ago
7 0

9514 1404 393

Answer:

  8 people

Step-by-step explanation:

If 20 of 140 people play tennis, then in a group of 504 people, we expect ...

  (20/140) × 504 = 72 . . . . tennis players

If 1 of 9 tennis players has a coach, then in a group of 72 tennis players, we expect ...

  (1/9) × 72 = 8 . . . . have a coach

We expect 8 of the 504 people have a tennis coach. We made this prediction by applying the given rates to the larger group.

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There are 5 red balls, 4 blue balls, 6 yellow balls and 10 green balls in a box,
lys-0071 [83]

<h2 /><h2>Here we go ~ </h2>

According to given information there are :

  • 5 red balls

  • 4 blue balls

  • 6 yellow balls

  • 10 green balls

<h3>1. what is the probability that the ball chosen is red ?</h3>

-

\qquad \sf  \dashrightarrow \: p(red) =  \dfrac{total \: red \: balls}{total \: balls}

\qquad \sf  \dashrightarrow \: p(red) =  \dfrac{5}{5 + 4 + 6 + 10}

\qquad \sf  \dashrightarrow \: p(red) =  \dfrac{5}{25}

\qquad \sf  \dashrightarrow \: p(red) =  \dfrac{1}{5}

<h3>2. what is the probability that the ball chosen is blue ?</h3>

\qquad \sf  \dashrightarrow \: p(blue) =  \dfrac{total \: blue \: balls}{total \: balls}

\qquad \sf  \dashrightarrow \: p(blue) =  \dfrac{4}{5 + 4 + 6 + 10}

\qquad \sf  \dashrightarrow \: p(blue) =  \dfrac{4}{25}

<h3>3. what is the probability that the ball chosen is yellow ?</h3>

\qquad \sf  \dashrightarrow \: p(yellow) =  \dfrac{total \: yellow\: balls}{total \: balls}

\qquad \sf  \dashrightarrow \: p(yellow) =  \dfrac{6}{5 + 4 + 6 + 10}

\qquad \sf  \dashrightarrow \: p(yellow) =  \dfrac{6}{25}

<h3>4. what is the probability that the ball chosen is green ?</h3>

\qquad \sf  \dashrightarrow \: p(green) =  \dfrac{total \: green\: balls}{total \: balls}

\qquad \sf  \dashrightarrow \: p(green) =  \dfrac{10}{5 + 4 + 6 + 10}

\qquad \sf  \dashrightarrow \: p(green) =  \dfrac{10}{25}

\qquad \sf  \dashrightarrow \: p(green) =  \dfrac{2}{5}

<h3>5. what is the probability that the ball chosen is not green ?</h3>

\qquad \sf  \dashrightarrow \: p(not \: green) =  \dfrac{total \: non \: green\: balls}{total \: balls}

\qquad \sf  \dashrightarrow \: p(not \: green) =  \dfrac{5 + 4 + 6}{5 + 4 + 6 + 10}

\qquad \sf  \dashrightarrow \: p(not \: green) =  \dfrac{15}{25}

\qquad \sf  \dashrightarrow \: p(not \: green) =  \dfrac{3}{5}

3 0
2 years ago
En un aeropuerto dos aviones A1 y A1 se acercan para aterrizar, si la ecuación de la trayectoria del primero es -x+2y-100=0, mie
liraira [26]

No hay solución en la que ambas trayectorias estén en la misma posición, puesto que no existe el riesgo de que sufran un choque entre ellos.

-------------------------

La trayectoria del primero es:

-x + 2y - 100 = 0

x_1 = 2y - 100

Para el segundo, tiene-se que:

-x + 200 + 2y = 0

x_2 = 2y + 200

Igualando los valores de x:

x_1 = x_2

2y - 100 = 2y + 200

0y = 300

No hay solución en la que ambas trayectorias estén en la misma posición, puesto que no existe el riesgo de que sufran un choque entre ellos.

Un problema similar es dado en brainly.com/question/24653364

3 0
3 years ago
Help me please...I will mark brainliest
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The answer is A because playing sport is not affected by playing sports (for more detail go to https://www.mathsisfun.com/data/probability-events-independent.html)

If it's not A then it's B

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3 years ago
Which factors affect friction between two solid surfaces? Select two options. the weight of the objects the surface area of the
andreev551 [17]

Answer:

a and d

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2 years ago
Evaluate S(3)=10+20log4(3+1)
dimulka [17.4K]
Asssuming you mean

evaluate 10+20log_4(3+1)

remember log_a(a)=1
so
10+20log_4(3+1)=
10+20log_4(4)=
10+20(1)=
10+20=
30
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3 years ago
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