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xxMikexx [17]
2 years ago
12

How many times does 58 go into 556

Mathematics
2 answers:
alisha [4.7K]2 years ago
8 0
9.5862068966.............
ValentinkaMS [17]2 years ago
5 0
9.   ___9.586_ 
  58l 556
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Based on the tree diagram below, what is the probability that a student doesn't have lice, given that the student tested negativ
oee [108]

Answer:

Option D (APEX)

Step-by-step explanation:

L= Student has lice

NL= Student has no lice

PT=Test Shows positive

NT=Test Shows negative

P(\frac{NT}{NL}) =0.7708 (Probability that a student has no lice and the test shows negative)

P(\frac{NT}{L}) = 0.0108 ( Probability that a student has lice and the test shows negative)

  1. P\frac{NT}{NL} = \frac{P\frac{NT}{NL} }{P\frac{NT}{NL} +P\frac{NT}{L} }
  2. \frac{0.7708}{0.7708+00108}
  3. \frac{0.7708}{0.7816}
  4. 0.98618219 (Simplify)

<u> Terms Of Percentages</u>

0.98618219 × 100

=98.6

Therefore your answer is 98.6% (Option D)

APEX

3 0
2 years ago
Read 2 more answers
Una lancha que viaja a 10 m/s pasa por debajo de un puente 3 segundos después que ha pasado un bote que viaja a 7 m/s, ¿después
ExtremeBDS [4]

Answer:

La lancha y el bote se encontrarán a 70 metros de distancia del puente.

Step-by-step explanation:

Sea el punto debajo del puente el punto de referencia y que ambas lanchas se desplazan a velocidad a continuación, las ecuaciones cinemáticas para cada embarcación son presentadas a continuación:

Bote a 7 metros por segundo

x_{A} = x_{o}+v_{A}\cdot t (Ec. 1)

Lancha a 10 metros por segundo

x_{B} = x_{o}+v_{B}\cdot (t-3\,s) (Ec. 2)

Donde:

x_{o} - Posición debajo del puente, medido en metros.

x_{A}, x_{B} - Posición final de cada embarcación, medido en metros.

v_{A}, v_{B} - Velocidad de cada embarcación, medida en metros por segundo.

t - Tiempo, medido en segundos.

Para determinar la posición en la que ambas embarcaciones se encuentran, se debe determinar el instante en que ocurre a partir de la siguiente condición: x_{A} = x_{B}

Igualando (Ec. 1) y (Ec. 2) se tiene que:

v_{A}\cdot t = v_{B}\cdot (t-3\,s)

Ahora despejamos el tiempo:

3\cdot v_{B} = (v_{B}-v_{A})\cdot t

t = \frac{3\cdot v_{B}}{v_{B}-v_{A}}

Si sabemos que v_{B} = 10\,\frac{m}{s} y v_{A} = 7\,\frac{m}{s}, entonces:

t = \frac{3\cdot \left(10\,\frac{m}{s} \right)}{10\,\frac{m}{s}-7\,\frac{m}{s}}

t = 10\,s

Ahora, la posición de encuentro es: (x_{o} = 0\,m, v_{A} = 7\,\frac{m}{s} y t = 10\,s)

x_{A} = 0\,m + \left(7\,\frac{m}{s} \right)\cdot (10\,s)

x_{A} = 70\,m

La lancha y el bote se encontrarán a 70 metros de distancia del puente.

6 0
3 years ago
Need ASAP I’ll mark brainliest if right
katen-ka-za [31]
Area= 50.24 or 50.2 rounded to the nearest tenth
5 0
3 years ago
Please Help!!
GarryVolchara [31]

Answer:

D:) (2,2,) is the Answer

Step-by-step explanation:

Solve the following system:


{X - 2 Y = -2 | (equation 1)


{3 X - 2 Y = 2 | (equation 2)



Swap equation 1 with equation 2:


{3 X - 2 Y = 2 | (equation 1)


{X - 2 Y = -2 | (equation 2)



Subtract 1/3 × (equation 1) from equation 2:


{3 X - 2 Y = 2 | (equation 1)


{0 X - (4 Y)/3 = (-8)/3 | (equation 2)



Multiply equation 2 by -3/4:


{3 X - 2 Y = 2 | (equation 1)


{0 X+Y = 2 | (equation 2)



Add 2 × (equation 2) to equation 1:


{3 X+0 Y = 6 | (equation 1)


{0 X+Y = 2 | (equation 2)



Divide equation 1 by 3:


{X+0 Y = 2 | (equation 1)


{0 X+Y = 2 | (equation 2)



Collect results:


Answer:  {X = 2 , Y = 2

4 0
3 years ago
Read 2 more answers
What is 27+3÷ 3 +4-1+72.4-34.24+6-4+75÷5 <br> Answer this fast please!
Dimas [21]

Answer:

86.16

Step-by-step explanation:

27+3÷ 3 +4-1+72.4-34.24+6-4+75÷5

PEMDAS

Multiply and divide first

3/3 is 1   and 75/5 = 15

Replace these into the equation

27+1 +4-1+72.4-34.24+6-4+15

Then add and subtract from left to right

31 +72.4-34.24+6-4+15

69.16+6-4+15

86.16

7 0
3 years ago
Read 2 more answers
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